Chapter 4 · Additional Maths · A4 Polynomials and partial fractions

Polynomials, cubic identities and partial fractions

A polynomial is the kind of expression you can add, multiply and divide without leaving the family. By the end of this chapter you can divide one polynomial by another, find remainders and factors without dividing, factorise and solve cubic equations, and split an algebraic fraction into simpler pieces.

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Chapter
4

Figure 1 · long division at a glance

4 outcomes · 1 page · every step shown ✦

Divide, then check

Dividend = divisor $\times$ quotient + remainder, and the remainder's degree is smaller than the divisor's. Write a zero placeholder for every missing power.

Remainder and factor

Dividing by $(x - a)$ leaves $f(a)$ . A remainder of $0$ means $(x - a)$ is a factor. For $(ax + b)$ use $f\left(-\tfrac{b}{a}\right)$ .

Split the fraction

Factorise the denominator, pick the form, then find the constants by substituting roots or comparing coefficients. Equal degrees mean divide first.

By the end of this chapter

Decide whether an expression is a polynomial, state its degree, evaluate $f(a)$ and multiply polynomials.
Divide a polynomial by a linear or quadratic divisor using long division, with zero placeholders for missing powers, and write the division statement.
Use the remainder theorem and the factor theorem, including non-monic divisors, unknown constants and a quadratic divisor.
Factorise and solve cubic equations (trial root, division, quadratic formula, sum and difference of cubes) and answer the 'hence' questions that follow.
Split a proper or improper fraction into partial fractions for the three syllabus denominator forms, and check the result by substitution.

✶

Start here

Maths In Real Life

Two situations where a cubic polynomial appears. The pictures come first; the maths drawn over them is generated from the same numbers that the solutions use.

$A flat rectangular aluminium sheet seen from above on an HDB corridor workbench, with the four corner squares of side x marked by dashed lines to be cut out; the length 28 cm, the width 20 cm and the cut size x cm are marked$

28\text{ cm}

20\text{ cm}

x\text{ cm}

Maths In Real Life 1

The HDB herb tray

A resident along an HDB corridor makes a herb tray from a flat aluminium sheet: a square is cut from each corner and the sides are folded up. The size of the squares decides the volume of soil, and the volume is a cubic in that size: the same volume can often be reached with two different cuts.

A resident of an HDB block makes an open herb tray from a rectangular aluminium sheet $28\text{ cm}$ long and $20\text{ cm}$ wide. Squares of side $x$ cm are cut from each corner and the sides are folded up. Let $V$ be the volume of the tray in $\text{cm}^3$ .

(a)Show that $V = 4x^3 - 96x^2 + 560x$ .
(b)The tray must hold exactly $768\text{ cm}^3$ . Find the possible values of $x$ .
(c)Find $V$ when $x = 3$ , and say whether the tray then holds more than $768\text{ cm}^3$ .

Show full solutionHide solution

Colour keyterm subtracted or zero placeholderfactor or divisorvalue substitutedfinal result

(a)

Cutting a square of side

x

from every corner and folding the sides up gives a tray of height

x

. The base loses

x

at both ends of each side, so the length is

28 - 2x

and the width is

20 - 2x

Draw the sheet: each side of the base is shorter by two cuts.

\displaystyle V\;= x(28 - 2x)(20 - 2x)

Volume of a cuboid: length times width times height.

\displaystyle \phantom{V}\;x(-2x + 28) = \textcolor{#010106}{-2x^2 + 28x}

Multiply the single term into every term of the bracket.

Now multiply by the next bracket,

(-2x + 20)

\displaystyle \phantom{V}\;(-2x^2 + 28x)(-2x + 20) = -2x^2(-2x + 20) + 28x(-2x + 20)

Multiply every term of the first bracket by the whole second bracket.

\displaystyle \phantom{V}\;= (4x^3 - 40x^2) + (-56x^2 + 560x)

Expand each product, keeping its sign.

\displaystyle \phantom{V}\;= \textcolor{#010106}{4x^3 - 96x^2 + 560x}

Collect the terms that have the same power of

x

(b)

Bring every term to one side so that the other side is

0

, then call the result

f(x)

Subtract

768

from both sides: the equation becomes

4x^3 - 96x^2 + 560x - 768 = 0

So the equation is

f(x) = 4x^3 - 96x^2 + 560x - 768 = 0

Possible rational roots are a factor of

768

divided by a factor of

4

. Try them in turn:

f(1) = -300

f(-1) = -1428

, and neither is

0

\displaystyle f(2)\;= 4\textcolor{#010105}{(2)}^3 - 96\textcolor{#010105}{(2)}^2 + 560\textcolor{#010105}{(2)} - 768

Replace every

x

2

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(2)}\;= 32 - 384 + 1120 - 768

Powers first, then the multiplications.

\displaystyle \phantom{f(2)}\;= \textcolor{#010106}{0}

Now add and subtract.

f(2) = 0

, so

\textcolor{#010104}{(x - 2)}

divides

f(x)

exactly: it is a factor (factor theorem).

Zero remainder means the bracket divides

f(x)

exactly.

Step 1: divide the leading terms,

4x^3 \div x = 4x^2

. Write

4x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(2)}\;(4x^3 - 96x^2 + 560x - 768) - \textcolor{#010103}{(4x^3 - 8x^2)} = -88x^2 + 560x - 768

Subtract

4x^2

times the divisor, and change every sign of the product as you subtract.

Step 2: divide the leading terms,

-88x^2 \div x = -88x

. Write

-88x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(2)}\;(-88x^2 + 560x - 768) - \textcolor{#010103}{(-88x^2 + 176x)} = 384x - 768

Subtract

-88x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

384x \div x = 384

. Write

384

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(2)}\;(384x - 768) - \textcolor{#010103}{(384x - 768)} = 0

Subtract

384

times the divisor, and change every sign of the product as you subtract.

The remainder

0

has lower degree than the divisor, so the division stops.

\displaystyle f(x)\;= (x - 2)(4x^2 - 88x + 384)

Linear factor times quotient.

\displaystyle \phantom{f(x)}\;4x^2 - 88x + 384 = 4(x - 6)(x - 16)

Factorise the quadratic: find two numbers with the right sum and product.

\displaystyle f(x)\;= 4x^3 - 96x^2 + 560x - 768 = (x - 2)(4x^2 - 88x + 384) = \textcolor{#010106}{4(x - 2)(x - 6)(x - 16)}

The fully factorised form.

Set each factor equal to

0

(a product is zero only when a factor is zero).

\displaystyle x - 2 = 0 \quad\Rightarrow\quad x = 2

\displaystyle x - 6 = 0 \quad\Rightarrow\quad x = 6

\displaystyle x - 16 = 0 \quad\Rightarrow\quad x = 16

x = 16

is rejected: it does not lie between

0

and

10

as the situation requires.

Always check each root against the story.

(c)

\displaystyle V(3)\;= 4\textcolor{#010105}{(3)}^3 - 96\textcolor{#010105}{(3)}^2 + 560\textcolor{#010105}{(3)}

Replace every

x

3

and keep the brackets so signs cannot slip.

\displaystyle \phantom{V(3)}\;= 108 - 864 + 1680

Powers first, then the multiplications.

\displaystyle \phantom{V(3)}\;= \textcolor{#010106}{924}

Now add and subtract.

Compare with the target:

V(3) = 924

is greater than

768

, so a cut of

3

cm gives a tray with more than the required volume.

Answer

(a)

4x^3 - 96x^2 + 560x

(b)

x = 2\text{ cm}, x = 6\text{ cm}

(c)

V(3) = 924\text{ cm}^3

;

924 > 768

, so the tray holds more than

768\text{ cm}^3

$A wooden shipping crate on a pallet in a warehouse with its height marked as (2 added to x) decimetres and its length and width marked with the letters l and w$

(x + 2)\text{ dm}

\ell

w

Maths In Real Life 2

The shipping crate

The space inside a crate is length times width times height. When each of these is written with the same unknown $x$ , the volume is a cubic polynomial, and dividing it by one known dimension leaves the area of the base.

A wooden crate has height $(x + 2)$ dm, length $\ell$ dm and width $w$ dm, where $\ell > w$ . Its volume is $V = x^3 + 10x^2 + 31x + 30$ litres (one litre is one cubic decimetre), where $x$ is a positive number of decimetres.

(a)Divide $V$ by the height to find an expression for the area of the base.
(b)Hence factorise $V$ completely and write down $\ell$ and $w$ in terms of $x$ .
(c)Find the volume of the crate when its height is $6$ dm.

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Colour keyterm subtracted or zero placeholdervalue substitutedfinal result

(a)

Divide

V

by the height

(x + 2)

Step 1: divide the leading terms,

x^3 \div x = x^2

. Write

x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (x^3 + 10x^2 + 31x + 30) - \textcolor{#010103}{(x^3 + 2x^2)} = 8x^2 + 31x + 30

Subtract

x^2

times the divisor, and change every sign of the product as you subtract.

Step 2: divide the leading terms,

8x^2 \div x = 8x

. Write

8x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (8x^2 + 31x + 30) - \textcolor{#010103}{(8x^2 + 16x)} = 15x + 30

Subtract

8x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

15x \div x = 15

. Write

15

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (15x + 30) - \textcolor{#010103}{(15x + 30)} = 0

Subtract

15

times the divisor, and change every sign of the product as you subtract.

The remainder

0

has lower degree than the divisor, so the division stops.

(b)

Write

f(x)

for the volume

V

in this part.

From the earlier part,

f(x) = (x + 2)(x^2 + 8x + 15)

, so only the quadratic factor still needs attention.

\displaystyle f(x)\;= (x + 2)(x^2 + 8x + 15)

Linear factor times quotient.

\displaystyle \phantom{f(x)}\;x^2 + 8x + 15 = (x + 5)(x + 3)

Factorise the quadratic: find two numbers with the right sum and product.

\displaystyle f(x)\;= x^3 + 10x^2 + 31x + 30 = (x + 2)(x^2 + 8x + 15) = \textcolor{#010106}{(x + 5)(x + 3)(x + 2)}

The fully factorised form.

(c)

The height is

x + 2 = 6

, so

x = 4

. Put this value into

V

\displaystyle V(4)\;= \textcolor{#010105}{(4)}^3 + 10\textcolor{#010105}{(4)}^2 + 31\textcolor{#010105}{(4)} + 30

Replace every

x

4

and keep the brackets so signs cannot slip.

\displaystyle \phantom{V(4)}\;= 64 + 160 + 124 + 30

Powers first, then the multiplications.

\displaystyle \phantom{V(4)}\;= \textcolor{#010106}{378}

Now add and subtract.

Answer

(a) quotient

x^2 + 8x + 15

; remainder

0

; the area of the base is

(x^2 + 8x + 15)\text{ dm}^2

(b)

f(x) = (x + 5)(x + 3)(x + 2)

;

\ell = (x + 5)

dm and

w = (x + 3)

dm (the longer side is

\ell

(c)

V(4) = 378\text{ litres}

Supports K341-A4-01

What is a polynomial? Degree, $f(a)$ and multiplying

A polynomial in $x$ is a sum of terms, each a number times $x$ raised to a whole-number power: 0, 1, 2, 3, .... So $4x^3 - 2x + 9$ is a polynomial, but $\dfrac{1}{x} + 2$ (a power of $-1$ ), $\sqrt{x}$ (a power of $\tfrac{1}{2}$ ) and $2^x$ (the unknown sits in the exponent) are not. The degree is the highest power of $x$ that appears. The number multiplying that power is the leading coefficient.

Case table 1: polynomial or not?
Expression	Polynomial?	Degree	Reason
$x^{3} - 4 x + 9$	Yes	$3$	Every power of $x$ is a whole number, 0 or more.
$x^{2} + \frac{1}{x}$	No	none	$\frac{1}{x}$ has $x$ in a denominator (a negative power of $x$ ).
$7$	Yes	$0$	Every power of $x$ is a whole number, 0 or more.
$\sqrt{x} + 3$	No	none	$\sqrt{x}$ has a fractional power of $x$ (a root of $x$ ).
$2^{x} + x$	No	none	$x$ is in an exponent ( $2^{x}$ ).
$- x^{4} + 5$	Yes	$4$	Every power of $x$ is a whole number, 0 or more.

The Core Idea

Only whole-number powers allowed; the degree is the biggest one.

A constant such as $7$ is a polynomial of degree $0$ . The notation $f(x)$ names the polynomial, and $f(a)$ means 'put $x = a$ into it'.

Example 1.1

Is it a polynomial?

Decide whether each expression is a polynomial in $x$ . For each polynomial, state its degree.

(i) $p(x) = 3 x^{4} - 2 x + 7$
(ii) $q(x) = x^{2} + \frac{1}{x}$
(iii) $r(x) = - x^{3} + 5$
(iv) $s(x) = \sqrt{x} + 1$

Show solutionHide solution

(i)

Check every term: each must be a number times

x

to a whole-number power (0, 1, 2, ...). The powers of

x

present are

4, 1, 0

, all whole numbers.

So it is a polynomial. Its degree is the highest power of

x

, which is

4

The degree ignores the coefficients: only the highest power counts.

(ii)

Check every term: a polynomial allows only a number times

x

to a whole-number power (0, 1, 2, ...).

The term

\frac{1}{x}

breaks the rule:

\frac{1}{x}

has

x

in a denominator (a negative power of

x

). So it is not a polynomial.

One bad term is enough to fail the test.

(iii)

Check every term: each must be a number times

x

to a whole-number power (0, 1, 2, ...). The powers of

x

present are

3, 0

, all whole numbers.

So it is a polynomial. Its degree is the highest power of

x

, which is

3

The degree ignores the coefficients: only the highest power counts.

(iv)

Check every term: a polynomial allows only a number times

x

to a whole-number power (0, 1, 2, ...).

The term

\sqrt{x}

breaks the rule:

\sqrt{x}

has a fractional power of

x

(a root of

x

). So it is not a polynomial.

One bad term is enough to fail the test.

Answer

(i) a polynomial of degree

4

(ii) not a polynomial:

\frac{1}{x}

has

x

in a denominator (a negative power of

x

(iii) a polynomial of degree

3

(iv) not a polynomial:

\sqrt{x}

has a fractional power of

x

(a root of

x

Your Turn 1.1

Decide whether each expression is a polynomial in $x$ . For each polynomial, state its degree.

(i) $p(x) = x^{5} - 6 x^{2}$
(ii) $q(x) = \sqrt{x} + x^{3}$
(iii) $r(x) = 7$
(iv) $s(x) = 3^{x} - x$

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(i)

Check every term: each must be a number times

x

to a whole-number power (0, 1, 2, ...). The powers of

x

present are

5, 2

, all whole numbers.

So it is a polynomial. Its degree is the highest power of

x

, which is

5

The degree ignores the coefficients: only the highest power counts.

(ii)

Check every term: a polynomial allows only a number times

x

to a whole-number power (0, 1, 2, ...).

The term

\sqrt{x}

breaks the rule:

\sqrt{x}

has a fractional power of

x

(a root of

x

). So it is not a polynomial.

One bad term is enough to fail the test.

(iii)

Check every term: each must be a number times

x

to a whole-number power (0, 1, 2, ...). The powers of

x

present are

0

, all whole numbers.

So it is a polynomial. Its degree is the highest power of

x

, which is

0

The degree ignores the coefficients: only the highest power counts.

(iv)

Check every term: a polynomial allows only a number times

x

to a whole-number power (0, 1, 2, ...).

The term

3^{x}

breaks the rule:

x

is in an exponent (

3^{x}

), which a polynomial never allows. So it is not a polynomial.

One bad term is enough to fail the test.

Answer

(i) a polynomial of degree

5

(ii) not a polynomial:

\sqrt{x}

has a fractional power of

x

(a root of

x

(iii) a polynomial of degree

0

(iv) not a polynomial:

x

is in an exponent (

3^{x}

), which a polynomial never allows.

SGSchoolKaki Tip

To test a term, ask 'what power of $x$ is this?'. Powers that are negative, fractional or contain $x$ itself ( $2^x$ ) all fail.

Example 1.2

Evaluating $f(a)$

Let $f(x) = 2x^3 - 5x^2 + 3$ .

(i)Find $f(2)$ .
(ii)Find $f(-1)$ .
(iii)Find $f\left(\frac{1}{2}\right)$ .

Show solutionHide solution

Colour keyvalue substitutedfinal result

(i)

\displaystyle f(2)\;= 2\textcolor{#010105}{(2)}^3 - 5\textcolor{#010105}{(2)}^2 + 3

Replace every

x

2

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(2)}\;= 16 - 20 + 3

Powers first, then the multiplications.

\displaystyle \phantom{f(2)}\;= \textcolor{#010106}{-1}

Now add and subtract.

(ii)

\displaystyle f(-1)\;= 2\textcolor{#010105}{(-1)}^3 - 5\textcolor{#010105}{(-1)}^2 + 3

Replace every

x

-1

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(-1)}\;= -2 - 5 + 3

Powers first, then the multiplications.

\displaystyle \phantom{f(-1)}\;= \textcolor{#010106}{-4}

Now add and subtract.

(iii)

\displaystyle f(\frac{1}{2})\;= 2\textcolor{#010105}{\left(\frac{1}{2}\right)}^3 - 5\textcolor{#010105}{\left(\frac{1}{2}\right)}^2 + 3

Replace every

x

\frac{1}{2}

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(\frac{1}{2})}\;= \frac{1}{4} - \frac{5}{4} + 3

Powers first, then the multiplications.

\displaystyle \phantom{f(\frac{1}{2})}\;= \textcolor{#010106}{2}

Now add and subtract.

Answer

(i)

f(2) = -1

(ii)

f(-1) = -4

(iii)

f(\frac{1}{2}) = 2

Your Turn 1.2

Let $f(x) = 3x^3 - 4x^2 + x - 2$ .

(i)Find $f(3)$ .
(ii)Find $f(-2)$ .
(iii)Find $f\left(- \frac{1}{3}\right)$ .

Show full solutionHide solution

Colour keyvalue substitutedfinal result

(i)

\displaystyle f(3)\;= 3\textcolor{#010105}{(3)}^3 - 4\textcolor{#010105}{(3)}^2 + \textcolor{#010105}{(3)} - 2

Replace every

x

3

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(3)}\;= 81 - 36 + 3 - 2

Powers first, then the multiplications.

\displaystyle \phantom{f(3)}\;= \textcolor{#010106}{46}

Now add and subtract.

(ii)

\displaystyle f(-2)\;= 3\textcolor{#010105}{(-2)}^3 - 4\textcolor{#010105}{(-2)}^2 + \textcolor{#010105}{(-2)} - 2

Replace every

x

-2

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(-2)}\;= -24 - 16 - 2 - 2

Powers first, then the multiplications.

\displaystyle \phantom{f(-2)}\;= \textcolor{#010106}{-44}

Now add and subtract.

(iii)

\displaystyle f(- \frac{1}{3})\;= 3\textcolor{#010105}{\left(- \frac{1}{3}\right)}^3 - 4\textcolor{#010105}{\left(- \frac{1}{3}\right)}^2 + \textcolor{#010105}{\left(- \frac{1}{3}\right)} - 2

Replace every

x

- \frac{1}{3}

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(- \frac{1}{3})}\;= - \frac{1}{9} - \frac{4}{9} - \frac{1}{3} - 2

Powers first, then the multiplications.

\displaystyle \phantom{f(- \frac{1}{3})}\;= \textcolor{#010106}{- \frac{26}{9}}

Now add and subtract.

Answer

(i)

f(3) = 46

(ii)

f(-2) = -44

(iii)

f(- \frac{1}{3}) = - \frac{26}{9}

Example 1.3

Multiplying two polynomials

Expand $(x + 2)(x^2 - 3x + 5)$ and state the degree of the product.

Show solutionHide solution

Colour keyfinal result

\displaystyle (x + 2)(x^2 - 3x + 5) = x(x^2 - 3x + 5) + 2(x^2 - 3x + 5)

Multiply every term of the first bracket by the whole second bracket.

\displaystyle = (x^3 - 3x^2 + 5x) + (2x^2 - 6x + 10)

Expand each product, keeping its sign.

\displaystyle = \textcolor{#010106}{x^3 - x^2 - x + 10}

Collect the terms that have the same power of

x

The degree of a product is the sum of the degrees:

1 + 2 = 3

Only the leading terms multiply to give the highest power.

Answer

x^3 - x^2 - x + 10

; degree

3

Your Turn 1.3

Expand $(x - 1)(2x^2 + 3x - 4)$ and state the degree of the product.

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Colour keyfinal result

\displaystyle (x - 1)(2x^2 + 3x - 4) = x(2x^2 + 3x - 4) - 1(2x^2 + 3x - 4)

Multiply every term of the first bracket by the whole second bracket.

\displaystyle = (2x^3 + 3x^2 - 4x) + (-2x^2 - 3x + 4)

Expand each product, keeping its sign.

\displaystyle = \textcolor{#010106}{2x^3 + x^2 - 7x + 4}

Collect the terms that have the same power of

x

The degree of a product is the sum of the degrees:

1 + 2 = 3

Only the leading terms multiply to give the highest power.

Answer

2x^3 + x^2 - 7x + 4

; degree

3

Example 1.4

Three brackets

Expand $(x + 2)(x - 1)(3x + 1)$ .

Show solutionHide solution

Colour keyfinal result

\displaystyle (x + 2)(x - 1) = x(x - 1) + 2(x - 1)

Multiply every term of the first bracket by the whole second bracket.

\displaystyle = (x^2 - x) + (2x - 2)

Expand each product, keeping its sign.

\displaystyle = \textcolor{#010106}{x^2 + x - 2}

Collect the terms that have the same power of

x

Now multiply by the next bracket,

(3x + 1)

\displaystyle (x^2 + x - 2)(3x + 1) = x^2(3x + 1) + x(3x + 1) - 2(3x + 1)

Multiply every term of the first bracket by the whole second bracket.

\displaystyle = (3x^3 + x^2) + (3x^2 + x) + (-6x - 2)

Expand each product, keeping its sign.

\displaystyle = \textcolor{#010106}{3x^3 + 4x^2 - 5x - 2}

Collect the terms that have the same power of

x

Answer

3x^3 + 4x^2 - 5x - 2

Your Turn 1.4

Expand $(x - 3)(x + 1)(2x - 1)$ .

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Colour keyfinal result

\displaystyle (x - 3)(x + 1) = x(x + 1) - 3(x + 1)

Multiply every term of the first bracket by the whole second bracket.

\displaystyle = (x^2 + x) + (-3x - 3)

Expand each product, keeping its sign.

\displaystyle = \textcolor{#010106}{x^2 - 2x - 3}

Collect the terms that have the same power of

x

Now multiply by the next bracket,

(2x - 1)

\displaystyle (x^2 - 2x - 3)(2x - 1) = x^2(2x - 1) - 2x(2x - 1) - 3(2x - 1)

Multiply every term of the first bracket by the whole second bracket.

\displaystyle = (2x^3 - x^2) + (-4x^2 + 2x) + (-6x + 3)

Expand each product, keeping its sign.

\displaystyle = \textcolor{#010106}{2x^3 - 5x^2 - 4x + 3}

Collect the terms that have the same power of

x

Answer

2x^3 - 5x^2 - 4x + 3

Think About It

Two polynomials have degrees

2

and

3

. Why does their product always have degree exactly

5

, and never lower?

Show the full workingHide the working

Only the two leading terms can multiply to give the highest power of

x

, so they are the only terms that matter for the degree.

Lower terms can never reach the top power.

\displaystyle (2x^2)(-3x^3) = -6x^5

Multiply the coefficients and add the powers.

The new leading coefficient

-6

is a product of two non-zero numbers, so it is never

0

. The

x^{ 5 }

term cannot vanish: degree

= 2 + 3 = 5

Answer

The leading terms multiply to

-6x^5

, which is never zero, so the degree of the product is the sum

2 + 3 = 5

Supports K341-A4-01

Dividing polynomials: long division and the division statement

Dividing $17$ by $5$ gives quotient $3$ and remainder $2$ , and you can write $17 = 5 \times 3 + 2$ . Polynomials divide the same way. If you divide $f(x)$ by $g(x)$ you get a quotient $q(x)$ and a remainder $r(x)$ , linked by the division statement.

The Core Idea

dividend = divisor $\times$ quotient + remainder

In symbols $f(x) = g(x)\,q(x) + r(x)$ , where the degree of $r(x)$ is smaller than the degree of $g(x)$ . Divide by a linear divisor and the remainder is a plain number; divide by a quadratic and the remainder can be $ax + b$ .

The layout: each power of

x

has its own column; subtract line by line.

Each round of long division does three things: divide the leading terms to get the next quotient term, multiply that term by the whole divisor, and subtract (changing every sign). Stop when what is left has a smaller degree than the divisor. If a power of $x$ is missing, write it with a coefficient of $0$ so the columns stay lined up.

Example 2.1

A linear divisor, remainder not zero

Divide $f(x) = x^3 - 4x^2 + 2x + 7$ by $(x - 3)$ . State the quotient and the remainder, and write the result as a division statement.

Show solutionHide solution

Colour keyterm subtracted or zero placeholder

Step 1: divide the leading terms,

x^3 \div x = x^2

. Write

x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (x^3 - 4x^2 + 2x + 7) - \textcolor{#010103}{(x^3 - 3x^2)} = -x^2 + 2x + 7

Subtract

x^2

times the divisor, and change every sign of the product as you subtract.

Step 2: divide the leading terms,

-x^2 \div x = -x

. Write

-x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (-x^2 + 2x + 7) - \textcolor{#010103}{(-x^2 + 3x)} = -x + 7

Subtract

-x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

-x \div x = -1

. Write

-1

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (-x + 7) - \textcolor{#010103}{(-x + 3)} = 4

Subtract

-1

times the divisor, and change every sign of the product as you subtract.

The remainder

4

has lower degree than the divisor, so the division stops.

\displaystyle x^3 - 4x^2 + 2x + 7 = (x - 3)(x^2 - x - 1) + 4

dividend = divisor

\times

quotient + remainder.

Answer

quotient

x^2 - x - 1

; remainder

4

;

x^3 - 4x^2 + 2x + 7 = (x - 3)(x^2 - x - 1) + 4

Your Turn 2.1

Divide $f(x) = x^3 + 5x^2 - 3x - 8$ by $(x + 2)$ . State the quotient and the remainder, and write the result as a division statement.

Show full solutionHide solution

Colour keyterm subtracted or zero placeholder

Step 1: divide the leading terms,

x^3 \div x = x^2

. Write

x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (x^3 + 5x^2 - 3x - 8) - \textcolor{#010103}{(x^3 + 2x^2)} = 3x^2 - 3x - 8

Subtract

x^2

times the divisor, and change every sign of the product as you subtract.

Step 2: divide the leading terms,

3x^2 \div x = 3x

. Write

3x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (3x^2 - 3x - 8) - \textcolor{#010103}{(3x^2 + 6x)} = -9x - 8

Subtract

3x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

-9x \div x = -9

. Write

-9

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (-9x - 8) - \textcolor{#010103}{(-9x - 18)} = 10

Subtract

-9

times the divisor, and change every sign of the product as you subtract.

The remainder

10

has lower degree than the divisor, so the division stops.

\displaystyle x^3 + 5x^2 - 3x - 8 = (x + 2)(x^2 + 3x - 9) + 10

dividend = divisor

\times

quotient + remainder.

Answer

quotient

x^2 + 3x - 9

; remainder

10

;

x^3 + 5x^2 - 3x - 8 = (x + 2)(x^2 + 3x - 9) + 10

Example 2.2

A missing power: write $0x^2$

Find the quotient and remainder when $2x^3 - 7x + 5$ is divided by $(x + 2)$ .

Show solutionHide solution

Colour keyterm subtracted or zero placeholder

Before dividing, write a zero placeholder for every missing power (the dividend is missing

x^2

) so each power keeps its own column.

\displaystyle f(x)\;= 2x^3 + \textcolor{#010103}{0x^2} - 7x + 5

The placeholder changes nothing: adding

0

does not change the polynomial.

Step 1: divide the leading terms,

2x^3 \div x = 2x^2

. Write

2x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(2x^3 + \textcolor{#010103}{0x^2} - 7x + 5) - \textcolor{#010103}{(2x^3 + 4x^2)} = -4x^2 - 7x + 5

Subtract

2x^2

times the divisor, and change every sign of the product as you subtract. The placeholder column stays in the first row so every power lines up with the layout.

Step 2: divide the leading terms,

-4x^2 \div x = -4x

. Write

-4x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(-4x^2 - 7x + 5) - \textcolor{#010103}{(-4x^2 - 8x)} = x + 5

Subtract

-4x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

x \div x = 1

. Write

1

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(x + 5) - \textcolor{#010103}{(x + 2)} = 3

Subtract

1

times the divisor, and change every sign of the product as you subtract.

The remainder

3

has lower degree than the divisor, so the division stops.

Answer

quotient

2x^2 - 4x + 1

; remainder

3

Your Turn 2.2

Find the quotient and remainder when $x^3 - 6x^2 + 1$ is divided by $(x - 2)$ .

Show full solutionHide solution

Colour keyterm subtracted or zero placeholder

Before dividing, write a zero placeholder for every missing power (the dividend is missing

x

) so each power keeps its own column.

\displaystyle f(x)\;= x^3 - 6x^2 + \textcolor{#010103}{0x} + 1

The placeholder changes nothing: adding

0

does not change the polynomial.

Step 1: divide the leading terms,

x^3 \div x = x^2

. Write

x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(x^3 - 6x^2 + \textcolor{#010103}{0x} + 1) - \textcolor{#010103}{(x^3 - 2x^2)} = -4x^2 + 1

Subtract

x^2

times the divisor, and change every sign of the product as you subtract. The placeholder column stays in the first row so every power lines up with the layout.

Step 2: divide the leading terms,

-4x^2 \div x = -4x

. Write

-4x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(-4x^2 + 1) - \textcolor{#010103}{(-4x^2 + 8x)} = -8x + 1

Subtract

-4x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

-8x \div x = -8

. Write

-8

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(-8x + 1) - \textcolor{#010103}{(-8x + 16)} = -15

Subtract

-8

times the divisor, and change every sign of the product as you subtract.

The remainder

-15

has lower degree than the divisor, so the division stops.

Answer

quotient

x^2 - 4x - 8

; remainder

-15

Example 2.3

A divisor whose $x$ has a coefficient

Divide $6x^3 - 5x^2 - 7x + 4$ by $(2x - 1)$ . Use the remainder to say whether $(2x - 1)$ is a factor of $6x^3 - 5x^2 - 7x + 4$ .

Show solutionHide solution

Colour keyterm subtracted or zero placeholder

Step 1: divide the leading terms,

6x^3 \div 2x = 3x^2

. Write

3x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (6x^3 - 5x^2 - 7x + 4) - \textcolor{#010103}{(6x^3 - 3x^2)} = -2x^2 - 7x + 4

Subtract

3x^2

times the divisor, and change every sign of the product as you subtract.

Step 2: divide the leading terms,

-2x^2 \div 2x = -x

. Write

-x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (-2x^2 - 7x + 4) - \textcolor{#010103}{(-2x^2 + x)} = -8x + 4

Subtract

-x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

-8x \div 2x = -4

. Write

-4

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (-8x + 4) - \textcolor{#010103}{(-8x + 4)} = 0

Subtract

-4

times the divisor, and change every sign of the product as you subtract.

The remainder

0

has lower degree than the divisor, so the division stops.

\displaystyle 6x^3 - 5x^2 - 7x + 4 = (2x - 1)(3x^2 - x - 4)

dividend = divisor

\times

quotient + remainder.

The remainder is

0

, so

(2x - 1)

divides

6x^3 - 5x^2 - 7x + 4

exactly: it is a factor.

Answer

quotient

3x^2 - x - 4

; remainder

0

;

6x^3 - 5x^2 - 7x + 4 = (2x - 1)(3x^2 - x - 4)

; remainder

0

, so

(2x - 1)

is a factor

Your Turn 2.3

Divide $8x^3 - 2x^2 + 3x - 5$ by $(2x + 1)$ . Use the remainder to say whether $(2x + 1)$ is a factor of $8x^3 - 2x^2 + 3x - 5$ .

Show full solutionHide solution

Colour keyterm subtracted or zero placeholder

Step 1: divide the leading terms,

8x^3 \div 2x = 4x^2

. Write

4x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (8x^3 - 2x^2 + 3x - 5) - \textcolor{#010103}{(8x^3 + 4x^2)} = -6x^2 + 3x - 5

Subtract

4x^2

times the divisor, and change every sign of the product as you subtract.

Step 2: divide the leading terms,

-6x^2 \div 2x = -3x

. Write

-3x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (-6x^2 + 3x - 5) - \textcolor{#010103}{(-6x^2 - 3x)} = 6x - 5

Subtract

-3x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

6x \div 2x = 3

. Write

3

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (6x - 5) - \textcolor{#010103}{(6x + 3)} = -8

Subtract

3

times the divisor, and change every sign of the product as you subtract.

The remainder

-8

has lower degree than the divisor, so the division stops.

\displaystyle 8x^3 - 2x^2 + 3x - 5 = (2x + 1)(4x^2 - 3x + 3) - 8

dividend = divisor

\times

quotient + remainder.

The remainder is not

0

, so

(2x + 1)

does not divide exactly: it is not a factor.

Answer

quotient

4x^2 - 3x + 3

; remainder

-8

;

8x^3 - 2x^2 + 3x - 5 = (2x + 1)(4x^2 - 3x + 3) - 8

; remainder is not

0

, so

(2x + 1)

is not a factor

Example 2.4

Dividing by a quadratic

Divide $x^4 - 2x^2 + 5x + 3$ by $(x^2 + 1)$ . Give the quotient and the remainder, and say why the division stops where it does.

Show solutionHide solution

Colour keyterm subtracted or zero placeholder

Before dividing, write a zero placeholder for every missing power (the dividend is missing

x^3

) so each power keeps its own column.

\displaystyle f(x)\;= x^4 + \textcolor{#010103}{0x^3} - 2x^2 + 5x + 3

The placeholder changes nothing: adding

0

does not change the polynomial.

Step 1: divide the leading terms,

x^4 \div x^2 = x^2

. Write

x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(x^4 + \textcolor{#010103}{0x^3} - 2x^2 + 5x + 3) - \textcolor{#010103}{(x^4 + x^2)} = -3x^2 + 5x + 3

Subtract

x^2

times the divisor, and change every sign of the product as you subtract. The placeholder column stays in the first row so every power lines up with the layout.

Step 2: divide the leading terms,

-3x^2 \div x^2 = -3

. Write

-3

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(-3x^2 + 5x + 3) - \textcolor{#010103}{(-3x^2 - 3)} = 5x + 6

Subtract

-3

times the divisor, and change every sign of the product as you subtract.

The remainder

5x + 6

has lower degree than the divisor, so the division stops.

Degree of the divisor

2

, so the remainder has degree at most

1

(a constant when the divisor is linear).

Answer

quotient

x^2 - 3

; remainder

5x + 6

; remainder degree below

2

Your Turn 2.4

Divide $2x^4 + x^3 - 3x + 4$ by $(x^2 + x - 1)$ . Give the quotient and the remainder, and say why the division stops where it does.

Show full solutionHide solution

Colour keyterm subtracted or zero placeholder

Before dividing, write a zero placeholder for every missing power (the dividend is missing

x^2

) so each power keeps its own column.

\displaystyle f(x)\;= 2x^4 + x^3 + \textcolor{#010103}{0x^2} - 3x + 4

The placeholder changes nothing: adding

0

does not change the polynomial.

Step 1: divide the leading terms,

2x^4 \div x^2 = 2x^2

. Write

2x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(2x^4 + x^3 + \textcolor{#010103}{0x^2} - 3x + 4) - \textcolor{#010103}{(2x^4 + 2x^3 - 2x^2)} = -x^3 + 2x^2 - 3x + 4

Subtract

2x^2

times the divisor, and change every sign of the product as you subtract. The placeholder column stays in the first row so every power lines up with the layout.

Step 2: divide the leading terms,

-x^3 \div x^2 = -x

. Write

-x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(-x^3 + 2x^2 - 3x + 4) - \textcolor{#010103}{(-x^3 - x^2 + x)} = 3x^2 - 4x + 4

Subtract

-x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

3x^2 \div x^2 = 3

. Write

3

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(3x^2 - 4x + 4) - \textcolor{#010103}{(3x^2 + 3x - 3)} = -7x + 7

Subtract

3

times the divisor, and change every sign of the product as you subtract.

The remainder

-7x + 7

has lower degree than the divisor, so the division stops.

Degree of the divisor

2

, so the remainder has degree at most

1

(a constant when the divisor is linear).

Answer

quotient

2x^2 - x + 3

; remainder

-7x + 7

; remainder degree below

2

Example 2.5

Write it as a statement and check it

Divide $3x^3 - 2x - 9$ by $(x - 2)$ , write the division statement, and check it by multiplying back.

Show solutionHide solution

Colour keyterm subtracted or zero placeholder

Before dividing, write a zero placeholder for every missing power (the dividend is missing

x^2

) so each power keeps its own column.

\displaystyle f(x)\;= 3x^3 + \textcolor{#010103}{0x^2} - 2x - 9

The placeholder changes nothing: adding

0

does not change the polynomial.

Step 1: divide the leading terms,

3x^3 \div x = 3x^2

. Write

3x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(3x^3 + \textcolor{#010103}{0x^2} - 2x - 9) - \textcolor{#010103}{(3x^3 - 6x^2)} = 6x^2 - 2x - 9

Subtract

3x^2

times the divisor, and change every sign of the product as you subtract. The placeholder column stays in the first row so every power lines up with the layout.

Step 2: divide the leading terms,

6x^2 \div x = 6x

. Write

6x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(6x^2 - 2x - 9) - \textcolor{#010103}{(6x^2 - 12x)} = 10x - 9

Subtract

6x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

10x \div x = 10

. Write

10

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(10x - 9) - \textcolor{#010103}{(10x - 20)} = 11

Subtract

10

times the divisor, and change every sign of the product as you subtract.

The remainder

11

has lower degree than the divisor, so the division stops.

\displaystyle \phantom{f(x)}\;3x^3 - 2x - 9 = (x - 2)(3x^2 + 6x + 10) + 11

dividend = divisor

\times

quotient + remainder.

\displaystyle \phantom{f(x)}\;(x - 2)(3x^2 + 6x + 10) + 11 = 3x^3 - 2x - 9

Multiply back and add the remainder: you must land on the original.

Answer

3x^3 - 2x - 9 = (x - 2)(3x^2 + 6x + 10) + 11

; checked: divisor

\times

quotient + remainder = dividend

Your Turn 2.5

Divide $2x^3 - 3x^2 + 4$ by $(x + 1)$ , write the division statement, and check it by multiplying back.

Show full solutionHide solution

Colour keyterm subtracted or zero placeholder

Before dividing, write a zero placeholder for every missing power (the dividend is missing

x

) so each power keeps its own column.

\displaystyle f(x)\;= 2x^3 - 3x^2 + \textcolor{#010103}{0x} + 4

The placeholder changes nothing: adding

0

does not change the polynomial.

Step 1: divide the leading terms,

2x^3 \div x = 2x^2

. Write

2x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(2x^3 - 3x^2 + \textcolor{#010103}{0x} + 4) - \textcolor{#010103}{(2x^3 + 2x^2)} = -5x^2 + 4

Subtract

2x^2

times the divisor, and change every sign of the product as you subtract. The placeholder column stays in the first row so every power lines up with the layout.

Step 2: divide the leading terms,

-5x^2 \div x = -5x

. Write

-5x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(-5x^2 + 4) - \textcolor{#010103}{(-5x^2 - 5x)} = 5x + 4

Subtract

-5x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

5x \div x = 5

. Write

5

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(5x + 4) - \textcolor{#010103}{(5x + 5)} = -1

Subtract

5

times the divisor, and change every sign of the product as you subtract.

The remainder

-1

has lower degree than the divisor, so the division stops.

\displaystyle \phantom{f(x)}\;2x^3 - 3x^2 + 4 = (x + 1)(2x^2 - 5x + 5) - 1

dividend = divisor

\times

quotient + remainder.

\displaystyle \phantom{f(x)}\;(x + 1)(2x^2 - 5x + 5) - 1 = 2x^3 - 3x^2 + 4

Multiply back and add the remainder: you must land on the original.

Answer

2x^3 - 3x^2 + 4 = (x + 1)(2x^2 - 5x + 5) - 1

; checked: divisor

\times

quotient + remainder = dividend

Forgetting a placeholder is the most common slip. In

2x^3 - 7x + 5

the

x^2

term is missing: write

2x^3 + 0x^2 - 7x + 5

before you divide, or the columns will not line up.

Think About It

Why must the remainder have a smaller degree than the divisor? What would go wrong if it did not?

Show the full workingHide the working

Suppose we stopped early, with a remainder that still has

x

in it. We could divide the divisor into it once more and get a bigger quotient. That is why we keep going until the degree drops below the divisor's.

Stopping early would leave a quotient that is not the final one.

\displaystyle x^3 + 5 = (x + 1)(x^2 - x + 1) + 4

Here the remainder

4

has degree

0

, below the degree

1

of the divisor, so we stop.

The remainder is a number, with degree

0

, which is smaller than the divisor's degree

1

. That is the signal to stop.

Answer

Keep dividing until the remainder's degree is less than the divisor's; otherwise another round of division is still possible.

Supports K341-A4-02

The remainder theorem: the remainder without dividing

Long division finds the whole quotient. Often a question wants only the remainder, and there is a shortcut. Divide $f(x)$ by $(x - a)$ . The division statement is $f(x) = (x - a)\,q(x) + r$ , with $r$ a number. Put $x = a$ : the bracket becomes $0$ and the quotient disappears, so $f(a) = r$ .

The Core Idea

Remainder on dividing by $(x - a)$ is $f(a)$ .

For a divisor $(ax + b)$ set it to zero: $x = -\dfrac{b}{a}$ , and the remainder is $f\left(-\dfrac{b}{a}\right)$ . So dividing by $(2x - 1)$ means substituting $x = \tfrac{1}{2}$ , and dividing by $(2x + 1)$ means $x = -\tfrac{1}{2}$ . Watch the sign.

Example 3.1

Divisor $(x - a)$

Without dividing, work out the remainder left by $(x - 4)$ in $f(x) = x^3 - 5x^2 + 2x + 9$ .

Show solutionHide solution

Colour keyvalue substitutedfinal result

Set the divisor

x - 4

equal to zero:

x = 4

. By the remainder theorem the remainder is

f(4)

, so we never need to divide.

For

(x - a)

the value to substitute is

a

\displaystyle f(4)\;= \textcolor{#010105}{(4)}^3 - 5\textcolor{#010105}{(4)}^2 + 2\textcolor{#010105}{(4)} + 9

Replace every

x

4

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(4)}\;= 64 - 80 + 8 + 9

Powers first, then the multiplications.

\displaystyle \phantom{f(4)}\;= \textcolor{#010106}{1}

Now add and subtract.

Answer

remainder

1

Your Turn 3.1

Without dividing, work out the remainder left by $(x + 2)$ in $f(x) = 2x^3 + 3x^2 - x + 5$ .

Show full solutionHide solution

Colour keyvalue substitutedfinal result

Set the divisor

x + 2

equal to zero:

x = -2

. By the remainder theorem the remainder is

f(-2)

, so we never need to divide.

For

(x - a)

the value to substitute is

a

\displaystyle f(-2)\;= 2\textcolor{#010105}{(-2)}^3 + 3\textcolor{#010105}{(-2)}^2 - \textcolor{#010105}{(-2)} + 5

Replace every

x

-2

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(-2)}\;= -16 + 12 + 2 + 5

Powers first, then the multiplications.

\displaystyle \phantom{f(-2)}\;= \textcolor{#010106}{3}

Now add and subtract.

Answer

remainder

3

Example 3.2

Divisor $(ax + b)$ : substitute $-\tfrac{b}{a}$

Use the remainder theorem to find what is left over when $f(x) = 4x^3 - 2x^2 + 3x - 5$ is shared out by $(2x - 1)$ .

Show solutionHide solution

Colour keyvalue substitutedfinal result

Set the divisor

2x - 1

equal to zero:

x = \frac{1}{2}

. By the remainder theorem the remainder is

f(\frac{1}{2})

, so we never need to divide.

Divisor

= 0

gives the value to substitute; for

(ax + b)

that is

-\tfrac{b}{a}

\displaystyle f(\frac{1}{2})\;= 4\textcolor{#010105}{\left(\frac{1}{2}\right)}^3 - 2\textcolor{#010105}{\left(\frac{1}{2}\right)}^2 + 3\textcolor{#010105}{\left(\frac{1}{2}\right)} - 5

Replace every

x

\frac{1}{2}

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(\frac{1}{2})}\;= \frac{1}{2} - \frac{1}{2} + \frac{3}{2} - 5

Powers first, then the multiplications.

\displaystyle \phantom{f(\frac{1}{2})}\;= \textcolor{#010106}{- \frac{7}{2}}

Now add and subtract.

Answer

remainder

- \frac{7}{2}

Your Turn 3.2

Use the remainder theorem to find what is left over when $f(x) = 6x^3 + x^2 - 4x + 2$ is shared out by $(2x + 3)$ .

Show full solutionHide solution

Colour keyvalue substitutedfinal result

Set the divisor

2x + 3

equal to zero:

x = - \frac{3}{2}

. By the remainder theorem the remainder is

f(- \frac{3}{2})

, so we never need to divide.

Divisor

= 0

gives the value to substitute; for

(ax + b)

that is

-\tfrac{b}{a}

\displaystyle f(- \frac{3}{2})\;= 6\textcolor{#010105}{\left(- \frac{3}{2}\right)}^3 + \textcolor{#010105}{\left(- \frac{3}{2}\right)}^2 - 4\textcolor{#010105}{\left(- \frac{3}{2}\right)} + 2

Replace every

x

- \frac{3}{2}

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(- \frac{3}{2})}\;= - \frac{81}{4} + \frac{9}{4} + 6 + 2

Powers first, then the multiplications.

\displaystyle \phantom{f(- \frac{3}{2})}\;= \textcolor{#010106}{-10}

Now add and subtract.

Answer

remainder

-10

SGSchoolKaki Tip

Divisor $(2x + 1)$ gives $x = -\tfrac{1}{2}$ : the sign flips. A quick test is to say 'what value makes the divisor zero?' before you substitute.

Example 3.3

Two methods, one answer

Let $f(x) = x^3 + 3x^2 - 2x + 4$ .

(i)Use the remainder theorem for the divisor $(x - 1)$ : what remainder does $f(x)$ leave?
(ii)Check your answer by long division.

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Colour keyterm subtracted or zero placeholdervalue substitutedfinal result

(i)

Set the divisor

x - 1

equal to zero:

x = 1

. By the remainder theorem the remainder is

f(1)

, so we never need to divide.

For

(x - a)

the value to substitute is

a

\displaystyle f(1)\;= \textcolor{#010105}{(1)}^3 + 3\textcolor{#010105}{(1)}^2 - 2\textcolor{#010105}{(1)} + 4

Replace every

x

1

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(1)}\;= 1 + 3 - 2 + 4

Powers first, then the multiplications.

\displaystyle \phantom{f(1)}\;= \textcolor{#010106}{6}

Now add and subtract.

(ii)

Step 1: divide the leading terms,

x^3 \div x = x^2

. Write

x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (x^3 + 3x^2 - 2x + 4) - \textcolor{#010103}{(x^3 - x^2)} = 4x^2 - 2x + 4

Subtract

x^2

times the divisor, and change every sign of the product as you subtract.

Step 2: divide the leading terms,

4x^2 \div x = 4x

. Write

4x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (4x^2 - 2x + 4) - \textcolor{#010103}{(4x^2 - 4x)} = 2x + 4

Subtract

4x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

2x \div x = 2

. Write

2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (2x + 4) - \textcolor{#010103}{(2x - 2)} = 6

Subtract

2

times the divisor, and change every sign of the product as you subtract.

The remainder

6

has lower degree than the divisor, so the division stops.

Answer

(i) remainder

6

(ii) remainder

6

Your Turn 3.3

Let $f(x) = 2x^3 - x^2 + 5$ .

(i)Use the remainder theorem for the divisor $(x + 2)$ : what remainder does $f(x)$ leave?
(ii)Check your answer by long division.

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Colour keyterm subtracted or zero placeholdervalue substitutedfinal result

(i)

Set the divisor

x + 2

equal to zero:

x = -2

. By the remainder theorem the remainder is

f(-2)

, so we never need to divide.

For

(x - a)

the value to substitute is

a

\displaystyle f(-2)\;= 2\textcolor{#010105}{(-2)}^3 - \textcolor{#010105}{(-2)}^2 + 5

Replace every

x

-2

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(-2)}\;= -16 - 4 + 5

Powers first, then the multiplications.

\displaystyle \phantom{f(-2)}\;= \textcolor{#010106}{-15}

Now add and subtract.

(ii)

Before dividing, write a zero placeholder for every missing power (the dividend is missing

x

) so each power keeps its own column.

\displaystyle f(x)\;= 2x^3 - x^2 + \textcolor{#010103}{0x} + 5

The placeholder changes nothing: adding

0

does not change the polynomial.

Step 1: divide the leading terms,

2x^3 \div x = 2x^2

. Write

2x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(2x^3 - x^2 + \textcolor{#010103}{0x} + 5) - \textcolor{#010103}{(2x^3 + 4x^2)} = -5x^2 + 5

Subtract

2x^2

times the divisor, and change every sign of the product as you subtract. The placeholder column stays in the first row so every power lines up with the layout.

Step 2: divide the leading terms,

-5x^2 \div x = -5x

. Write

-5x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(-5x^2 + 5) - \textcolor{#010103}{(-5x^2 - 10x)} = 10x + 5

Subtract

-5x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

10x \div x = 10

. Write

10

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(x)}\;(10x + 5) - \textcolor{#010103}{(10x + 20)} = -15

Subtract

10

times the divisor, and change every sign of the product as you subtract.

The remainder

-15

has lower degree than the divisor, so the division stops.

Answer

(i) remainder

-15

(ii) remainder

-15

Think About It

Dividing

f(x) = x^3 - 3x^2 + x + 2

(x - 2)

leaves remainder

0

. What does that tell you about

f(2)

, and about the bracket?

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Colour keyvalue substituted

By the remainder theorem the remainder equals

f(2)

. Check it by substituting.

Remainder and value at

a

are the same number.

\displaystyle f(2)\;= \textcolor{#010105}{(2)}^3 - 3\textcolor{#010105}{(2)}^2 + \textcolor{#010105}{(2)} + 2 = 8 - 12 + 2 + 2 = 0

Substitute, then evaluate.

A remainder of

0

means

f(2) = 0

, so

(x - 2)

divides

f(x)

exactly: it is a factor. That observation is the factor theorem, coming next.

Answer

The remainder is

f(2) = 0

, so

(x - 2)

divides

f(x)

exactly, which makes it a factor.

Supports K341-A4-02

The factor theorem, unknown constants and more

A zero remainder is special enough to deserve its own name. The factor theorem says: $(ax + b)$ is a factor of $f(x)$ exactly when $f\left(-\dfrac{b}{a}\right) = 0$ . To show a bracket is a factor, substitute and get $0$ . To find a factor, look for a value that makes $f$ zero.

The Core Idea

$f(a) = 0$ if and only if $(x - a)$ is a factor.

Once you know one factor, divide it out to leave a quadratic, and the quadratic can be factorised or solved in the usual way.

Example 4.1

Show it is a factor, then factorise

Let $f(x) = x^3 - 3x^2 - 4x + 12$ .

(i)Use the factor theorem to confirm that $(x - 2)$ divides $f(x)$ exactly.
(ii)Hence factorise $f(x)$ completely.

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Colour keyterm subtracted or zero placeholdervalue substitutedfinal result

(i)

Set the divisor

x - 2

equal to zero:

x = 2

. By the remainder theorem the remainder is

f(2)

, so we never need to divide.

For

(x - a)

the value to substitute is

a

\displaystyle f(2)\;= \textcolor{#010105}{(2)}^3 - 3\textcolor{#010105}{(2)}^2 - 4\textcolor{#010105}{(2)} + 12

Replace every

x

2

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(2)}\;= 8 - 12 - 8 + 12

Powers first, then the multiplications.

\displaystyle \phantom{f(2)}\;= \textcolor{#010106}{0}

Now add and subtract.

The remainder is

0

, so

x - 2

divides

f(x)

exactly: it is a factor (factor theorem).

(ii)

The earlier part already showed that

(x - 2)

divides

f(x)

exactly, so go straight to dividing it out.

Step 1: divide the leading terms,

x^3 \div x = x^2

. Write

x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (x^3 - 3x^2 - 4x + 12) - \textcolor{#010103}{(x^3 - 2x^2)} = -x^2 - 4x + 12

Subtract

x^2

times the divisor, and change every sign of the product as you subtract.

Step 2: divide the leading terms,

-x^2 \div x = -x

. Write

-x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (-x^2 - 4x + 12) - \textcolor{#010103}{(-x^2 + 2x)} = -6x + 12

Subtract

-x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

-6x \div x = -6

. Write

-6

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (-6x + 12) - \textcolor{#010103}{(-6x + 12)} = 0

Subtract

-6

times the divisor, and change every sign of the product as you subtract.

The remainder

0

has lower degree than the divisor, so the division stops.

\displaystyle f(x)\;= (x - 2)(x^2 - x - 6)

Linear factor times quotient.

\displaystyle \phantom{f(x)}\;x^2 - x - 6 = (x + 2)(x - 3)

Factorise the quadratic: find two numbers with the right sum and product.

\displaystyle f(x)\;= x^3 - 3x^2 - 4x + 12 = (x - 2)(x^2 - x - 6) = \textcolor{#010106}{(x + 2)(x - 2)(x - 3)}

The fully factorised form.

Answer

(i) remainder

0

, so

x - 2

is a factor (it divides exactly)

(ii)

f(x) = (x + 2)(x - 2)(x - 3)

Your Turn 4.1

Let $f(x) = x^3 + 2x^2 - 5x - 6$ .

(i)Use the factor theorem to confirm that $(x + 1)$ divides $f(x)$ exactly.
(ii)Hence factorise $f(x)$ completely.

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Colour keyterm subtracted or zero placeholdervalue substitutedfinal result

(i)

Set the divisor

x + 1

equal to zero:

x = -1

. By the remainder theorem the remainder is

f(-1)

, so we never need to divide.

For

(x - a)

the value to substitute is

a

\displaystyle f(-1)\;= \textcolor{#010105}{(-1)}^3 + 2\textcolor{#010105}{(-1)}^2 - 5\textcolor{#010105}{(-1)} - 6

Replace every

x

-1

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(-1)}\;= -1 + 2 + 5 - 6

Powers first, then the multiplications.

\displaystyle \phantom{f(-1)}\;= \textcolor{#010106}{0}

Now add and subtract.

The remainder is

0

, so

x + 1

divides

f(x)

exactly: it is a factor (factor theorem).

(ii)

The earlier part already showed that

(x + 1)

divides

f(x)

exactly, so go straight to dividing it out.

Step 1: divide the leading terms,

x^3 \div x = x^2

. Write

x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (x^3 + 2x^2 - 5x - 6) - \textcolor{#010103}{(x^3 + x^2)} = x^2 - 5x - 6

Subtract

x^2

times the divisor, and change every sign of the product as you subtract.

Step 2: divide the leading terms,

x^2 \div x = x

. Write

x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (x^2 - 5x - 6) - \textcolor{#010103}{(x^2 + x)} = -6x - 6

Subtract

x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

-6x \div x = -6

. Write

-6

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle (-6x - 6) - \textcolor{#010103}{(-6x - 6)} = 0

Subtract

-6

times the divisor, and change every sign of the product as you subtract.

The remainder

0

has lower degree than the divisor, so the division stops.

\displaystyle f(x)\;= (x + 1)(x^2 + x - 6)

Linear factor times quotient.

\displaystyle \phantom{f(x)}\;x^2 + x - 6 = (x + 3)(x - 2)

Factorise the quadratic: find two numbers with the right sum and product.

\displaystyle f(x)\;= x^3 + 2x^2 - 5x - 6 = (x + 1)(x^2 + x - 6) = \textcolor{#010106}{(x + 3)(x + 1)(x - 2)}

The fully factorised form.

Answer

(i) remainder

0

, so

x + 1

is a factor (it divides exactly)

(ii)

f(x) = (x + 3)(x + 1)(x - 2)

Example 4.2

Two unknown constants

The polynomial $f(x) = 2x^3 + ax^2 - 5x + b$ , where $a$ and $b$ are constants, has $(x - 2)$ as a factor and leaves a remainder of $12$ when divided by $(x + 1)$ . Find $a$ and $b$ .

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Colour keyvalue substitutedfinal result

Divisor

x - 2

f(2) = 0

because it is a factor, so the remainder is

0

Remainder theorem: put the value that makes the divisor zero into

f(x)

\displaystyle f(2)\;= 2\textcolor{#010105}{(2)}^3 + a\textcolor{#010105}{(2)}^2 - 5\textcolor{#010105}{(2)} + b = 4 a + b + 6

Substitute, then simplify: a linear equation in the unknown constants.

\displaystyle 4 a + b = -6\quad (1)

Move the plain number to the right-hand side.

Divisor

x + 1

f(-1) = 12

because the remainder is

12

Remainder theorem: put the value that makes the divisor zero into

f(x)

\displaystyle f(-1)\;= 2\textcolor{#010105}{(-1)}^3 + a\textcolor{#010105}{(-1)}^2 - 5\textcolor{#010105}{(-1)} + b = a + b + 3

Substitute, then simplify: a linear equation in the unknown constants.

\displaystyle a + b = 9\quad (2)

Move the plain number to the right-hand side.

Solve the equations together.

\displaystyle b = - 4 a - 6

Make

b

the subject of equation (1).

\displaystyle \left(- 4 a - 6\right) + a = 9

Substitute into equation (2).

\displaystyle - 3 a = 15

Expand and collect: one unknown is left.

\displaystyle a = \textcolor{#010106}{-5}

Solve the one-unknown equation.

\displaystyle b = \textcolor{#010106}{14}

Put

a = -5

back into

b = - 4 a - 6

Put the values back:

f(x) = 2x^3 - 5x^2 - 5x + 14

Always write the completed polynomial; later parts use it.

Answer

a = -5

b = 14

Your Turn 4.2

The polynomial $f(x) = 3x^3 + ax^2 + bx - 6$ , where $a$ and $b$ are constants, has $(x - 1)$ as a factor and leaves a remainder of $-12$ when divided by $(x + 2)$ . Find $a$ and $b$ .

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Colour keyvalue substitutedfinal result

Divisor

x - 1

f(1) = 0

because it is a factor, so the remainder is

0

Remainder theorem: put the value that makes the divisor zero into

f(x)

\displaystyle f(1)\;= 3\textcolor{#010105}{(1)}^3 + a\textcolor{#010105}{(1)}^2 + b\textcolor{#010105}{(1)} - 6 = a + b - 3

Substitute, then simplify: a linear equation in the unknown constants.

\displaystyle a + b = 3\quad (1)

Move the plain number to the right-hand side.

Divisor

x + 2

f(-2) = -12

because the remainder is

-12

Remainder theorem: put the value that makes the divisor zero into

f(x)

\displaystyle f(-2)\;= 3\textcolor{#010105}{(-2)}^3 + a\textcolor{#010105}{(-2)}^2 + b\textcolor{#010105}{(-2)} - 6 = 4 a - 2 b - 30

Substitute, then simplify: a linear equation in the unknown constants.

\displaystyle 4 a - 2 b = 18\quad (2)

Move the plain number to the right-hand side.

Solve the equations together.

\displaystyle a = - b + 3

Make

a

the subject of equation (1).

\displaystyle 4 \left(- b + 3\right) - 2 b = 18

Substitute into equation (2).

\displaystyle - 6 b = 6

Expand and collect: one unknown is left.

\displaystyle b = \textcolor{#010106}{-1}

Solve the one-unknown equation.

\displaystyle a = \textcolor{#010106}{4}

Put

b = -1

back into

a = - b + 3

Put the values back:

f(x) = 3x^3 + 4x^2 - x - 6

Always write the completed polynomial; later parts use it.

Answer

a = 4

b = -1

Example 4.3

A repeated factor: how many different roots?

Show that the equation $x^3 - 5x^2 + 8x - 4 = 0$ has a repeated root. Find all the roots and say how many different real roots there are.

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Call the left-hand side

f(x)

\displaystyle f(1)\;= \textcolor{#010105}{(1)}^3 - 5\textcolor{#010105}{(1)}^2 + 8\textcolor{#010105}{(1)} - 4

Replace every

x

1

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(1)}\;= 1 - 5 + 8 - 4

Powers first, then the multiplications.

\displaystyle \phantom{f(1)}\;= \textcolor{#010106}{0}

Now add and subtract.

f(1) = 0

, so

\textcolor{#010104}{(x - 1)}

divides

f(x)

exactly: it is a factor (factor theorem).

Zero remainder means the bracket divides

f(x)

exactly.

Step 1: divide the leading terms,

x^3 \div x = x^2

. Write

x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(1)}\;(x^3 - 5x^2 + 8x - 4) - \textcolor{#010103}{(x^3 - x^2)} = -4x^2 + 8x - 4

Subtract

x^2

times the divisor, and change every sign of the product as you subtract.

Step 2: divide the leading terms,

-4x^2 \div x = -4x

. Write

-4x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(1)}\;(-4x^2 + 8x - 4) - \textcolor{#010103}{(-4x^2 + 4x)} = 4x - 4

Subtract

-4x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

4x \div x = 4

. Write

4

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(1)}\;(4x - 4) - \textcolor{#010103}{(4x - 4)} = 0

Subtract

4

times the divisor, and change every sign of the product as you subtract.

The remainder

0

has lower degree than the divisor, so the division stops.

\displaystyle f(x)\;= (x - 1)(x^2 - 4x + 4)

Linear factor times quotient.

\displaystyle \phantom{f(x)}\;x^2 - 4x + 4 = (x - 2)^2

Factorise the quadratic: find two numbers with the right sum and product.

\displaystyle f(x)\;= x^3 - 5x^2 + 8x - 4 = (x - 1)(x^2 - 4x + 4) = \textcolor{#010106}{(x - 1)(x - 2)^2}

The fully factorised form.

Set each factor equal to

0

(a product is zero only when a factor is zero).

\displaystyle x - 1 = 0 \quad\Rightarrow\quad x = 1

\displaystyle x - 2 = 0 \quad\Rightarrow\quad x = 2

(x - 2)^2

is a squared factor, so

x = 2

is a repeated root: it counts as one value of

x

So the equation has exactly 2 distinct real roots, because a squared factor gives its root only once, so the three linear factors supply just two different values.

Answer

f(x) = (x - 1)(x - 2)^2

;

x = 1, x = 2

;

x = 2

is a repeated root; exactly

2

distinct real roots

Your Turn 4.3

Show that the equation $x^3 - 4x^2 + 5x - 2 = 0$ has a repeated root. Find all the roots and say how many different real roots there are.

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Colour keyterm subtracted or zero placeholderfactor or divisorvalue substitutedfinal result

Call the left-hand side

f(x)

\displaystyle f(1)\;= \textcolor{#010105}{(1)}^3 - 4\textcolor{#010105}{(1)}^2 + 5\textcolor{#010105}{(1)} - 2

Replace every

x

1

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(1)}\;= 1 - 4 + 5 - 2

Powers first, then the multiplications.

\displaystyle \phantom{f(1)}\;= \textcolor{#010106}{0}

Now add and subtract.

f(1) = 0

, so

\textcolor{#010104}{(x - 1)}

divides

f(x)

exactly: it is a factor (factor theorem).

Zero remainder means the bracket divides

f(x)

exactly.

Step 1: divide the leading terms,

x^3 \div x = x^2

. Write

x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(1)}\;(x^3 - 4x^2 + 5x - 2) - \textcolor{#010103}{(x^3 - x^2)} = -3x^2 + 5x - 2

Subtract

x^2

times the divisor, and change every sign of the product as you subtract.

Step 2: divide the leading terms,

-3x^2 \div x = -3x

. Write

-3x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(1)}\;(-3x^2 + 5x - 2) - \textcolor{#010103}{(-3x^2 + 3x)} = 2x - 2

Subtract

-3x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

2x \div x = 2

. Write

2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(1)}\;(2x - 2) - \textcolor{#010103}{(2x - 2)} = 0

Subtract

2

times the divisor, and change every sign of the product as you subtract.

The remainder

0

has lower degree than the divisor, so the division stops.

\displaystyle f(x)\;= (x - 1)(x^2 - 3x + 2)

Linear factor times quotient.

\displaystyle \phantom{f(x)}\;x^2 - 3x + 2 = (x - 1)(x - 2)

Factorise the quadratic: find two numbers with the right sum and product.

\displaystyle f(x)\;= x^3 - 4x^2 + 5x - 2 = (x - 1)(x^2 - 3x + 2) = \textcolor{#010106}{(x - 1)^2(x - 2)}

The fully factorised form.

Set each factor equal to

0

(a product is zero only when a factor is zero).

\displaystyle x - 1 = 0 \quad\Rightarrow\quad x = 1

\displaystyle x - 2 = 0 \quad\Rightarrow\quad x = 2

(x - 1)^2

is a squared factor, so

x = 1

is a repeated root: it counts as one value of

x

So the equation has exactly 2 distinct real roots, because a squared factor gives its root only once, so the three linear factors supply just two different values.

Answer

f(x) = (x - 1)^2(x - 2)

;

x = 1, x = 2

;

x = 1

is a repeated root; exactly

2

distinct real roots

Example 4.4

Dividing by a quadratic: a linear remainder

When $f(x) = x^3 - 2x^2 + 3x + 5$ is divided by $(x^2 + x - 6)$ the remainder is of the form $Ax + B$ . Find $A$ and $B$ .

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Colour keyvalue substitutedfinal result

\displaystyle x^2 + x - 6 = (x + 3)(x - 2)

Factorise the divisor first: its roots are the values to substitute.

The divisor has degree

2

, so the remainder has degree at most

1

: write it as

Ax + B

. Then

f(x) = (x + 3)(x - 2)\,Q(x) + Ax + B

The product

(x - p)(x - q)

is zero at

x = p

and at

x = q

, which removes

Q(x)

\displaystyle f(-3)\;= \textcolor{#010105}{(-3)}^3 - 2\textcolor{#010105}{(-3)}^2 + 3\textcolor{#010105}{(-3)} + 5 = -49

Put

x = -3

into

f(x)

\displaystyle -3A + B = -49\quad (1)

Put

x = -3

into

Ax + B

: the bracket is zero, so

Q(x)

disappears.

\displaystyle f(2)\;= \textcolor{#010105}{(2)}^3 - 2\textcolor{#010105}{(2)}^2 + 3\textcolor{#010105}{(2)} + 5 = 11

Put

x = 2

into

f(x)

\displaystyle 2A + B = 11\quad (2)

Put

x = 2

into

Ax + B

\displaystyle \left(2 - (-3)\right)A =11 - (-49)

Subtract (1) from (2) to remove

B

\displaystyle A = \textcolor{#010106}{12}

Divide by the number in front of

A

\displaystyle B = -49 - (-3)\left(12\right) = \textcolor{#010106}{-13}

Put

A

back into (1).

The remainder is

12x - 13

Answer

A = 12

B = -13

, so the remainder is

12x - 13

Your Turn 4.4

When $f(x) = 2x^3 - 5x + 4$ is divided by $(x^2 - 2x - 3)$ the remainder is of the form $Ax + B$ . Find $A$ and $B$ .

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Colour keyvalue substitutedfinal result

\displaystyle x^2 - 2x - 3 = (x + 1)(x - 3)

Factorise the divisor first: its roots are the values to substitute.

The divisor has degree

2

, so the remainder has degree at most

1

: write it as

Ax + B

. Then

f(x) = (x + 1)(x - 3)\,Q(x) + Ax + B

The product

(x - p)(x - q)

is zero at

x = p

and at

x = q

, which removes

Q(x)

\displaystyle f(-1)\;= 2\textcolor{#010105}{(-1)}^3 - 5\textcolor{#010105}{(-1)} + 4 = 7

Put

x = -1

into

f(x)

\displaystyle -A + B = 7\quad (1)

Put

x = -1

into

Ax + B

: the bracket is zero, so

Q(x)

disappears.

\displaystyle f(3)\;= 2\textcolor{#010105}{(3)}^3 - 5\textcolor{#010105}{(3)} + 4 = 43

Put

x = 3

into

f(x)

\displaystyle 3A + B = 43\quad (2)

Put

x = 3

into

Ax + B

\displaystyle \left(3 - (-1)\right)A =43 - 7

Subtract (1) from (2) to remove

B

\displaystyle A = \textcolor{#010106}{9}

Divide by the number in front of

A

\displaystyle B = 7 - (-1)\left(9\right) = \textcolor{#010106}{16}

Put

A

back into (1).

The remainder is

9x + 16

Answer

A = 9

B = 16

, so the remainder is

9x + 16

Example 4.5

Build the polynomial from its roots

A cubic polynomial $f(x)$ is zero at $x = -2$ , $x = 1$ and $x = 3$ , and $f(2) = -24$ . Find $f(x)$ , in expanded form.

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Colour keyfinal result

A polynomial with roots

-2, 1, 3

has factors

(x + 2)(x - 1)(x - 3)

. The leading coefficient is not known to be

1

, so put an unknown constant

k

in front.

Never assume the leading coefficient is

1

: the extra given value is there to find

k

\displaystyle f(x)\;= k(x + 2)(x - 1)(x - 3)

k

is the leading coefficient.

\displaystyle -24 = k(2 + 2)(2 - 1)(2 - 3)

Use

f(2) = -24

\displaystyle -24 = -4k

Work out each bracket, then multiply them.

\displaystyle k = \textcolor{#010106}{6}

Divide by

-4

Multiply out two brackets at a time to get the expanded form.

\displaystyle \phantom{f(x)}\;(x + 2)(x - 1) = x(x - 1) + 2(x - 1)

Multiply every term of the first bracket by the whole second bracket.

\displaystyle \phantom{f(x)}\;= (x^2 - x) + (2x - 2)

Expand each product, keeping its sign.

\displaystyle \phantom{f(x)}\;= \textcolor{#010106}{x^2 + x - 2}

Collect the terms that have the same power of

x

\displaystyle \phantom{f(x)}\;(x^2 + x - 2)(x - 3) = x^2(x - 3) + x(x - 3) - 2(x - 3)

Multiply every term of the first bracket by the whole second bracket.

\displaystyle \phantom{f(x)}\;= (x^3 - 3x^2) + (x^2 - 3x) + (-2x + 6)

Expand each product, keeping its sign.

\displaystyle \phantom{f(x)}\;= \textcolor{#010106}{x^3 - 2x^2 - 5x + 6}

Collect the terms that have the same power of

x

\displaystyle f(x)\;= 6(x^3 - 2x^2 - 5x + 6) = \textcolor{#010106}{6x^3 - 12x^2 - 30x + 36}

Multiply every term by

k

at the end.

Answer

k = 6

f(x) = 6x^3 - 12x^2 - 30x + 36

Your Turn 4.5

A cubic polynomial $f(x)$ is zero at $x = -1$ , $x = 2$ and $x = 4$ , and $f(1) = -12$ . Find $f(x)$ , in expanded form.

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Colour keyfinal result

A polynomial with roots

-1, 2, 4

has factors

(x + 1)(x - 2)(x - 4)

. The leading coefficient is not known to be

1

, so put an unknown constant

k

in front.

Never assume the leading coefficient is

1

: the extra given value is there to find

k

\displaystyle f(x)\;= k(x + 1)(x - 2)(x - 4)

k

is the leading coefficient.

\displaystyle -12 = k(1 + 1)(1 - 2)(1 - 4)

Use

f(1) = -12

\displaystyle -12 = 6k

Work out each bracket, then multiply them.

\displaystyle k = \textcolor{#010106}{-2}

Divide by

6

Multiply out two brackets at a time to get the expanded form.

\displaystyle \phantom{f(x)}\;(x + 1)(x - 2) = x(x - 2) + 1(x - 2)

Multiply every term of the first bracket by the whole second bracket.

\displaystyle \phantom{f(x)}\;= (x^2 - 2x) + (x - 2)

Expand each product, keeping its sign.

\displaystyle \phantom{f(x)}\;= \textcolor{#010106}{x^2 - x - 2}

Collect the terms that have the same power of

x

\displaystyle \phantom{f(x)}\;(x^2 - x - 2)(x - 4) = x^2(x - 4) - x(x - 4) - 2(x - 4)

Multiply every term of the first bracket by the whole second bracket.

\displaystyle \phantom{f(x)}\;= (x^3 - 4x^2) + (-x^2 + 4x) + (-2x + 8)

Expand each product, keeping its sign.

\displaystyle \phantom{f(x)}\;= \textcolor{#010106}{x^3 - 5x^2 + 2x + 8}

Collect the terms that have the same power of

x

\displaystyle f(x)\;= -2(x^3 - 5x^2 + 2x + 8) = \textcolor{#010106}{-2x^3 + 10x^2 - 4x - 16}

Multiply every term by

k

at the end.

Answer

k = -2

f(x) = -2x^3 + 10x^2 - 4x - 16

Do not assume the leading coefficient is

1

. A cubic with roots

p

q

r

k(x - p)(x - q)(x - r)

, and the extra given value is there to find

k

Example 4.6

Part (b) does not need part (a)

The polynomial $f(x) = x^3 + ax^2 - 4x - 12$ has $(x + 2)$ as a factor, where $a$ is a constant.

(i)Show that $a = 3$ .
(ii)Using $a = 3$ (whether or not you managed the first part), factorise $f(x)$ completely.

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Colour keyterm subtracted or zero placeholderfactor or divisorvalue substitutedfinal result

(i)

Divisor

x + 2

f(-2) = 0

because it is a factor, so the remainder is

0

Remainder theorem: put the value that makes the divisor zero into

f(x)

\displaystyle f(-2)\;= \textcolor{#010105}{(-2)}^3 + a\textcolor{#010105}{(-2)}^2 - 4\textcolor{#010105}{(-2)} - 12 = 4 a - 12

Substitute, then simplify: a linear equation in the unknown constants.

\displaystyle 4 a = 12\quad (1)

Move the plain number to the right-hand side.

Solve the equations together.

\displaystyle 4 a = 12

One unknown left: a one-step equation.

\displaystyle a = \textcolor{#010106}{3}

Divide by the coefficient.

Put the values back:

f(x) = x^3 + 3x^2 - 4x - 12

Always write the completed polynomial; later parts use it.

(ii)

The problem tells us

x + 2

is a factor, so test it: the remainder must be

0

\displaystyle f(-2)\;= \textcolor{#010105}{(-2)}^3 + 3\textcolor{#010105}{(-2)}^2 - 4\textcolor{#010105}{(-2)} - 12

Replace every

x

-2

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(-2)}\;= -8 + 12 + 8 - 12

Powers first, then the multiplications.

\displaystyle \phantom{f(-2)}\;= \textcolor{#010106}{0}

Now add and subtract.

f(-2) = 0

, so

\textcolor{#010104}{(x + 2)}

divides

f(x)

exactly: it is a factor (factor theorem).

Zero remainder means the bracket divides

f(x)

exactly.

Step 1: divide the leading terms,

x^3 \div x = x^2

. Write

x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(-2)}\;(x^3 + 3x^2 - 4x - 12) - \textcolor{#010103}{(x^3 + 2x^2)} = x^2 - 4x - 12

Subtract

x^2

times the divisor, and change every sign of the product as you subtract.

Step 2: divide the leading terms,

x^2 \div x = x

. Write

x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(-2)}\;(x^2 - 4x - 12) - \textcolor{#010103}{(x^2 + 2x)} = -6x - 12

Subtract

x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

-6x \div x = -6

. Write

-6

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(-2)}\;(-6x - 12) - \textcolor{#010103}{(-6x - 12)} = 0

Subtract

-6

times the divisor, and change every sign of the product as you subtract.

The remainder

0

has lower degree than the divisor, so the division stops.

\displaystyle f(x)\;= (x + 2)(x^2 + x - 6)

Linear factor times quotient.

\displaystyle \phantom{f(x)}\;x^2 + x - 6 = (x + 3)(x - 2)

Factorise the quadratic: find two numbers with the right sum and product.

\displaystyle f(x)\;= x^3 + 3x^2 - 4x - 12 = (x + 2)(x^2 + x - 6) = \textcolor{#010106}{(x + 3)(x + 2)(x - 2)}

The fully factorised form.

Answer

(i)

a = 3

(ii)

f(x) = (x + 3)(x + 2)(x - 2)

Your Turn 4.6

The polynomial $f(x) = 2x^3 + ax^2 - x - 6$ has $(x - 2)$ as a factor, where $a$ is a constant.

(i)Show that $a = -2$ .
(ii)Using $a = -2$ (whether or not you managed the first part), factorise $f(x)$ completely.

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Colour keyterm subtracted or zero placeholderfactor or divisorvalue substitutedfinal resultdiscriminant D

(i)

Divisor

x - 2

f(2) = 0

because it is a factor, so the remainder is

0

Remainder theorem: put the value that makes the divisor zero into

f(x)

\displaystyle f(2)\;= 2\textcolor{#010105}{(2)}^3 + a\textcolor{#010105}{(2)}^2 - \textcolor{#010105}{(2)} - 6 = 4 a + 8

Substitute, then simplify: a linear equation in the unknown constants.

\displaystyle 4 a = -8\quad (1)

Move the plain number to the right-hand side.

Solve the equations together.

\displaystyle 4 a = -8

One unknown left: a one-step equation.

\displaystyle a = \textcolor{#010106}{-2}

Divide by the coefficient.

Put the values back:

f(x) = 2x^3 - 2x^2 - x - 6

Always write the completed polynomial; later parts use it.

(ii)

The problem tells us

x - 2

is a factor, so test it: the remainder must be

0

\displaystyle f(2)\;= 2\textcolor{#010105}{(2)}^3 - 2\textcolor{#010105}{(2)}^2 - \textcolor{#010105}{(2)} - 6

Replace every

x

2

and keep the brackets so signs cannot slip.

\displaystyle \phantom{f(2)}\;= 16 - 8 - 2 - 6

Powers first, then the multiplications.

\displaystyle \phantom{f(2)}\;= \textcolor{#010106}{0}

Now add and subtract.

f(2) = 0

, so

\textcolor{#010104}{(x - 2)}

divides

f(x)

exactly: it is a factor (factor theorem).

Zero remainder means the bracket divides

f(x)

exactly.

Step 1: divide the leading terms,

2x^3 \div x = 2x^2

. Write

2x^2

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(2)}\;(2x^3 - 2x^2 - x - 6) - \textcolor{#010103}{(2x^3 - 4x^2)} = 2x^2 - x - 6

Subtract

2x^2

times the divisor, and change every sign of the product as you subtract.

Step 2: divide the leading terms,

2x^2 \div x = 2x

. Write

2x

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(2)}\;(2x^2 - x - 6) - \textcolor{#010103}{(2x^2 - 4x)} = 3x - 6

Subtract

2x

times the divisor, and change every sign of the product as you subtract.

Step 3: divide the leading terms,

3x \div x = 3

. Write

3

in the quotient.

Only the two leading terms decide the next quotient term.

\displaystyle \phantom{f(2)}\;(3x - 6) - \textcolor{#010103}{(3x - 6)} = 0

Subtract

3

times the divisor, and change every sign of the product as you subtract.

The remainder

0

has lower degree than the divisor, so the division stops.

\displaystyle f(x)\;= (x - 2)(2x^2 + 2x + 3)

Linear factor times quotient.

Check the quadratic for further factors with its discriminant

\textcolor{#010102}{D} = b^2 - 4ac

\displaystyle \textcolor{#010102}{D}\;= 2^2 - 4(2)(3) = -20

\textcolor{#010102}{D} < 0

, so the quadratic has no real roots and cannot be factorised over the real numbers.

\displaystyle f(x)\;= 2x^3 - 2x^2 - x - 6 = \textcolor{#010106}{(x - 2)(2x^2 + 2x + 3)}

The fully factorised form.

Answer

(i)

a = -2

(ii)

f(x) = (x - 2)(2x^2 + 2x + 3)

(the quadratic factor has

\textcolor{#010102}{D} < 0

, so it cannot be factorised further)

SGSchoolKaki Tip

In a 'show that' part the answer is given, so you can use it in the next part even if you could not prove it. Always read on: later marks do not depend on the earlier proof.

Links To

The quadratic that is left after taking out a factor is handled with the discriminant, exactly as in Chapter 1: Quadratic functions

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Notes version 1.0 · Last updated 3 Nov 2026 · Chapter 4 of 17

Polynomials, cubic identities and partial fractions

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What is a polynomial? Degree, f(a)f(a)f(a) and multiplying

Dividing polynomials: long division and the division statement

The remainder theorem: the remainder without dividing

The factor theorem, unknown constants and more

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What is a polynomial? Degree, $f(a)$ and multiplying