Mathematics 9709 · Pure Mathematics 2

The factor theorem

The chapter's best shortcut: to test whether $x-c$ is a factor, don't divide. Just plug a number in.
A factor divides exactly, with no remainder. The remainder (next section) is just $P(c)$ . So a factor is a value of x that makes the polynomial equal 0.

$P(c)=0 \iff (x-c)\text{ is a factor}$ . To test a factor, plug in its root and check for 0. Plug in, never divide.

A factor ⟺ the polynomial is zero there

$P(c)=0 \iff (x-c)\text{ is a factor of }P(x)$

For a slanted factor $ax-b$ , test its root $x=\tfrac{b}{a}$ : if $P\!\left(\tfrac{b}{a}\right)=0$ then $ax-b$ is a factor.

Worked example

Show that $x-3$ is a factor of $x^3+2x^2-16x+3$ (Worked example 1.9, textbook)

The root of $x-3$ is $x=3$ , so work out $f(3)$ :

$f(3)=27+18-48+3=0$

Answer

f(3)=0\ \Rightarrow\ (x-3)\text{ is a factor}

Worked example

Real question: show $x+3$ is a factor of $x^3+5x^2+31x+75$

9709/31 May/June 2023 Q10(a) — the factor test is all of part (a).

The root of $x+3$ is $x=-3$ , so evaluate $p(-3)$ .
$p(-3)=(-27)+5(9)+31(-3)+75=-27+45-93+75=0$ .

Answer

p(-3)=0\ \Rightarrow\ (x+3)\text{ is a factor}

Use it to factorise, or to find unknowns

To factorise a cubic: find one root by trying small factors of the last number $(\pm1,\pm2,\dots)$ , divide that linear factor out, then factorise the quadratic that is left.
To find unknown coefficients, each factor you are given gives one equation $P(\text{root})=0$ .

Worked example

Solve $x^3+4x^2-19x+14=0$ (Worked example 1.11, textbook)

Try factors of 14: $f(1)=1+4-19+14=0$ , so $x-1$ is a factor.
Divide it out: $x^3+4x^2-19x+14=(x-1)(x^2+5x-14)$ .
Factorise the quadratic: $x^2+5x-14=(x-2)(x+7)$ , giving $(x-1)(x-2)(x+7)=0$ .

Answer

x=1,\ 2,\ -7

Worked example

Real question: factorise $f(x)=8x^3+54x^2-17x-21$ given $x+7$ is a factor

9709/33 May/June 2024 Q7 — part (a) is the factor test; part (b) divides to get the quotient, then factorises it.

Part (a): $f(-7)=8(-343)+54(49)-17(-7)-21=-2744+2646+119-21=0$ ✓.
Part (b): divide out $x+7$ to get the quotient $8x^2-2x-3$ .
Factorise the quotient: $8x^2-2x-3=(2x+1)(4x-3)$ .

Answer

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

Common mistake

Check the sign of the root.

x-3

tests

x=+3

;

x+3

tests

x=-3

;

2x-1

tests

x=\tfrac12

. Set the bracket equal to 0 to read off the right value.

Now you try

The polynomial $p(x)=ax^3-10x^2+bx+8$ has $x-2$ as a factor of both $p(x)$ and its derivative $p'(x)$ . Find $a$ and $b$ . (9709/31 May/June 2022 Q5)

Your turn— tap to reveal the worked answer (9709/31 May/June 2022 Q5)

A factor of $p$ means $p(2)=0$ ; a factor of $p'$ means $p'(2)=0$ . Two clues, two equations:

$p(2)=8a-40+2b+8=0\Rightarrow 8a+2b=32$

$p'(x)=3ax^2-20x+b,\ \ p'(2)=12a-40+b=0$

Solving the pair gives:

Answer

a=3,\quad b=4

PreviousDivision of polynomials NextThe remainder theorem

The factor theorem

The chapter's best shortcut: to test whether $x-c$ is a factor, don't divide. Just plug a number in.
A factor divides exactly, with no remainder. The remainder (next section) is just $P(c)$ . So a factor is a value of x that makes the polynomial equal 0.

$P(c)=0 \iff (x-c)\text{ is a factor}$ . To test a factor, plug in its root and check for 0. Plug in, never divide.

A factor ⟺ the polynomial is zero there

$P(c)=0 \iff (x-c)\text{ is a factor of }P(x)$

For a slanted factor $ax-b$ , test its root $x=\tfrac{b}{a}$ : if $P\!\left(\tfrac{b}{a}\right)=0$ then $ax-b$ is a factor.

Worked example

Show that $x-3$ is a factor of $x^3+2x^2-16x+3$ (Worked example 1.9, textbook)

The root of $x-3$ is $x=3$ , so work out $f(3)$ :

$f(3)=27+18-48+3=0$

Answer

f(3)=0\ \Rightarrow\ (x-3)\text{ is a factor}

Worked example

Real question: show $x+3$ is a factor of $x^3+5x^2+31x+75$

9709/31 May/June 2023 Q10(a) — the factor test is all of part (a).

The root of $x+3$ is $x=-3$ , so evaluate $p(-3)$ .
$p(-3)=(-27)+5(9)+31(-3)+75=-27+45-93+75=0$ .

Answer

p(-3)=0\ \Rightarrow\ (x+3)\text{ is a factor}

Use it to factorise, or to find unknowns

To factorise a cubic: find one root by trying small factors of the last number $(\pm1,\pm2,\dots)$ , divide that linear factor out, then factorise the quadratic that is left.
To find unknown coefficients, each factor you are given gives one equation $P(\text{root})=0$ .

Worked example

Solve $x^3+4x^2-19x+14=0$ (Worked example 1.11, textbook)

Try factors of 14: $f(1)=1+4-19+14=0$ , so $x-1$ is a factor.
Divide it out: $x^3+4x^2-19x+14=(x-1)(x^2+5x-14)$ .
Factorise the quadratic: $x^2+5x-14=(x-2)(x+7)$ , giving $(x-1)(x-2)(x+7)=0$ .

Answer

x=1,\ 2,\ -7

Worked example

Real question: factorise $f(x)=8x^3+54x^2-17x-21$ given $x+7$ is a factor

9709/33 May/June 2024 Q7 — part (a) is the factor test; part (b) divides to get the quotient, then factorises it.

Part (a): $f(-7)=8(-343)+54(49)-17(-7)-21=-2744+2646+119-21=0$ ✓.
Part (b): divide out $x+7$ to get the quotient $8x^2-2x-3$ .
Factorise the quotient: $8x^2-2x-3=(2x+1)(4x-3)$ .

Answer

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

Common mistake

Check the sign of the root.

x-3

tests

x=+3

;

x+3

tests

x=-3

;

2x-1

tests

x=\tfrac12

. Set the bracket equal to 0 to read off the right value.

Now you try

The polynomial $p(x)=ax^3-10x^2+bx+8$ has $x-2$ as a factor of both $p(x)$ and its derivative $p'(x)$ . Find $a$ and $b$ . (9709/31 May/June 2022 Q5)

Your turn— tap to reveal the worked answer (9709/31 May/June 2022 Q5)

A factor of $p$ means $p(2)=0$ ; a factor of $p'$ means $p'(2)=0$ . Two clues, two equations:

$p(2)=8a-40+2b+8=0\Rightarrow 8a+2b=32$

$p'(x)=3ax^2-20x+b,\ \ p'(2)=12a-40+b=0$

Solving the pair gives:

Answer

a=3,\quad b=4