Factor Theorem and Remainder TheoremEdexcel International A Level Maths: Revision notes

Section 1

The Remainder Theorem

When a polynomial $f(x)$ is divided by $(ax-b)$ , the remainder is $f\left(\frac{b}{a}\right)$ . This lets you find a remainder by substitution, with no division.

Divide by $(x-3)$ : remainder $f(3)$ .
Divide by $(x+1)$ : remainder $f(-1)$ .
Divide by $(2x-1)$ : remainder $f\left(\frac12\right)$ .

For $f(x)=x^3-7x+6$ , the remainder on division by $(x-3)$ is $f(3)=27-21+6=12$ .

Key termsremainder theoremremainder

Common mistake

Using the wrong sign: the remainder for $(x+2)$ is $f(-2)$ , not $f(2)$ .

Section 2

The Factor Theorem

If the remainder is zero then the divisor is a factor. The Factor Theorem says: if $f\left(\frac{b}{a}\right)=0$ then $(ax-b)$ is a factor of $f(x)$ . The converse also holds: if $(ax-b)$ is a factor then $f\left(\frac{b}{a}\right)=0$ .

To show that $(x+3)$ is a factor of $f(x)=x^3-7x+6$ : $f(-3)=-27+21+6=0,$ so $(x+3)$ is a factor by the Factor Theorem.

Always state both the value and the conclusion: ' $f(-3)=0$ , so $(x+3)$ is a factor'.

Key termsFactor Theoremroot

Exam tip

A factor $(ax-b)$ corresponds to the root $x=\frac{b}{a}$ . Match the sign carefully.

Section 3

Factorising a cubic

To factorise a cubic $f(x)$ :

Find a value $x=\frac{b}{a}$ with $f\left(\frac{b}{a}\right)=0$ by trial. Try $\pm1$ , $\pm2$ first. For other values, $b$ divides the constant term and $a$ divides the leading coefficient.
Write down the linear factor $(ax-b)$ .
Divide, or compare coefficients, to find the quadratic factor.
Factorise the quadratic.

Example: $f(x)=x^3+3x^2-4$ . $f(1)=0$ , so $(x-1)$ is a factor. Then $f(x)=(x-1)(x^2+4x+4)=(x-1)(x+2)^2$ .

Example: $f(x)=6x^3+11x^2-x-6$ . $f(-1)=-6+11+1-6=0$ , so $(x+1)$ is a factor. Then $f(x)=(x+1)(6x^2+5x-6)=(x+1)(3x-2)(2x+3)$ .

Key termslinear factorquadratic factor

Exam tip

Check your factorisation by expanding, or by substituting a simple value such as $x=1$ .

Section 4

Solving cubic equations

To solve $f(x)=0$ , factorise $f(x)$ completely and set each factor to zero. For $f(x)=6x^3+11x^2-3x-2$ : $f(-2)=0$ , so $f(x)=(x+2)(6x^2-x-1)=(x+2)(3x+1)(2x-1)$ .

The solutions are $x=-2$ , $x=-\frac13$ and $x=\frac12$ .

If the quadratic factor does not factorise, use the discriminant $b^2-4ac$ to decide whether it has real roots. A negative discriminant means no real roots, so the cubic then has just one real root.

Key termsdiscriminant

Common mistake

Stopping after finding one factor. Factorise the quadratic as well, or show that it has no real roots.

Section 5

Finding unknown coefficients

When a cubic contains unknown constants, use factor and remainder information to form equations.

Example: $p(x)=2x^3+ax^2+bx-6$ has factor $(x-2)$ and leaves remainder $-12$ on division by $(x+1)$ .

$p(2)=0$ : $16+4a+2b-6=0$ , so $2a+b=-5$ . $p(-1)=-12$ : $-2+a-b-6=-12$ , so $a-b=-4$ .

Solving gives $a=-3$ and $b=1$ . Check: $p(2)=16-12+2-6=0$ and $p(-1)=-2-3-1-6=-12$ .

Key termsunknown coefficient

Common mistake

Setting $p(-1)=0$ when the question gives a non-zero remainder. A remainder of $-12$ means $p(-1)=-12$ .

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Factor Theorem and Remainder Theorem

Let $f(x)=x^3-7x+6$ .
Show that $(x+3)$ is a factor of $f(x)$ .2 marks
Let $g(x)=2x^3+x^2-13x+6$ .
Show that $(x+3)$ is a factor of $g(x)$ .2 marks
Let $f(x)=6x^3+11x^2-3x-2$ .
Show that $(x+2)$ is a factor of $f(x)$ , and find the quadratic $q(x)$ such that $f(x)=(x+2)q(x)$ .3 marks

See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).