Polynomials, factor theorem and algebraic divisionEdexcel A-Level Maths: Revision notes

Section 1

Manipulating polynomials

A polynomial is a sum of terms of the form $ax^n$ with $n$ a non-negative integer, such as $f(x)=x^3+3x^2-4$ . Its degree is the highest power. To expand brackets, multiply every term in one bracket by every term in the other, then collect like terms (same power of $x$ ). Example: $(x-1)(x+2)^2=(x-1)(x^2+4x+4)=x^3+4x^2+4x-x^2-4x-4=x^3+3x^2-4$ . To factorise, take out any common factor first, then look for a quadratic factorisation (sum, product or difference of squares). Cubics need the factor theorem, below.

Key termspolynomialdegreelike terms

Common mistake

Squaring a bracket as $(x+2)^2=x^2+4$ . It is $x^2+4x+4$ .

Section 2

Algebraic division

You can divide a polynomial by a linear expression $(ax+b)$ using long division. Divide the leading terms each time, multiply back, subtract, and bring down the next term. Example: $(2x^3+x^2-13x+6)\div(x-2)$ .

$2x^3\div x=2x^2$ ; subtract $2x^3-4x^2$ , leaving $5x^2-13x$ .
$5x^2\div x=5x$ ; subtract $5x^2-10x$ , leaving $-3x+6$ .
$-3x\div x=-3$ ; subtract $-3x+6$ , leaving $0$ . The quotient is $2x^2+5x-3$ with remainder $0$ , so $2x^3+x^2-13x+6=(x-2)(2x^2+5x-3)$ . Division by $(2x+3)$ works the same way: $(6x^2+5x-6)\div(2x+3)=3x-2$ . A zero remainder means $(ax+b)$ is a factor.

Key termsquotientremainder

Exam tip

Write a $0x^2$ or $0x$ term for any missing power so the columns line up.

Common mistake

Forgetting to change every sign when subtracting the line below.

Section 3

The factor theorem

Factor theorem: if $f\left(\frac ba\right)=0$ then $(ax-b)$ is a factor of $f(x)$ . The converse is also true: if $(ax-b)$ is a factor then $f\left(\frac ba\right)=0$ . Example: $f(x)=x^3+3x^2-4$ . $f(1)=1+3-4=0$ , so $(x-1)$ is a factor. $f(-1)=-2$ , so $(x+1)$ is not. For $(2x-1)$ test $x=\frac12$ ; for $(2x+3)$ test $x=-\frac32$ . The sign flips: the factor $(x+p)$ gives the root $x=-p$ .

Key termsfactor theoremroot

Common mistake

Testing $f(2)$ to check the factor $(x+2)$ . Test $f(-2)$ .

Section 4

Factorising cubics

Try $x=\pm1,\pm2,\ldots$ (factors of the constant term, divided by factors of the leading coefficient) until $f(x)=0$ .
Divide by the factor you found to get a quadratic.
Factorise the quadratic. Example: $x^3+3x^2-4$ . $f(1)=0$ , and dividing by $(x-1)$ gives $x^2+4x+4$ , so $f(x)=(x-1)(x+2)^2$ . The root $x=-2$ is repeated. Example: $6x^3+11x^2-x-6$ . $f(-1)=-6+11+1-6=0$ , so $(x+1)$ is a factor. Dividing gives $6x^2+5x-6=(3x-2)(2x+3)$ , so $f(x)=(x+1)(3x-2)(2x+3)$ and the roots are $-1,\ \frac23,\ -\frac32$ .

Key termsrepeated root

Exam tip

Check your factorisation by expanding, or by testing one value of $x$ in both forms.

Section 5

Finding unknown constants

When a polynomial contains an unknown constant and you are told it has a factor, use the factor theorem to form an equation. Example: $p(x)=2x^3+ax^2-x+6$ has factor $(x-2)$ . Then $p(2)=16+4a-2+6=0$ , so $a=-5$ and $p(x)=(x-2)(2x-3)(x+1)$ . With two unknowns you need two factors, giving simultaneous equations. For $6x^3+11x^2+ax+b$ with factors $(x+1)$ and $(x-1)$ : $b-a=-5$ and $a+b=-17$ , so $a=-6$ and $b=-11$ .

Key termsconstant

Exam tip

Number of unknown constants = number of equations you need.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Polynomials, factor theorem and algebraic division

The polynomial $f(x)=x^3+3x^2-4$ .
Hence solve $f(x)=0$ , stating which root is repeated.2 marks
The polynomial $g(x)=2x^3+x^2-13x+6$ .
Given that $(x-2)$ is a factor, factorise $g(x)$ completely.2 marks
The polynomial $p(x)=2x^3+ax^2-x+6$ , where $a$ is a constant, has a factor $(x-2)$ .
Find the value of $a$ .3 marks

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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).