Factorisation Of Polynomials
Understand factorising polynomials for Queensland Year 11 Mathematical Methods (QCAA). Factorising rewrites a polynomial as a product of simpler factors, which reveals its zeros and makes equations far easier to solve.
You will learn to factorise a sum or difference of two cubes, use the factor theorem to break a cubic into a linear factor and a quadratic, group terms, and recognise an irreducible quadratic — foundational QCAA algebra.
Every question with a fully worked solution.
- Factorisation Of Polynomials - Video - Factorisation of polynomials Watch
Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), to factorise a polynomial is to write it as a product of simpler factors. This page covers the sum and difference of two cubes, factorising by grouping, the factor theorem, fully factorising a cubic, factorising a quartic as a quadratic in \(x^2\), and spotting an irreducible quadratic factor.
To factorise a polynomial is to rewrite it as a product of lower-degree factors. Two useful identities are the difference of cubes \(a^3-b^3=(a-b)(a^2+ab+b^2)\) and the sum of cubes \(a^3+b^3=(a+b)(a^2-ab+b^2)\).
Factorising by grouping pairs terms that share a common factor. The factor theorem gives linear factors of a cubic: if \(P(a)=0\) then \((x-a)\) is a factor, so test small values \(x=\pm1,\pm2,\dots\) (the rational-zero candidates divide the constant term).
A quadratic with a negative discriminant \((\Delta=b^2-4ac\lt 0)\) is irreducible over the real numbers: it cannot be broken down further, and you report the polynomial as "fully factorised over the reals".
Difference and sum of two cubes:
Factor theorem (linear factor of \(P\)):
Discriminant test for an irreducible quadratic:
How to fully factorise a polynomial
- Common factor: take out any common numerical or variable factor first, and check for a special form (difference/sum of cubes, or a quadratic in \(x^2\)).
- Find a factor: for a cubic with no obvious grouping, use the factor theorem — test \(x=\pm1,\pm2,\dots\) until \(P(a)=0\), giving the factor \((x-a)\).
- Divide and finish: divide out the factor, factorise the remaining quadratic, and stop when every factor is linear or an irreducible quadratic.
Write each term as a cube: \(27x^3=(3x)^3\) and \(8=(2)^3\):
| \(27x^3-8\) | \(=\) | \((3x)^3-(2)^3\) |
Apply \(a^3-b^3=(a-b)(a^2+ab+b^2)\) with \(a=3x,\ b=2\):
| \(=\) | \((3x-2)\big((3x)^2+(3x)(2)+(2)^2\big)\) | |
| \(=\) | \((3x-2)(9x^2+6x+4)\) |
\(27x^3-8=(3x-2)(9x^2+6x+4)\).
Pair the terms and take a common factor from each pair:
| \(x^3+3x^2-4x-12\) | \(=\) | \(x^2(x+3)-4(x+3)\) |
Both pairs share \((x+3)\); factor it out:
| \(=\) | \((x+3)(x^2-4)\) |
\(x^2-4\) is a difference of squares:
| \(=\) | \((x+3)(x-2)(x+2)\) |
\(x^3+3x^2-4x-12=(x+3)(x-2)(x+2)\).
Test candidates for a factor. Try \(x=3\):
| \(P(3)\) | \(=\) | \(2(3)^3-3(3)^2-11(3)+6\) |
| \(=\) | \(54-27-33+6\) | |
| \(=\) | \(0\) |
So \(P(3)=0\) and \((x-3)\) is a factor. Dividing gives the quotient \(2x^2+3x-2\).
Factorise the quadratic \(2x^2+3x-2\):
| \(2x^2+3x-2\) | \(=\) | \(2x^2+4x-x-2\) |
| \(=\) | \(2x(x+2)-(x+2)\) | |
| \(=\) | \((2x-1)(x+2)\) |
\(2x^3-3x^2-11x+6=(x-3)(2x-1)(x+2)\).
Treat it as a quadratic in \(x^2\) and factorise:
| \(x^4-5x^2-36\) | \(=\) | \((x^2-9)(x^2+4)\) |
\(x^2-9\) is a difference of squares; test \(x^2+4\):
| \(x^2-9\) | \(=\) | \((x-3)(x+3)\) |
| \(\Delta\text{ of }x^2+4\) | \(=\) | \(0-4(1)(4)=-16\) |
Since \(\Delta\lt 0\), \(x^2+4\) is irreducible over the reals and stays as a factor.
\(x^4-5x^2-36=(x-3)(x+3)(x^2+4)\).
Common pitfalls
Frequently asked questions
How do you factorise a difference of two cubes?
Use \(a^3-b^3=(a-b)(a^2+ab+b^2)\). Write each term as a cube to read off \(a\) and \(b\); for example \(27x^3-8=(3x-2)(9x^2+6x+4)\).
How do you factorise a cubic polynomial?
Find one linear factor with the factor theorem (a value \(a\) with \(P(a)=0\)), divide it out to leave a quadratic, then factorise that quadratic.
What is the factor theorem used for in factorising?
It locates linear factors: if \(P(a)=0\) then \((x-a)\) is a factor. You test the values \(x=\pm1,\pm2,\dots\) that divide the constant term.
What does it mean for a quadratic to be irreducible?
Its discriminant \(\Delta=b^2-4ac\) is negative, so it has no real roots and cannot be factorised over the real numbers. You leave it as a single factor.
How do you factorise a quartic like \(x^4-5x^2-36\)?
Treat it as a quadratic in \(x^2\): \(x^4-5x^2-36=(x^2-9)(x^2+4)\), then factorise each part over the reals, giving \((x-3)(x+3)(x^2+4)\).
How do you factorise by grouping?
Pair the terms, take a common factor from each pair, and if both pairs leave the same bracket, factor that bracket out. Then factorise anything remaining.