Factorising a Cubic with the Factor Theorem

Factorising a cubic runs in two stages: confirm or find a linear factor (factor theorem), then divide to leave a quadratic you can factorise the usual way. “Completely” means finish the job, all the way down to linear factors.

(i) Show that $(x - 2)$ is a factor of $f(x) = x^3 - 3x^2 - 4x + 12$. [2] (ii) Hence factorise $f(x)$ completely. [3]

The working

(i) By the factor theorem, $(x - 2)$ is a factor if $f(2) = 0$: $f(2) = 2^3 - 3(2)^2 - 4(2) + 12 = 8 - 12 - 8 + 12 = 0 \quad \text{(M1, A1)}$

Since $f(2) = 0$, $(x - 2)$ is a factor. ✓

(ii) Divide $f(x)$ by $(x - 2)$ (long division or comparing coefficients) to get the quadratic factor: $x^3 - 3x^2 - 4x + 12 = (x - 2)(x^2 - x - 6) \quad \text{(M1)}$

Factorise the quadratic $x^2 - x - 6 = (x - 3)(x + 2)$ (M1): $f(x) = (x - 2)(x - 3)(x + 2) \quad \text{(A1)}$

A quick expansion check confirms $(x-2)(x^2 - x - 6) = x^3 - 3x^2 - 4x + 12$.

Where the marks are won and lost

• The factor theorem uses $f(2) = 0$ for the factor $(x - 2)$, note the sign: the factor $(x - a)$ tests $x = +a$.
• The division must be accurate: the quotient $x^2 - x - 6$ carries the middle terms. A slip here breaks the final factorisation.
• “Completely” means factorise the quadratic too. Leaving $(x - 2)(x^2 - x - 6)$ loses the last accuracy mark.

Common mistakes

• Testing $f(-2)$ for the factor $(x - 2)$.
• Errors in the long division giving a wrong quadratic.
• Stopping at the quadratic instead of splitting it into $(x-3)(x+2)$.

Full method: Factor Theorem notes. See also Factorising & Solving Cubics. Topic home: Factors of Polynomials pillar.