Polynomial Long Division

Once you know one linear factor of a cubic, long division gives the remaining quadratic, which you then factorise. Here the factor theorem confirms the factor, and division finds the rest.

Divide $2x^3 + 3x^2 - 11x - 6$ by $(x - 2)$, and hence factorise $2x^3 + 3x^2 - 11x - 6$ completely. [5]

The working

Step 1, confirm $(x-2)$ is a factor with the factor theorem, $f(2) = 0$: $f(2) = 2(8) + 3(4) - 11(2) - 6 = 16 + 12 - 22 - 6 = 0 \ \checkmark \quad \text{(B1)}$

Step 2, long division of $2x^3 + 3x^2 - 11x - 6$ by $(x - 2)$:

• $2x^3 \div x = 2x^2$; then $2x^2(x-2) = 2x^3 - 4x^2$. Subtract: $\ 7x^2 - 11x - 6$.
• $7x^2 \div x = 7x$; then $7x(x-2) = 7x^2 - 14x$. Subtract: $\ 3x - 6$.
• $3x \div x = 3$; then $3(x-2) = 3x - 6$. Subtract: $\ 0$. (M1, A1)

The quotient is $2x^2 + 7x + 3$, remainder $0$.

Step 3, factorise the quadratic quotient: $2x^2 + 7x + 3 = (2x + 1)(x + 3) \quad \text{(M1)}$

Step 4, write the full factorisation: $2x^3 + 3x^2 - 11x - 6 = (x - 2)(2x + 1)(x + 3) \quad \text{(A1)}$

Where the marks are won and lost

• Keep terms aligned by degree during division; a missing term should be held as $0x$ to avoid slips.
• Each subtraction changes signs: subtracting $2x^3 - 4x^2$ from $2x^3 + 3x^2$ gives $+7x^2$, not $-x^2$.
• Finish the job: the question says completely, so factorise the quotient too.

Common mistakes

• Sign errors in the subtraction steps (the most frequent division slip).
• Stopping at the quotient $2x^2 + 7x + 3$ without factorising it.
• Mis-factorising the quadratic: check $(2x+1)(x+3) = 2x^2 + 7x + 3$ by expanding.

Topic home: Factors of Polynomials pillar. More: Worked examples.