Factorising & Solving Cubic Equations

Every 0606 cubic surrenders to the same four-step routine. It’s among the most predictable multi-mark questions in the syllabus, a drilled student should treat it as banked marks.

The routine

Solve $x^3 - 2x^2 - 5x + 6 = 0$.

Step 1, find one root by trial. Test factors of the constant 6: $p(1) = 1 - 2 - 5 + 6 = 0$ ✓ State the inference: $(x - 1)$ is a factor (factor theorem).

Step 2, extract the quadratic by comparing coefficients. $x^3 - 2x^2 - 5x + 6 = (x - 1)(x^2 + bx - 6)$ Expand mentally to match the $x^2$ term: $b - 1 = -2 \to$ $b = -1$. Verify with the $x$ term: $-b - 6$ should equal $-5$: $-(-1) - 6 = -5$ ✓ (Long division earns the same marks; coefficient comparison is faster and self-checking.)

Step 3, factorise the quadratic. $x^2 - x - 6 = (x - 3)(x + 2)$ Full factorisation: $(x - 1)(x - 3)(x + 2)$

Step 4, solve. $x = 1, 3, -2$, all roots stated, because “solve” means every solution.

Mark anatomy: root found + theorem stated (M, A), quadratic extracted (M, A), full factorisation/solutions (A). The middle-term verification in step 2 is free error-detection, most coefficient slips die there instead of in your final answer.

Variations to expect

• Repeated roots: the quadratic factors as a square, say so; the graph touches the axis there, which matters if a sketch or inequality follows.
• Irrational leftovers: the quadratic may not factorise, finish it with the quadratic formula or discriminant, and if $b^2 - 4ac < 0$, the cubic has exactly one real root: state that.
• $2x^3$ leading coefficient: the quadratic factor starts $2x^2$; trial roots may include fractions like $\pm\frac{1}{2}, \pm\frac{3}{2}$, still drawn from (factors of constant)/(factors of leading coefficient).
• “Hence” follow-ups: the factorised form feeds cubic inequalities and graph sketches, don’t re-derive what part (i) handed you.

Common mistakes

• Trial roots not drawn from the constant’s factors (slow, looks like guessing)
• Quadratic extracted without verifying the middle term
• “Solve” answered with the factorisation only
• The negative root dropped when stating solutions
• $p(a)$ computed with sign slips on odd powers of negatives

On the non-calculator paper the numbers always cooperate, an ugly quadratic factor means a wrong first root. Full topic context: Factors of Polynomials notes.