Solving a Cubic Equation

Solving a cubic means finding its roots, and the reliable route is: hunt one integer root (factor theorem), divide to get a quadratic, then solve the quadratic. The roots of the cubic are the values that make each factor zero.

Solve the equation $x^3 + 2x^2 - 5x - 6 = 0$. [6]

The working

Step 1, find one root. Test small factors of the constant $-6$. Try $x = -1$: $(-1)^3 + 2(-1)^2 - 5(-1) - 6 = -1 + 2 + 5 - 6 = 0 \quad \text{(M1, A1)}$

So $x = -1$ is a root and $(x + 1)$ is a factor.

Step 2, divide $x^3 + 2x^2 - 5x - 6$ by $(x + 1)$ to get the quadratic factor: $x^3 + 2x^2 - 5x - 6 = (x + 1)(x^2 + x - 6) \quad \text{(M1)}$

Step 3, factorise the quadratic $x^2 + x - 6 = (x + 3)(x - 2)$ (M1): $(x + 1)(x + 3)(x - 2) = 0$

Step 4, read off the roots from each factor: $x = -1, \quad x = -3, \quad x = 2 \quad \text{(A1)}$

Where the marks are won and lost

• The trial root is the first method mark. Testing divisors of the constant term ($\pm1, \pm2, \pm3, \pm6$) is systematic; guessing blindly wastes time.
• $x = -1$ is a root of the equation and gives the factor $(x + 1)$, mind the sign relationship between root and factor.
• All three roots are required. Factorising correctly but then reading a sign wrong (for example $x = +3$ from $(x + 3)$) drops the final mark.

Common mistakes

• Reading roots with the wrong sign: $(x + 3) = 0$ gives $x = -3$, not $+3$.
• Division errors producing a wrong quadratic.
• Stopping after finding one root instead of solving fully.

Full method: Factorising & Solving Cubics notes. Topic home: Factors of Polynomials pillar.