How to Solve Cubic Equations Using the Factor Theorem

To solve a cubic $ax^3 + bx^2 + cx + d = 0$, first find one root by testing likely values (the rational root theorem). If $x = r$ is a root, then $(x - r)$ is a factor. Divide it out to leave a quadratic, then solve that with the quadratic formula.

Step 1: Try likely roots

Any rational root $\frac{p}{q}$ has $p$ dividing the constant $d$ and $q$ dividing the leading coefficient $a$. For $x^3 - 6x^2 + 11x - 6 = 0$, try $\pm 1, \pm 2, \pm 3, \pm 6$.

$f(1) = 1 - 6 + 11 - 6 = 0$, so $x = 1$ is a root.

Step 2: Divide out the factor

Divide by $(x - 1)$ (long or synthetic division):

Step 3: Solve the quadratic

$x^2 - 5x + 6 = (x - 2)(x - 3)$, so the roots are

Worked example 2

$2x^3 - 3x^2 - 3x + 2 = 0$. Testing $x = -1$: $-2 - 3 + 3 + 2 = 0$ ✓. Dividing gives $(x + 1)(2x^2 - 5x + 2) = (x + 1)(2x - 1)(x - 2)$, so $x = -1, \frac{1}{2}, 2$.

When there is no nice root

$x^3 - 2x - 5 = 0$ has no rational root (none of $\pm1, \pm5$ work). Its only real root is irrational, about $x \approx 2.0946$, found numerically. The other two roots are complex.

Common mistakes

• Testing only positive candidates.
• Arithmetic slips in synthetic division.
• Forgetting complex roots when the quadratic's discriminant is negative.

Practice questions

1. Show $x = 2$ is a root of $x^3 - 8$.
2. Solve $x^3 - x = 0$.
3. Solve $x^3 + x^2 + x + 1 = 0$.

Answers: 1) $8 - 8 = 0$ 2) $x = -1, 0, 1$ 3) $x = -1$ and $x = \pm i$

The cubic equation solver does all three steps and graphs the curve. To factor any polynomial use the polynomial factoring calculator, and see how to factor polynomials.

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