Solve cubic and higher polynomial equations by factorising fully and applying the zero-product principle to find all real roots

What this dot point is asking

SEAB wants you to solve polynomial equations, most often cubics, by factorising the polynomial completely and then setting each factor to zero. It builds directly on the factor theorem: find one root, reduce to a quadratic, and finish with familiar quadratic techniques.

The answer

The zero-product principle

The whole method rests on one fact: a product is zero only if one of its factors is zero. So if

then $x = a$, $x = b$ or $x = c$. Solving an equation therefore means writing it as a product equal to zero and reading off the roots.

Step one: find a first root

For a cubic, test small values that are factors of the constant term (divided by factors of the leading coefficient). When $f(a) = 0$, the factor theorem gives the linear factor $(x - a)$.

Step two: reduce to a quadratic

Divide the cubic by the linear factor to leave a quadratic. The original cubic is now (linear factor) times (quadratic).

Step three: solve the quadratic

Factorise the quadratic if it factorises, or use the quadratic formula. Combine its roots with the first root for the complete solution set. If the quadratic has no real roots, the cubic has just one real root.

Choosing values to test

The rational root theorem narrows the search: any rational root is a factor of the constant term divided by a factor of the leading coefficient. For $2x^3 + \dots - 6$, test $\pm 1, \pm 2, \pm 3, \pm 6$ and their halves. Trying the smallest whole numbers first usually finds a root quickly.

Checking your factorisation

After finding the linear and quadratic factors, expand them back mentally or on paper to confirm they reproduce the original cubic. A quick check of the constant term (the product of the constants in each factor) catches most arithmetic slips before they cost marks.

Examples in context

Example 1. Where a curve meets the axis. Solving $f(x) = 0$ finds the $x$-intercepts of $y = f(x)$. A cubic crossing the axis three times corresponds to three distinct real roots, the algebraic and graphical pictures matching.

Example 2. Volume problems. A box of volume $V$ with dimensions expressed in one variable often leads to a cubic in that variable; solving it and discarding negative or unphysical roots gives the realistic dimension, a classic applied use of the method.

Try this

Q1. Solve $(x - 1)(x + 4)(x - 2) = 0$. [1 mark]

• Cue. Zero-product principle: $x = 1, -4, 2$.

Q2. Given that $x = 1$ is a root, solve $x^3 - 3x + 2 = 0$. [3 marks]

• Cue. $(x - 1)(x^2 + x - 2) = (x - 1)(x - 1)(x + 2)$, so $x = 1$ (repeated) or $x = -2$.

Q3. Solve $x^3 - 7x - 6 = 0$. [4 marks]

• Cue. $x = -1$ works; $(x + 1)(x^2 - x - 6) = (x + 1)(x - 3)(x + 2)$, so $x = -1, 3, -2$.