Math Core

Quadratic equations have the quadratic formula. Cubics and quartics have formulas too, but they're so long that nobody uses them by hand. Instead, you solve higher-degree equations the same way you first solved quadratics: get $0$ on one side, factor, and use the zero product property. This lesson collects the factoring tools that make that possible.

The plan

Solving by factoring

Rewrite the equation as $P(x) = 0$ .
Factor $P(x)$ as completely as you can.
Set each factor equal to $0$ and solve.

This works because of the zero product property: if a product is $0$ , at least one factor is $0$ .

Always start by pulling out a greatest common factor. In $x^3 - 9x = 0$ , the common factor $x$ gives $x(x^2 - 9) = x(x - 3)(x + 3) = 0$ , so $x = 0, 3, -3$ .

Common mistake

Never divide both sides by a variable to "simplify." Dividing $x^3 = 9x$ by $x$ leaves $x^2 = 9$ and silently throws away the solution $x = 0$ . Move everything to one side and factor out the $x$ instead.

Factoring by grouping

A four-term polynomial can often be factored by splitting it into two pairs, factoring each pair, and pulling out the common binomial.

Worked example: Grouping

Solve $x^3 - 2x^2 - 9x + 18 = 0$ .

Group the first two terms and the last two:

\begin{aligned} x^2(x - 2) - 9(x - 2) &= 0 \\ (x - 2)(x^2 - 9) &= 0 \\ (x - 2)(x - 3)(x + 3) &= 0. \end{aligned}

The solutions are $x = 2$ , $x = 3$ and $x = -3$ .

Grouping works when the two pairs share a binomial factor. Watch the sign when the second pair starts with a minus: $-9x + 18 = -9(x - 2)$ .

Sums and differences of cubes

You know $a^2 - b^2 = (a - b)(a + b)$ . Cubes have their own patterns.

Sum and difference of cubes

\begin{aligned} a^3 + b^3 &= (a + b)(a^2 - ab + b^2) \\ a^3 - b^3 &= (a - b)(a^2 + ab + b^2) \end{aligned}

You can verify either one by multiplying out the right side: the middle terms cancel. A memory aid for the signs is SOAP: the binomial has the Same sign as the original, the next sign is the Opposite, and the last is Always Positive.

The quadratic factor $a^2 \pm ab + b^2$ never factors over the real numbers, so it produces a pair of complex solutions.

Worked example: Difference of cubes

Solve $x^3 = 8$ over the complex numbers.

Rewrite as $x^3 - 8 = 0$ and use $a = x$ , $b = 2$ :

(x - 2)(x^2 + 2x + 4) = 0.

From $x - 2 = 0$ you get $x = 2$ . For $x^2 + 2x + 4 = 0$ , use the quadratic formula:

x = \frac{-2 \pm \sqrt{4 - 16}}{2} = \frac{-2 \pm \sqrt{-12}}{2} = \frac{-2 \pm 2i\sqrt{3}}{2} = -1 \pm i\sqrt{3}.

The equation has one real solution, $2$ , and two complex solutions, $-1 + i\sqrt{3}$ and $-1 - i\sqrt{3}$ .

Equations in quadratic form

A quartic like $x^4 - 5x^2 - 36$ only has even powers. If you let $u = x^2$ , it becomes the quadratic $u^2 - 5u - 36$ . Any equation of the form $a(\text{something})^2 + b(\text{something}) + c = 0$ is in quadratic form.

Worked example: A quartic in quadratic form

Solve $x^4 - 5x^2 - 36 = 0$ .

Let $u = x^2$ : $u^2 - 5u - 36 = 0$ , so $(u - 9)(u + 4) = 0$ . Substitute back:

(x^2 - 9)(x^2 + 4) = 0.

From $x^2 = 9$ : $x = \pm 3$ . From $x^2 = -4$ : $x = \pm 2i$ .

The real solutions are $3$ and $-3$ ; all four complex solutions are $3, -3, 2i, -2i$ .

Common mistake

After solving for $u$ , don't stop. The question asks for $x$ . Each value of $u$ gives two values of $x$ (possibly complex).

The rational root theorem

When nothing obvious factors, you need a starting point: one zero. The rational root theorem narrows the search to a short list.

Rational root theorem

If a polynomial with integer coefficients has a rational zero $\dfrac{p}{q}$ in lowest terms, then $p$ divides the constant term and $q$ divides the leading coefficient.

So the candidates are $\pm \dfrac{\text{factor of the constant term}}{\text{factor of the leading coefficient}}$ . Test candidates with the remainder theorem (or synthetic division) until one gives $0$ , divide it out, and solve what's left.

Worked example: Finding a first root

Solve $2x^3 - x^2 - 13x - 6 = 0$ .

The constant term $-6$ has factors $1, 2, 3, 6$ ; the leading coefficient $2$ has factors $1, 2$ . The candidates are

\pm 1, \ \pm 2, \ \pm 3, \ \pm 6, \ \pm \tfrac{1}{2}, \ \pm \tfrac{3}{2}.

Test a few with $P(x) = 2x^3 - x^2 - 13x - 6$ : $P(1) = -18$ and $P(-1) = 4$ , so neither is a root. $P(-2) = -16 - 4 + 26 - 6 = 0$ , so $-2$ is a root. Divide by $x + 2$ :

\begin{array}{r|rrrr} -2 & 2 & -1 & -13 & -6 \\ & & -4 & 10 & 6 \\ \hline & 2 & -5 & -3 & 0 \end{array}

Factor the quotient: $2x^2 - 5x - 3 = (2x + 1)(x - 3)$ . So

2x^3 - x^2 - 13x - 6 = (x + 2)(2x + 1)(x - 3) = 0,

and the solutions are $x = -2$ , $x = -\tfrac{1}{2}$ and $x = 3$ .

Tip

A sign change between two values, like $P(1) = -18$ and $P(-1) = 4$ above, tells you a real zero lies between them. That can point you to which candidates are worth testing first. And once the quotient is a quadratic, you can always finish with the quadratic formula even if it doesn't factor.

Practice

Practice 1

Solve $x^3 - 9x = 0$ .

Separate answers with commas, e.g. 2, -5

Practice 2

Solve $x^3 + 4x^2 - x - 4 = 0$ .

Separate answers with commas, e.g. 2, -5

Practice 3

Solve $x^4 - 13x^2 + 36 = 0$ .

Separate answers with commas, e.g. 2, -5

Practice 4

Find all complex solutions of $x^3 + 27 = 0$ .

Practice 5

Which number is not a possible rational root of $3x^3 + 2x^2 - 7x + 2 = 0$ , according to the rational root theorem?

Practice 6

Solve $2x^3 + 3x^2 - 8x + 3 = 0$ .

Separate answers with commas, e.g. 2, -5

Practice 7

Find all real solutions of $x^3 - 3x^2 - 2x + 6 = 0$ .

Separate answers with commas, e.g. 2, -5

Practice 8

A shipping box has dimensions $x$ , $x + 1$ and $x + 2$ inches, and its volume is $210$ cubic inches. Find $x$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.