Solving Polynomial Equations

Once you can factor a polynomial, you can solve equations like $x^3 - 4x^2 - 9x + 36 = 0$. The roots you find are exactly where the graph of the polynomial crosses or touches the $x$-axis. This page shows how to solve polynomial equations of degree up to $4$, what it means when some roots aren’t real, and how to use these equations in real problems.

Key ideas

Roots and x-intercepts

The roots of the equation $P(x) = 0$ are the values of $x$ that make it true. They are the same numbers as the zeros of the function $f(x) = P(x)$, and the real roots are the $x$-intercepts of the graph of $y = P(x)$.

For example, the equation $x^4 - 13x^2 + 36 = 0$ has roots $-3$, $-2$, $2$ and $3$, and the graph of $y = x^4 - 13x^2 + 36$ crosses the $x$-axis at exactly those four points (left graph below).

This is why graphing technology such as Desmos is a good way to check your roots: graph $y = P(x)$ and look at the $x$-intercepts.

Solving by factoring

1. Rearrange so one side is $0$: $P(x) = 0$.
2. Factor $P(x)$ fully, using the factoring strategy (including the factor theorem when needed).
3. Use the zero product property: if a product is $0$, at least one factor is $0$. Set each factor equal to $0$ and solve.
4. If a quadratic factor won’t factor, use the quadratic formula on it.

How many roots?

A polynomial equation of degree $n$ has at most $n$ real roots. A cubic has at most $3$ and a quartic at most $4$. It can have fewer, for two reasons:

• Repeated roots. In $(x - 2)^2(x + 1) = 0$, the root $2$ comes from two factors. The graph touches the $x$-axis at $x = 2$ instead of crossing it.
• Non-real roots. A quadratic factor with a negative discriminant gives no real roots. (Its roots are non-real numbers, which you’ll meet in later courses. You don’t need to calculate them here.) These roots don’t appear as $x$-intercepts.

For example, $x^3 + x^2 + 2x - 4 = (x - 1)(x^2 + 2x + 4)$. The discriminant of $x^2 + 2x + 4$ is $2^2 - 4(1)(4) = -12 \lt 0$, so the only real root is $x = 1$. The graph (right below) crosses the $x$-axis just once.

Left: four real roots, four $x$-intercepts. Right: one real root and two non-real roots, so only one $x$-intercept.

Applications

In a word problem, the equation often has more roots than make sense. Always check each root against the situation: lengths must be positive, a cut can’t be bigger than the sheet, and so on. Then answer in context with units.

Worked examples

Example 1: Solving by grouping

Solve $x^3 - 4x^2 - 9x + 36 = 0$.

Solution. Four terms, so try grouping:

$$\begin{aligned} x^2(x - 4) - 9(x - 4) &= 0 \\ (x - 4)(x^2 - 9) &= 0 \\ (x - 4)(x - 3)(x + 3) &= 0 \end{aligned}$$

So $x - 4 = 0$, $x - 3 = 0$ or $x + 3 = 0$. The roots are $x = 4$, $x = 3$ and $x = -3$.

Check $x = -3$: $-27 - 36 + 27 + 36 = 0$. ✓

Example 2: A quadratic-type quartic

Solve $x^4 - 13x^2 + 36 = 0$, and connect the roots to the graph.

Solution. Treat $x^2$ as the unknown:

$$\begin{aligned} (x^2 - 4)(x^2 - 9) &= 0 \\ (x - 2)(x + 2)(x - 3)(x + 3) &= 0 \end{aligned}$$

The roots are $x = \pm 2$ and $x = \pm 3$. These are the four $x$-intercepts of $y = x^4 - 13x^2 + 36$ in the left graph above.

Example 3: Factor theorem, then the quadratic formula

Solve $x^3 - 3x^2 - 2x + 4 = 0$. Give exact roots, and decimals to two decimal places.

Solution. The possible integer roots are $\pm 1, \pm 2, \pm 4$. Try $x = 1$: $1 - 3 - 2 + 4 = 0$. So $x - 1$ is a factor:

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} 1 & 1 & -3 & -2 & 4 \\ & & 1 & -2 & -4 \\ \hline & 1 & -2 & -4 & \boxed{0} \end{array}$$

So $(x - 1)(x^2 - 2x - 4) = 0$. The quadratic doesn’t factor over the integers, so use the quadratic formula:

$$x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-4)}}{2} = \frac{2 \pm \sqrt{20}}{2} = \frac{2 \pm 2\sqrt{5}}{2} = 1 \pm \sqrt{5}$$

The roots are $x = 1$, $x = 1 + \sqrt{5} \approx 3.24$ and $x = 1 - \sqrt{5} \approx -1.24$.

Example 4: Volume of a box

A box with no lid is made from a $30$ cm by $20$ cm sheet of cardboard by cutting a square of side $x$ cm from each corner and folding up the sides. For what values of $x$ is the volume $1000$ cm³?

Solution. After cutting, the base is $(30 - 2x)$ by $(20 - 2x)$ and the height is $x$:

$$V(x) = x(30 - 2x)(20 - 2x)$$

The cut must be positive and less than half of $20$ cm, so $0 \lt x \lt 10$.

Set $V(x) = 1000$ and expand:

$$\begin{aligned} x(30 - 2x)(20 - 2x) &= 1000 \\ 4x^3 - 100x^2 + 600x &= 1000 \\ x^3 - 25x^2 + 150x - 250 &= 0 && \text{divide by 4} \end{aligned}$$

Test factors of $250$. Try $x = 5$: $125 - 625 + 750 - 250 = 0$. ✓ Divide by $x - 5$:

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} 5 & 1 & -25 & 150 & -250 \\ & & 5 & -100 & 250 \\ \hline & 1 & -20 & 50 & \boxed{0} \end{array}$$

Now solve $x^2 - 20x + 50 = 0$:

$$x = \frac{20 \pm \sqrt{400 - 200}}{2} = \frac{20 \pm \sqrt{200}}{2} = 10 \pm 5\sqrt{2}$$

So the roots are $5$, $10 + 5\sqrt{2} \approx 17.07$ and $10 - 5\sqrt{2} \approx 2.93$. The root $17.07$ is outside $0 \lt x \lt 10$ (you can’t cut $17$ cm squares from a $20$ cm side), so reject it.

The volume is $1000$ cm³ when the squares have side $5$ cm or about $2.93$ cm.

Check $x = 5$: the box is $20$ cm by $10$ cm by $5$ cm, and $20 \times 10 \times 5 = 1000$. ✓

Common mistakes

Dividing both sides by x. In $3x^3 = 12x$, dividing by $x$ gives $x^2 = 4$ and loses the root $x = 0$. Instead, move everything to one side and factor: $3x(x - 2)(x + 2) = 0$.

Not setting the equation equal to zero first. The zero product property only works for a product equal to $0$. $(x - 1)(x + 2) = 4$ does not mean $x - 1 = 4$ or $x + 2 = 4$.

Missing negative roots. $x^2 - 9 = 0$ gives $x = 3$ and $x = -3$. Factoring as $(x - 3)(x + 3)$ makes both roots visible.

Expecting every root to be an x-intercept. $x^4 - 5x^2 - 36 = 0$ has four roots: two real ones ($\pm 3$) and two non-real ones from $x^2 + 4 = 0$. Only the two real roots show up as $x$-intercepts. When a quadratic factor has a negative discriminant, say it has “no real roots”, not “no roots”.

Keeping roots that don’t fit the context. In Example 4, $x \approx 17.07$ solves the equation but not the problem. Always check each root against the restrictions.

Practice

1. (Warm-up) Solve $(x - 2)(x + 5)(2x - 1) = 0$.

Solution

$x - 2 = 0$, $x + 5 = 0$ or $2x - 1 = 0$, so $x = 2$, $x = -5$ or $x = \tfrac{1}{2}$.

2. (Warm-up) Solve $3x^3 = 12x$.

Solution

$$\begin{aligned} 3x^3 - 12x &= 0 \\ 3x(x^2 - 4) &= 0 \\ 3x(x - 2)(x + 2) &= 0 \end{aligned}$$

$x = 0$, $x = 2$ or $x = -2$.

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

Solution

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

$x = -2$, $x = 3$ or $x = -3$.

4. (Core) Solve $x^4 - 5x^2 - 36 = 0$. How many $x$-intercepts does the graph of $y = x^4 - 5x^2 - 36$ have?

Solution

$$(x^2 - 9)(x^2 + 4) = 0 \quad\Rightarrow\quad (x - 3)(x + 3)(x^2 + 4) = 0$$

$x^2 + 4 = 0$ has no real roots (since $x^2 \ge 0$, $x^2 + 4 \ge 4$). So the real roots are $x = 3$ and $x = -3$, and the other two roots are non-real.

The graph has two $x$-intercepts, at $(3, 0)$ and $(-3, 0)$.

5. (Core) Solve $2x^3 + x^2 - 13x + 6 = 0$.

Solution

Try $x = 2$: $16 + 4 - 26 + 6 = 0$, so $x - 2$ is a factor.

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} 2 & 2 & 1 & -13 & 6 \\ & & 4 & 10 & -6 \\ \hline & 2 & 5 & -3 & \boxed{0} \end{array}$$

$2x^2 + 5x - 3 = (2x - 1)(x + 3)$, so $(x - 2)(2x - 1)(x + 3) = 0$.

$x = 2$, $x = \tfrac{1}{2}$ or $x = -3$.

6. (Core) Solve $x^3 - 6x^2 + 12x - 8 = 0$. What does the graph of $y = x^3 - 6x^2 + 12x - 8$ look like at its $x$-intercept?

Solution

Try $x = 2$: $8 - 24 + 24 - 8 = 0$. Dividing by $x - 2$ gives $x^2 - 4x + 4 = (x - 2)^2$, so

$$(x - 2)^3 = 0$$

The only root is $x = 2$ (a triple root). The graph has just one $x$-intercept, $(2, 0)$, where it flattens out as it passes through the axis.

7. (Core) A storage crate has a volume of $30$ m³. Its length is $2$ m more than its width, and its height is $1$ m less than its width. Find its dimensions.

Solution

Let the width be $w$ metres. Then

$$\begin{aligned} w(w + 2)(w - 1) &= 30 \\ w^3 + w^2 - 2w - 30 &= 0 \end{aligned}$$

Try $w = 3$: $27 + 9 - 6 - 30 = 0$. ✓ Dividing by $w - 3$ gives $w^2 + 4w + 10$, whose discriminant is $16 - 40 = -24 \lt 0$. So $w = 3$ is the only real root.

The crate is $5$ m long, $3$ m wide and $2$ m high. Check: $5 \times 3 \times 2 = 30$. ✓

8. (Challenge) Find where the graphs of $y = x^3$ and $y = 7x + 6$ intersect.

Solution

Set the expressions equal: $x^3 = 7x + 6$, so $x^3 - 7x - 6 = 0$.

Try $x = -1$: $-1 + 7 - 6 = 0$. Dividing by $x + 1$ gives $x^2 - x - 6 = (x - 3)(x + 2)$, so

$$(x + 1)(x - 3)(x + 2) = 0$$

$x = -1$, $3$ or $-2$. Using $y = x^3$, the points are $(-1, -1)$, $(3, 27)$ and $(-2, -8)$.

Check with the line: $7(3) + 6 = 27$ ✓, $7(-2) + 6 = -8$ ✓, $7(-1) + 6 = -1$ ✓.

9. (Challenge) One root of $x^3 + kx^2 + x + 6 = 0$ is $x = 2$. Find $k$ and the other roots.

Solution

Substitute $x = 2$:

$$8 + 4k + 2 + 6 = 0 \quad\Rightarrow\quad 4k = -16 \quad\Rightarrow\quad k = -4$$

The equation is $x^3 - 4x^2 + x + 6 = 0$. Divide by $x - 2$:

$$\def\arraystretch{1.3} \begin{array}{r|rrrr} 2 & 1 & -4 & 1 & 6 \\ & & 2 & -4 & -6 \\ \hline & 1 & -2 & -3 & \boxed{0} \end{array}$$

$x^2 - 2x - 3 = (x - 3)(x + 1)$, so the other roots are $x = 3$ and $x = -1$.