Polynomial Inequalities

An equation like $x^3 - 9x = 0$ asks where a polynomial equals zero. An inequality like $x^3 - 9x \ge 0$ asks where it is positive or zero, and the answer is usually whole intervals of numbers, not just a few values. Inequalities answer questions such as “for which production levels is the profit positive?”, and the method here works again for rational inequalities later.

Key ideas

Equation or inequality?

• The solution to an equation such as $(x + 2)(x - 1)(x - 3) = 0$ is a list of numbers: $x = -2$, $1$ or $3$.
• The solution to an inequality such as $(x + 2)(x - 1)(x - 3) \ge 0$ is a set of intervals: $-2 \le x \le 1$ or $x \ge 3$.

The roots of the equation are still important: they are the boundary points where the polynomial can change sign. To show that a value is a solution, substitute it. For example, $x = 4$ gives $(6)(3)(1) = 18 \ge 0$, so $x = 4$ is a solution. $x = 2$ gives $(4)(1)(-1) = -4$, so it isn’t.

On this page, solutions are written as inequalities joined by “or”, like $-2 \le x \le 1$ or $x \ge 3$.

Linear inequalities

Solve a linear inequality like a linear equation, with one extra rule: when you multiply or divide both sides by a negative number, reverse the inequality sign. For example, $-2x \lt 6$ becomes $x \gt -3$.

Reading the answer from a graph

For $f(x) \gt 0$, find where the graph of $y = f(x)$ is above the $x$-axis. For $f(x) \lt 0$, find where it’s below. For $\ge$ or $\le$, include the $x$-intercepts as well.

$f(x) = (x + 2)(x - 1)(x - 3)$ is positive between $-2$ and $1$, and to the right of $3$.

Solving algebraically: intervals and test points

1. Move everything to one side so the other side is $0$.
2. Factor and find the roots. They split the number line into intervals.
3. Pick a test point in each interval and find the sign of the polynomial there. The sign can’t change inside an interval, because the polynomial can only change sign at a root.
4. Choose the intervals with the sign you want. Include the roots if the inequality is $\le$ or $\ge$, and leave them out for $\lt$ or $\gt$.

A sign chart (a table of the signs of each factor) is a tidy way to do step 3. The product is positive when there’s an even number of negative factors, and negative when there’s an odd number.

A squared factor like $(x - 2)^2$ is never negative, so the sign doesn’t change at a root that comes from a squared factor.

Showing the solution on a number line

Shade the intervals that are solutions. Use a closed dot for an endpoint that’s included ($\le$, $\ge$) and an open dot for one that’s left out ($\lt$, $\gt$). An arrow shows that an interval goes on forever.

Top: the solution to Example 2. Bottom: the solution to Example 3.

Worked examples

Example 1: A linear inequality

Solve $5 - 2x \gt 11$ and describe the solution on a number line.

Solution.

$$\begin{aligned} 5 - 2x &\gt 11 \\ -2x &\gt 6 \\ x &\lt -3 && \text{divide by } -2 \text{, reverse the sign} \end{aligned}$$

On a number line: an open dot at $-3$ and shading to the left, with an arrow.

Check with $x = -4$: $5 + 8 = 13 \gt 11$. ✓ And $x = 0$ gives $5 \gt 11$, which is false, as expected.

Example 2: Reading from a graph

Use the graph of $f(x) = (x + 2)(x - 1)(x - 3)$ in Key ideas to solve $f(x) \ge 0$.

Solution. The graph is on or above the $x$-axis from $x = -2$ to $x = 1$, and from $x = 3$ onwards. The inequality is $\ge$, so the $x$-intercepts are included:

$$-2 \le x \le 1 \quad \text{or} \quad x \ge 3$$

This is the top number line in the figure above. Check $x = 0$: $f(0) = (2)(-1)(-3) = 6 \ge 0$. ✓

Example 3: Intervals and test points

Solve $x^4 - 10x^2 + 9 \lt 0$.

Solution. Factor (it’s a quadratic-type quartic):

$$x^4 - 10x^2 + 9 = (x^2 - 1)(x^2 - 9) = (x + 3)(x + 1)(x - 1)(x - 3)$$

The roots $-3$, $-1$, $1$ and $3$ split the number line into five intervals. Test one point in each:

Interval	Test point	$x + 3$	$x + 1$	$x - 1$	$x - 3$	Product
$x \lt -3$	$-4$	$-$	$-$	$-$	$-$	$+$
$-3 \lt x \lt -1$	$-2$	$+$	$-$	$-$	$-$	$-$
$-1 \lt x \lt 1$	$0$	$+$	$+$	$-$	$-$	$+$
$1 \lt x \lt 3$	$2$	$+$	$+$	$+$	$-$	$-$
$x \gt 3$	$4$	$+$	$+$	$+$	$+$	$+$

We want the product to be negative, and the inequality is strict, so the roots are not included:

$$-3 \lt x \lt -1 \quad \text{or} \quad 1 \lt x \lt 3$$

Check with the original polynomial: at $x = 2$, $16 - 40 + 9 = -15 \lt 0$. ✓ At $x = 0$, the value is $9$, which is not less than $0$, as expected.

Example 4: Rearranging first

Solve $x^3 + 2x^2 \ge 9x + 18$.

Solution. Move everything to one side:

$$x^3 + 2x^2 - 9x - 18 \ge 0$$

Factor by grouping:

$$x^2(x + 2) - 9(x + 2) = (x + 2)(x^2 - 9) = (x + 2)(x - 3)(x + 3)$$

The roots are $-3$, $-2$ and $3$. Let $f(x) = x^3 + 2x^2 - 9x - 18$ and test a point in each interval:

Interval	Test point	Value of $f$	Sign
$x \lt -3$	$-4$	$-64 + 32 + 36 - 18 = -14$	$-$
$-3 \lt x \lt -2$	$-2.5$	$1.375$	$+$
$-2 \lt x \lt 3$	$0$	$-18$	$-$
$x \gt 3$	$4$	$64 + 32 - 36 - 18 = 42$	$+$

We want $f(x) \ge 0$, so take the positive intervals and include the roots:

$$-3 \le x \le -2 \quad \text{or} \quad x \ge 3$$

(For the test point $-2.5$, it’s easier to use the factored form: $(-0.5)(-5.5)(0.5) = 1.375$.)

Common mistakes

Forgetting to reverse the sign. When you divide by a negative, the inequality flips: $-2x \gt 6$ gives $x \lt -3$, not $x \gt -3$.

Not moving everything to one side. For $x^3 + 2x^2 \ge 9x + 18$, you can’t factor each side separately and compare. Get $0$ on one side first.

Dividing by a variable. Dividing $x^3 \ge 4x^2$ by $x^2$ gives $x \ge 4$ and loses the solution $x = 0$ (and you can’t divide by $x$ if it might be negative or zero). Instead, write $x^2(x - 4) \ge 0$ and use a sign chart: the solution is $x = 0$ or $x \ge 4$.

Assuming the signs always alternate. Signs switch at a root only if its factor appears an odd number of times. At a root from a squared factor like $(x - 2)^2$, the sign stays the same. Test every interval rather than guessing.

Mixing up open and closed endpoints. For $\lt$ and $\gt$, the roots are not solutions, so use open dots. For $\le$ and $\ge$ they are, so use closed dots.

Practice

1. (Warm-up) Show whether each value is a solution of $x^3 - 5x \gt 0$: (a) $x = 2$ (b) $x = 3$.

Solution

(a) $2^3 - 5(2) = 8 - 10 = -2$, which is not greater than $0$. So $x = 2$ is not a solution.

(b) $3^3 - 5(3) = 27 - 15 = 12 \gt 0$. So $x = 3$ is a solution.

2. (Warm-up) Solve $-3x + 4 \le 10$, and describe the solution on a number line.

Solution

$$-3x \le 6 \quad\Rightarrow\quad x \ge -2$$

(The sign reverses because we divided by $-3$.) On a number line: a closed dot at $-2$ and shading to the right, with an arrow.

3. (Warm-up) Solve $x^2 - x - 12 \le 0$.

Solution

$x^2 - x - 12 = (x - 4)(x + 3)$, with roots $-3$ and $4$. The parabola opens up, so it’s below the $x$-axis between the roots. Including the roots:

$$-3 \le x \le 4$$

4. (Core) Solve $(x - 1)(x + 2)(x - 4) \gt 0$.

Solution

Roots: $-2$, $1$, $4$. Test points:

• $x = -3$: $(-4)(-1)(-7) = -28$, negative.
• $x = 0$: $(-1)(2)(-4) = 8$, positive.
• $x = 2$: $(1)(4)(-2) = -8$, negative.
• $x = 5$: $(4)(7)(1) = 28$, positive.

$$-2 \lt x \lt 1 \quad \text{or} \quad x \gt 4$$

5. (Core) Solve $x^3 - 9x \ge 0$, and show the solution on a number line.

Solution

$x^3 - 9x = x(x - 3)(x + 3)$, with roots $-3$, $0$, $3$. Test points:

• $x = -4$: $-64 + 36 = -28$, negative.
• $x = -1$: $-1 + 9 = 8$, positive.
• $x = 1$: $1 - 9 = -8$, negative.
• $x = 4$: $64 - 36 = 28$, positive.

$$-3 \le x \le 0 \quad \text{or} \quad x \ge 3$$

On a number line: closed dots at $-3$ and $0$ with the segment between them shaded, and a closed dot at $3$ with shading to the right and an arrow.

6. (Core) Solve $x^4 - 13x^2 + 36 \le 0$.

Solution

$x^4 - 13x^2 + 36 = (x^2 - 4)(x^2 - 9) = (x + 3)(x + 2)(x - 2)(x - 3)$, with roots $-3$, $-2$, $2$, $3$.

Test points: $x = -4$ gives $(-1)(-2)(-6)(-7) = 84$ (positive); $x = -2.5$ gives $(0.5)(-0.5)(-4.5)(-5.5) = -6.1875$ (negative); $x = 0$ gives $36$ (positive); $x = 2.5$ gives $(5.5)(4.5)(0.5)(-0.5) = -6.1875$ (negative); $x = 4$ gives $84$ (positive).

$$-3 \le x \le -2 \quad \text{or} \quad 2 \le x \le 3$$

7. (Core) A small company’s weekly profit, in thousands of dollars, is modelled by $P(x) = -x^3 + 12x^2 - 20x$, where $x$ is the number of items made, in hundreds, and $0 \le x \le 12$. For what production levels is the profit positive?

Solution

Factor:

$$P(x) = -x(x^2 - 12x + 20) = -x(x - 2)(x - 10)$$

The roots in $0 \le x \le 12$ are $0$, $2$ and $10$. Test points:

• $x = 1$: $-1 + 12 - 20 = -9$, negative.
• $x = 5$: $-125 + 300 - 100 = 75$, positive.
• $x = 11$: $-1331 + 1452 - 220 = -99$, negative.

So $P(x) \gt 0$ for $2 \lt x \lt 10$. The company makes a profit when it produces more than $200$ and fewer than $1000$ items per week.

8. (Challenge) Solve $x^4 \le 2x^3 + 3x^2$.

Solution

$$x^4 - 2x^3 - 3x^2 \le 0 \quad\Rightarrow\quad x^2(x^2 - 2x - 3) \le 0 \quad\Rightarrow\quad x^2(x - 3)(x + 1) \le 0$$

The roots are $-1$, $0$ and $3$. Since $x^2 \ge 0$, the sign doesn’t change at $0$. Test points:

• $x = -2$: $4(-5)(-1) = 20$, positive.
• $x = -0.5$: $0.25(-3.5)(0.5) = -0.4375$, negative.
• $x = 1$: $1(-2)(2) = -4$, negative.
• $x = 4$: $16(1)(5) = 80$, positive.

The polynomial is negative on $-1 \lt x \lt 0$ and $0 \lt x \lt 3$, and zero at $-1$, $0$ and $3$. Putting it together:

$$-1 \le x \le 3$$

9. (Challenge) Solve $x^3 + 3x^2 - 4 \ge 0$.

Solution

Try $x = 1$: $1 + 3 - 4 = 0$, so $x - 1$ is a factor. Dividing gives $x^2 + 4x + 4 = (x + 2)^2$, so

$$x^3 + 3x^2 - 4 = (x - 1)(x + 2)^2$$

$(x + 2)^2 \ge 0$ always, so the sign of the product matches the sign of $x - 1$, except at $x = -2$, where the product is $0$.

• For $x \gt 1$: positive.
• For $x \lt 1$ (other than $-2$): negative.

Including the roots (since the inequality is $\ge$):

$$x = -2 \quad \text{or} \quad x \ge 1$$

Check $x = -3$: $-27 + 27 - 4 = -4 \lt 0$, so it’s correctly left out.