Solve $(x - 1)(x + 4) > 0$.

$x 1$, that is $(-\infty, -4) \cup (1, \infty)$.

Solve $x^2 - 5x + 6 \le 0$.

$2 \le x \le 3$, that is $[2, 3]$.

Solve $x^3 - 4x < 0$.

$x < -2$ or $0 < x < 2$, that is $(-\infty, -2) \cup (0, 2)$.

Solve $(x + 1)^2(x - 3) \ge 0$.

$x \ge 3$ or $x = -1$, that is $\{-1\} \cup [3, \infty)$.

Solve $x^3 + 2x^2 \le 5x + 6$.

$x \le -3$ or $-1 \le x \le 2$, that is $(-\infty, -3] \cup [-1, 2]$.

Grade 12 · MHF4U · Advanced Functions

MHF4U: Polynomial Inequalities

A polynomial inequality asks where the graph is above or below the $x$ -axis. The zeros split the number line into intervals, the sign is constant on each, and you only need one test point per interval to find it.

How It Works

Get zero on one side first, so the question is $P(x) > 0$ , $P(x) \ge 0$ , $P(x) < 0$ or $P(x) \le 0$ . Moving terms across is fine; multiplying or dividing both sides by a negative flips the inequality, so avoid doing that with an unknown expression.

Factor $P(x)$ fully and mark its zeros on a number line. Between consecutive zeros a polynomial cannot change sign, since it would have to cross the axis to do so. So pick one test value in each interval, substitute it (into the factored form, it is quicker) and record $+$ or $-$ .

A sketch does the same job: once you know the end behaviour and the zeros, read off where the graph is above the axis. Either way, the answer is the union of the intervals with the sign the question asks for.

Endpoints: a strict inequality ( $<$ or $>$ ) excludes the zeros, so use round brackets; $\le$ or $\ge$ includes them, so use square brackets. Write $x < -2$ or $0 < x < 2$ as $(-\infty, -2) \cup (0, 2)$ .

The double-root trap. A factor with an even power, like $(x + 1)^2$ , is never negative, so the sign does not change at its zero. For $(x + 1)^2(x - 3) \ge 0$ the zero at $-1$ is a single point in the solution, not an endpoint of an interval, and for $> 0$ it is excluded altogether. Test every interval; do not just alternate signs.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Solve $(x - 1)(x + 4) > 0$ .
Show solution
1. The zeros are $x = -4$ and $x = 1$ , which split the line into three intervals.
2. Test $x = -5$ : $(-6)(-1) = 6 > 0$ . Test $x = 0$ : $(-1)(4) = -4 < 0$ . Test $x = 2$ : $(1)(6) = 6 > 0$ .
3. Strict inequality, so the zeros are excluded.
Answer: $x < -4$ or $x > 1$ , that is $(-\infty, -4) \cup (1, \infty)$ .
Easy
Solve $x^2 - 5x + 6 \le 0$ .
Show solution
1. Factor: $(x - 2)(x - 3) \le 0$ . Zeros at $2$ and $3$ .
2. Test $x = 1$ : $(-1)(-2) = 2$ , positive. Test $x = 2.5$ : $(0.5)(-0.5) = -0.25$ , negative. Test $x = 4$ : $(2)(1) = 2$ , positive.
3. The inequality allows equality, so both zeros are included.
Answer: $2 \le x \le 3$ , that is $[2, 3]$ .
Medium
Solve $x^3 - 4x < 0$ .
Show solution
1. Factor: $x(x^2 - 4) = x(x - 2)(x + 2) < 0$ . Zeros at $-2$ , $0$ and $2$ .
2. Test $x = -3$ : $(-3)(-5)(-1) = -15$ , negative. Test $x = -1$ : $(-1)(-3)(1) = 3$ , positive. Test $x = 1$ : $(1)(-1)(3) = -3$ , negative. Test $x = 3$ : $(3)(1)(5) = 15$ , positive.
3. Negative on the first and third intervals, zeros excluded.
Answer: $x < -2$ or $0 < x < 2$ , that is $(-\infty, -2) \cup (0, 2)$ .
Medium
Solve $(x + 1)^2(x - 3) \ge 0$ .
Show solution
1. Zeros at $x = -1$ (double) and $x = 3$ .
2. Test $x = -2$ : $(1)(-5) = -5$ , negative. Test $x = 0$ : $(1)(-3) = -3$ , negative. Test $x = 4$ : $(25)(1) = 25$ , positive. The sign did not change at $-1$ because $(x + 1)^2$ is never negative.
3. $\ge$ includes the zeros: $x = 3$ starts the positive interval, and $x = -1$ makes the expression $0$ , so it is a solution on its own.
Answer: $x \ge 3$ or $x = -1$ , that is $\{-1\} \cup [3, \infty)$ .
Test level
Solve $x^3 + 2x^2 \le 5x + 6$ .
Show solution
1. Zero on one side: $x^3 + 2x^2 - 5x - 6 \le 0$ .
2. Factor theorem: $P(-1) = -1 + 2 + 5 - 6 = 0$ , so $x + 1$ is a factor. Dividing gives $x^2 + x - 6 = (x + 3)(x - 2)$ , so $P(x) = (x + 1)(x + 3)(x - 2)$ .
3. Zeros at $-3$ , $-1$ , $2$ . Test $x = -4$ : $(-3)(-1)(-6) = -18$ , negative. Test $x = -2$ : $(-1)(1)(-4) = 4$ , positive. Test $x = 0$ : $(1)(3)(-2) = -6$ , negative. Test $x = 3$ : $(4)(6)(1) = 24$ , positive.
4. $\le 0$ : the negative intervals, zeros included.
Answer: $x \le -3$ or $-1 \le x \le 2$ , that is $(-\infty, -3] \cup [-1, 2]$ .

Learn It on Martian Lab

Martian Lab teaches this in the Polynomial Inequalities lesson of its MHF4U Polynomial Equations unit. Watch it, then take the Polynomial Equations practice tests to see if it stuck.

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Free Polynomial Equations practice test, with solutions

More Polynomial Equations topics: Long and Synthetic Division · Remainder and Factor Theorem · Solving Polynomial Equations · Sum and Difference of Cubes

More MHF4U topics: Vertical and Horizontal Asymptotes · Proving Trig Identities · Solving Trig Equations · Laws of Logarithms · Solving Logarithmic Equations · End Behaviour and Degree · Finding the Equation of a Polynomial · Even and Odd Functions · Holes in Rational Functions · Graphing Rational Functions · Rational Equations and Inequalities · Radian Measure · Unit Circle and Special Angles · Compound Angle Formulas · Double Angle Formulas · Graphing Sine and Cosine with Transformations · Graphs of Tan, Sec, Csc and Cot · Sinusoidal Word Problems · Solving Exponential Equations · Graphing Logarithmic Functions · Exponential and Log Word Problems · Sums, Products and Quotients of Functions · Composition of Functions · Average vs Instantaneous Rate of Change

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