C. Polynomial and Rational Functions · Polynomial Functions and Equations

Solving Factorable Polynomial Inequalities Algebraically

C4.3 C4.1About 70 minutes

Learning goals

I can solve a factorable polynomial inequality by finding the zeros and testing the sign on each interval.
I can use a sign chart (factor table) or test values, and decide whether boundary points are included.
I can handle negative leading coefficients, repeated factors and inequalities that must be rearranged first.
I can solve inequalities that come from real situations and state the answer within the domain.

Words to know

test value
A number chosen inside an interval to find the sign of a polynomial on the whole interval.

sign chart (factor table)
A table of the sign of each factor on each interval; multiplying the signs gives the sign of the polynomial.

critical values
The zeros of the polynomial: the only places where its sign can change.

Algebra instead of a picture

Last lesson you read inequalities from graphs. Graphs are quick, but they can hide exact values such as 1 − √5. Algebra gives exact answers.

Method for P(x) > 0, ≥ 0, < 0 or ≤ 0:

Rearrange so one side is 0. (Never divide both sides by an expression containing x: its sign is unknown, and it might be 0.)
Factor P(x) fully.
Find the zeros (critical values). They split the number line into intervals.
Find the sign of P(x) on each interval, with a sign chart or one test value per interval.
Choose the intervals with the right sign. Include the zeros for ≤ or ≥; exclude them for < or >.
Write the answer in interval notation (and check it against a graph if you can).

Worked example

Example 1: A cubic with a sign chart

Solve x³ − 2x² − 5x + 6 ≤ 0.

Factor (lesson 11): (x + 2)(x − 1)(x − 3) ≤ 0. Zeros: −2, 1, 3.
Sign chart: each factor is negative to the left of its zero and positive to the right. Multiply the signs in each column (see the chart below): −, +, −, +.
We want ≤ 0: the negative intervals plus the zeros.

(−∞, −2] ∪ [1, 3]

Sign chart for (x + 2)(x − 1)(x − 3). The polynomial is negative on x < −2 and 1 < x < 3.

Worked example

Example 2: A quartic with test values

Solve x⁴ − 5x² + 4 > 0.

Factor: (x² − 1)(x² − 4) = (x + 2)(x + 1)(x − 1)(x − 2). Zeros: −2, −1, 1, 2.
Test one value per interval: P(−3) = 81 − 45 + 4 = 40 > 0; P(−1.5) = 5.0625 − 11.25 + 4 = −2.1875 < 0; P(0) = 4 > 0; P(1.5) = −2.1875 < 0; P(3) = 40 > 0.
We want > 0, so exclude the zeros.

(−∞, −2) ∪ (−1, 1) ∪ (2, ∞)

Solutions of x⁴ − 5x² + 4 > 0 (open dots: rings with a lightly shaded centre, because the zeros −2, −1, 1, 2 are excluded) and of x⁴ − 5x² + 4 ≤ 0 (solid, filled-in dots, because the zeros are included). Together they cover the whole number line.

Worked example

Example 3: A negative leading coefficient

Solve −2x³ + 2x² + 4x ≥ 0.

Factor: −2x(x² − x − 2) = −2x(x − 2)(x + 1). Zeros: −1, 0, 2.
Signs, starting from the right: degree 3 with a = −2 < 0, so P(x) < 0 for x > 2. Every zero has order 1, so the sign switches at each: + on (0, 2), − on (−1, 0), + on (−∞, −1).
Or divide both sides by −2 and reverse the sign: x(x − 2)(x + 1) ≤ 0. Same answer.
Include the zeros (≥).

(−∞, −1] ∪ [0, 2]

y = −2x³ + 2x² + 4x is on or above the x-axis for x ≤ −1 and 0 ≤ x ≤ 2 (shaded), matching Example 3.

Worked example

Example 4: Rearrange first

Solve x³ + 4 > x² + 4x.

Move every term to one side: x³ − x² − 4x + 4 > 0.
Group: x²(x − 1) − 4(x − 1) = (x − 1)(x − 2)(x + 2) > 0. Zeros: −2, 1, 2.
a > 0, degree 3: + for x > 2, − on (1, 2), + on (−2, 1), − for x < −2.

(−2, 1) ∪ (2, ∞)

Worked example

Example 5: A repeated factor

Solve (x − 3)²(x + 1) < 0.

Zeros: 3 (order 2) and −1 (order 1).
(x − 3)² is positive except at x = 3, where it is 0. So the sign of the product is the sign of x + 1, except that the product is 0 at x = 3.
x + 1 < 0 when x < −1. At x = 3 the product is 0, which is not < 0, but 3 is not in x < −1 anyway.

(−∞, −1)

Worked example

Example 6: An application

The roller-coaster track h(x) = 0.005x(x − 20)(x − 40), 0 ≤ x ≤ 20, from lesson 14. For which x is the track at least 15 m high?

h(x) ≥ 15 becomes x³ − 60x² + 800x − 3000 ≥ 0 (multiply by 200 and rearrange).
Factor (lesson 14): (x − 10)(x² − 50x + 300) ≥ 0. Zeros: 10 and 25 ± 5√13 (about 6.97 and 43.03).
Cubic with a > 0: positive between 6.97 and 10 and after 43.03; negative elsewhere.
Keep only the part inside 0 ≤ x ≤ 20.

25 − 5√13 ≤ x ≤ 10, about 6.97 m to 10 m.

Two quick checks

1. The sign pattern must match the end behaviour: the rightmost interval has the sign of the leading coefficient. 2. Signs alternate only across zeros of odd order; across a zero of even order, the sign stays the same.

Just for fun

A sign chart is a weather forecast for a polynomial: positive, negative, positive again, with a 100% chance of zero at the boundaries.

Hands-on activity

Two-method challenge

You need: a partner, paper, and a device with a free graphing tool (such as Desmos) or a graphing calculator.

Inequalities: (a) x³ − 4x² − x + 4 ≥ 0 (b) x⁴ − 10x² + 9 < 0 (c) −x³ + 3x² > 0.

One person solves each inequality algebraically with a sign chart; the other graphs it and reads the answer.
Compare. If you disagree, find the step where the two methods split.
Swap roles and invent a new inequality for each other.

What to look for: (a) (x − 1)(x + 1)(x − 4) ≥ 0: [−1, 1] ∪ [4, ∞). (b) (x² − 1)(x² − 9) < 0: (−3, −1) ∪ (1, 3). (c) −x²(x − 3) > 0: positive when x < 3 except x = 0: (−∞, 0) ∪ (0, 3).

Your turn

Solve (x + 4)(x − 1) > 0.

Solve x(x − 2)(x + 3) ≤ 0.

Solve x³ − 9x > 0.

Solve x³ − 7x + 6 ≥ 0.

Solve (x − 2)²(x + 5) ≥ 0.

Solve −x³ + 4x² < 0.

Solve x⁴ − 13x² + 36 ≤ 0.

More practice: warm-up

What are the critical values of (x − 5)(x + 1) < 0?

For ≤ or ≥, are the zeros included?

What is the sign of (x − 2)(x + 3) at x = 0?

Can (x − 1)² ever be negative?

To find the sign of x(x − 4) on the interval x > 4, which test value could you use?

More practice: on your own

Solve 2x³ + 3x² − 8x + 3 < 0.

Solve x³ + 2x² − 9x − 18 ≥ 0.

Solve x³ ≤ x.

Solve x⁴ + 3x² − 4 > 0.

A student solves x³ > 4x by dividing both sides by x and getting x² > 4, so x < −2 or x > 2. Explain the mistake and give the correct solution.

Challenge

Solve (x² − 4)(x² − 2x − 3) < 0.

Solve x³ − 3x² − 2x + 4 ≥ 0 exactly.

For the box V(x) = x(30 − 2x)(20 − 2x), 0 < x < 10, solve V(x) ≥ 1000 algebraically.

Check your understanding

Quiz

What is the solution of (x − 1)(x + 2)(x − 4) > 0?

What is the solution of x³ − x² − 6x > 0?

Why should you not divide both sides of x³ ≥ 9x by x?

What is the solution of (x + 1)²(x − 2) ≤ 0?

For −x(x − 3)(x + 2), what is the sign on the interval x > 3?

What is the solution of x⁴ − 16 < 0?

For parents and teachers

Big idea (C4.3, C4.1): to solve a polynomial inequality algebraically, get 0 on one side, factor, find the zeros, and check the sign of the polynomial between them (a sign chart or test values). Include the zeros only for ≤ and ≥.

How to help: Ask your teen to show the sign chart and to check that the rightmost sign matches the leading coefficient. Graphing the polynomial in Desmos is a good final check.

Common difficulties:

Dividing both sides by x or by a negative expression.
Not rearranging to 0 on one side first.
Switching signs across a double zero.
Including endpoints for strict inequalities, or using brackets with ∞.