Polynomial & Rational Inequalities

You’ll be able to

Solve a polynomial inequality using a sign chart built from its zeros
Solve a rational inequality without multiplying through by the denominator
Decide correctly whether each endpoint belongs in the solution set

A product changes sign only at a zero

A polynomial is continuous, so it can only pass from positive to negative by passing through zero. That single fact reduces every polynomial inequality to bookkeeping: find the zeros, which cut the number line into intervals, and determine the sign on each interval. Within an interval the sign cannot change, so one test point settles the whole interval.

Sign chart method

find all zeros → mark them on a number line → test one point per interval → read off the intervals with the sign you want

For rational expressions, mark the zeros of the numerator AND the zeros of the denominator. Both are places the sign can flip.

Multiplicity decides whether the sign flips

Crossing a zero of odd multiplicity flips the sign; crossing a zero of even multiplicity does not, because the squared factor is positive on both sides. So (x − 1)²(x + 3) is negative for x < −3 and positive everywhere else — the sign does not change at x = 1 even though the function touches zero there. Sign charts that ignore multiplicity get these problems wrong in a way that is invisible unless you test.

Worked example

Solve (x − 1)²(x + 3) > 0.

1.Zeros are x = 1 (multiplicity 2) and x = −3 (multiplicity 1). These split the line into x < −3, −3 < x < 1, and x > 1.
2.Test x = −4: (−5)²(−1) = 25(−1) = −25, negative.
3.Test x = 0: (−1)²(3) = 3, positive.
4.Test x = 2: (1)²(5) = 5, positive.
5.The inequality is strict, so exclude both x = −3 and x = 1, where the expression equals 0.

Answer: The solution is −3 < x < 1 or x > 1, which can be written as (−3, 1) ∪ (1, ∞). Note that x = 1 is punched out even though the expression is positive on both sides of it, because at x = 1 the value is exactly 0 and the inequality is strict.

Watch out

Never multiply both sides of a rational inequality by the denominator. Its sign is unknown, and multiplying by a negative reverses the inequality — so the step is invalid on part of the domain and silently produces wrong answers.

Checkpoint

Solve (x − 4)/(x + 1) ≥ 0.

Getting one side to zero first

Inequalities like (x + 2)/(x − 3) < 1 must be rearranged before a sign chart applies, and subtracting is the only safe move: (x + 2)/(x − 3) − 1 < 0. Combining over the common denominator gives [(x + 2) − (x − 3)]/(x − 3) = 5/(x − 3) < 0. Now the sign chart is trivial — 5 is always positive, so the quotient is negative exactly when x − 3 < 0.

Worked example

Solve x³ ≤ 4x.

1.Move everything to one side: x³ − 4x ≤ 0. Do not divide by x — its sign is unknown and dividing loses the zero.
2.Factor: x(x² − 4) = x(x − 2)(x + 2) ≤ 0.
3.Zeros at x = −2, 0, 2 give four intervals.
4.Test x = −3: (−3)(−5)(−1) = −15, negative. Test x = −1: (−1)(−3)(1) = 3, positive.
5.Test x = 1: (1)(−1)(3) = −3, negative. Test x = 3: (3)(1)(5) = 15, positive.
6.The inequality allows equality, so all three zeros are included.

Answer: x ≤ −2 or 0 ≤ x ≤ 2, that is (−∞, −2] ∪ [0, 2]. Dividing the original by x would have destroyed the solution x = 0 and reversed the inequality for negative x — two errors in one step.

Checkpoint

Why must the zeros of the denominator appear on the sign chart for a rational inequality?

Answer the 2 checkpoints as you read.