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Lesson 5 of 5Polynomial Functions
Unit 1 · Lesson 5 of 5

Polynomial Inequalities

Solving polynomial inequalities with a factored sign chart, handling multiplicity correctly, and writing the answer in interval notation.

4 min read10 practice questionsAsk AI about this →

Key things to remember

  1. 1Always rearrange to compare with before factoring. Never divide both sides by an expression containing .
  2. 2The sign of a polynomial can change only at its zeros, so the zeros cut the number line into test intervals.
  3. 3The sign flips across a zero of odd multiplicity and does not flip across a zero of even multiplicity.
  4. 4Strict inequalities (, ) exclude the zeros and use round brackets; non-strict (, ) include them and use square brackets.
  5. 5 and always take round brackets: , never .
  6. 6Separate pieces of a solution are joined with a union symbol, as in .

Solving finds the handful of places where a graph touches the -axis. Solving asks a bigger question: on which stretches of the number line is the graph above the axis? Because a polynomial graph is one smooth unbroken curve, it can only switch from positive to negative by passing through zero. That single observation is the whole method: find the zeros, and the number line splits into intervals on which the sign never changes. Then you only need to test one point per interval.

Read as "the -values where the graph lies below the -axis", and as "on or above". Your answer is therefore never a list of numbers — it is a set of intervals. The very first move, every time, is to get everything onto one side so the other side is exactly . You cannot sign-chart as written; you must rewrite it as first.

Never divide both sides by a variable expression
It is tempting to turn into by dividing by . That is wrong twice over: if is negative the inequality sign must flip, and if is zero you have divided by zero. Move everything to one side and factor instead.

The sign chart method

  1. Move every term to one side so the inequality compares to .
  2. Factor the polynomial completely.
  3. Find the zeros and mark them on a number line — these are the only places the sign can change.
  4. Choose one convenient test value inside each interval and determine the sign of each factor there; multiply the signs.
  5. Select the intervals matching the inequality, then decide whether the endpoints belong: include them for and , exclude them for and .
P(x) = (x + 2)(x − 1)(x − 3)The sign of P can only change where a factor is zero, so the zeros chop the number line into test intervals.−213x(x + 2)−+++(x − 1)−−++(x − 3)−−−+P(x)−+−+test x = −3test x = 0test x = 2test x = 4Count the negative factors in each column: an odd number of − gives −, an even number gives +. P(x) > 0 on (−2, 1) ∪ (3, ∞)
Fig 1A number line split at the zeros, with the sign of each factor and of the product recorded on every interval.

Work through . Factor first: pull out the common , then recognize a difference of squares. The zeros , and cut the number line into four intervals, and a single test point in each settles the sign.

IntervalTest valueSign of xSign of (x − 2)Sign of (x + 2)Product
−−−− (below)
−−++ (above)
+−+− (below)
++++ (above)
Sign chart for x(x − 2)(x + 2). Count the negative factors: an odd count gives a negative product.

The inequality asked for , so keep the intervals whose product is positive. The inequality is strict, so the zeros themselves are excluded and every bracket is round: the solution is .

VideoFactor, mark the zeros on a number line, test one value in each interval and write the answer in interval notation, worked through for quadratic and cubic inequalities.Watch on YouTube

How multiplicity affects the sign

In the chart above the sign alternated neatly, because every zero had multiplicity . That is not guaranteed. A factor raised to an even power, such as , is positive on both sides of its zero, so the product keeps the same sign as you walk past . Only odd multiplicities flip the sign — exactly the same rule that decided whether a graph crosses or bounces. So never assume the signs alternate: count the multiplicities first.

Isolated solutions from even multiplicity
Consider . The expression is negative only on , but it also equals zero at and at . Since the inequality is non-strict, both zeros qualify, and the solution is — an interval plus one lone point. Even multiplicities create these isolated solutions; strict inequalities never do.

Interval notation, strict and non-strict

InequalityInterval notationEndpoint
2 excluded — round bracket
2 included — square bracket
−3 included, 4 excluded
or Two separate pieces joined by a union
Infinity is never reachable, so it always takes a round bracket, even in a non-strict inequality.
Sanity checks before you write the final answer
Pick one number from inside your answer and one from outside, and substitute both into the original inequality. The first should satisfy it and the second should fail. Also check the far ends against the end behaviour you already know: for an odd-degree polynomial with a positive leading coefficient, the far right must always be positive.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1
Solve and give the answer in interval notation.
Example 2
Solve .
Example 3
Solve .
Example 4
Solve .

Where marks get lost

  • Dividing both sides by (or by any factor containing ). You lose solutions and may need to flip the sign — move everything to one side instead.
  • Factoring before rearranging. Writing as achieves nothing; you need first.
  • Including the endpoints on a strict inequality. For the zeros make , which is not greater than , so they are excluded.
  • Assuming the signs alternate across every zero. At a zero of even multiplicity the sign stays the same, and skipping that check flips half your answer.
  • Writing or joining separate intervals with "and". Infinity always gets a round bracket, and disjoint pieces are joined with a union.

Flashcards 7

First step in any polynomial inequality
Why do zeros split the number line?
Sign change rule
Round vs square bracket
+ 3 more

Practice

1
What is the solution to ?
2
Solve .
3
A polynomial has 4 distinct real zeros. How many intervals does its sign chart have?
4
For , at which zero does the sign of NOT change?
5
Write in interval notation.
6
What is the smallest integer in the solution set of ?
7
Why is it invalid to solve by dividing both sides by ?
8
How many separate intervals make up the solution set of ?
9
Order the steps of the sign chart method:
  1. 1Shade the intervals that satisfy the inequality and decide on the endpoints
  2. 2Test one value inside each interval
  3. 3Mark the zeros on a number line
  4. 4Move every term to one side so the other side is 0
  5. 5Factor the polynomial completely
10
Match each inequality to its interval notation.
x > 2
x ≥ 2
x < 2
x ≤ 2

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