Precalculus · A.CED.A.1

Polynomial and rational inequalities with sign charts

You can solve a quadratic inequality by finding its roots and testing the stretches between them. Does the same idea work when the expression is a cubic, a quartic, or a fraction with x in the denominator?

You need

Graphing rational functionsAlgebra 2 Quadratic inequalitiesAlgebra 2

Where you are 7 in Inequalities

Algebra9

Inequalities7

▴ 2 above Writing an inequality for a condition Solving linear inequalities3 Compound inequalities: and, or Absolute value inequalities

Polynomial and rational inequalities with sign charts

When you reach for it

Quadratic inequalities rest on one fact: a quadratic can change sign only where it equals 0. Between two neighboring roots the graph has no way to get from above the axis to below it, so one test number speaks for the whole stretch.

That fact is not special to quadratics. The graph of any polynomial is one unbroken curve, so a polynomial of any degree changes sign only at its real zeros. A rational function, a quotient of two polynomials, has one more way to change sign: its graph can jump across a vertical asymptote, as you saw in graphing rational functions. Those two kinds of points, the zeros and the places the expression is undefined, cut the number line into stretches, and on each stretch the sign stays fixed.

You reach for this move whenever an inequality has a polynomial of degree three or more, or a variable in a denominator: where is (x + 3)(x - 2)^2(x - 6) negative, for which x is \displaystyle \frac{x + 2}{x - 5} at least 1, which parts of a curved track sit above a given height. Day 1 handles polynomials. Day 2 adds denominators, and with them a step that cannot be skipped.

The procedure

Compare with zero. Move every term to one side, so the question is about the sign of a single expression P(x): is it positive, negative, or zero? For a rational inequality, also combine everything into one fraction \displaystyle \frac{N(x)}{D(x)} over a common denominator, as in adding and subtracting rational expressions.

Do not multiply by an expression in x. Multiplying both sides of an inequality by a negative number reverses it. An expression such as x - 5 is negative for some x and positive for others, so multiplying by it is correct on part of the line and wrong on the rest. Subtracting to get zero on one side is always allowed; multiplying by an unknown sign is not. For the same reason, never divide both sides by a factor such as x or x + 4.

Find the critical numbers. Factor completely. The critical numbers are the real zeros of P, or for a fraction the real zeros of N together with the real zeros of D. A factor with no real zeros, such as x^2 + 5, is positive for every x, so it never changes the sign and adds no critical number.

Decide the sign on each stretch. The critical numbers split the line into open intervals. On each one, no factor is zero, so every factor keeps one sign and so does their product or quotient, by the sign rules for products. One test number per interval settles it. A faster route reads the signs from the factors. A factor (x - r)^k changes sign at r when the exponent k is odd and keeps its sign when k is even, because an even power is never negative; the other factors do not change sign near r at all. So: far to the right every factor x - r is positive, the sign there is the sign of the leading coefficients, and moving left the sign flips at each critical number of odd multiplicity and stays put at each one of even multiplicity. A test number is still the safest confirmation.

Decide the endpoints. For < and >, every critical number is left out. For \le and \ge, the zeros of the expression are included, because there it equals 0. A zero of the denominator is never included: the expression has no value there at all.

Read off the answer in interval notation, joining pieces with \cup.

Worked examples

Example 1

Day 1: a squared factor that does not flip the sign

Solve (x + 3)(x - 2)^2(x - 6) \le 0.

Solution

The expression is already factored and compared with 0. Its zeros are -3, 2 and 6. The factor (x - 2)^2 is never negative, and it is zero only at x = 2. The table shows the sign of each factor on each interval, and the product’s sign from the sign rules.

	x + 3	(x - 2)^2	x - 6	all
(-\infty, -3)	-	+	-	+
(-3, 2)	+	+	-	-
(2, 6)	+	+	-	-
(6, \infty)	+	+	+	+

The product is negative on (-3, 2) and on (2, 6). Because the inequality allows 0, the three zeros -3, 2 and 6 are all in. The point x = 2 joins the two negative pieces, so the answer is one interval:

[-3, 6].

Test numbers confirm two cells: at x = 0 the product is 3 \cdot 4 \cdot (-6) = -72, and at x = 4 it is 7 \cdot 4 \cdot (-2) = -56. The sign did not change at 2, which is what an even exponent predicts. On the graph, the curve touches the axis at 2 and turns back down.

Figure 1Example 1. The graph crosses the axis at -3 and 6 but only touches it at 2, so the sign is negative on both sides of 2.

Had the inequality been strict, (x + 3)(x - 2)^2(x - 6) < 0, the zero at 2 would be left out and the answer would split: (-3, 2) \cup (2, 6).

Example 2

Day 1: moving terms and factoring by grouping

Solve x^3 + 4x^2 \ge 9x + 36.

Solution

Subtract 9x + 36 from both sides, which keeps the direction:

x^3 + 4x^2 - 9x - 36 \ge 0.

Group the first two terms and the last two:

\begin{aligned} x^2(x + 4) - 9(x + 4) &= (x + 4)(x^2 - 9) \\ &= (x + 4)(x + 3)(x - 3). \end{aligned}

The critical numbers are -4, -3 and 3. Do not divide both sides by x + 4 to simplify: its sign changes at -4. Test one number in each interval.

interval	test	value	sign
(-\infty, -4)	-5	-16	-
(-4, -3)	-\tfrac{7}{2}	\tfrac{13}{8}	+
(-3, 3)	0	-36	-
(3, \infty)	4	56	+

The expression is positive on (-4, -3) and on (3, \infty), and the zeros are included because the inequality is \ge:

[-4, -3] \cup [3, \infty).

The signs alternate, as they must with three zeros of exponent 1 and a positive leading coefficient. Check x = 3 in the original: 27 + 36 = 63 and 27 + 36 = 63, equal, so it belongs.

Example 3

Day 1: a factor that is always positive

Solve x^5 + 4x^3 \le 5x.

Solution

Subtract 5x to get x^5 + 4x^3 - 5x \le 0. Dividing by x would lose the zero at 0 and ignore that x can be negative, so factor x out instead. What remains is a quadratic in x^2:

\begin{aligned} x^5 + 4x^3 - 5x &= x(x^4 + 4x^2 - 5) \\ &= x(x^2 + 5)(x^2 - 1) \\ &= x(x^2 + 5)(x + 1)(x - 1). \end{aligned}

The factor x^2 + 5 is at least 5 for every x, so it is positive everywhere and contributes no critical number. The critical numbers are -1, 0 and 1, each from a factor with exponent 1.

The leading coefficient is positive, so the sign is + on (1, \infty), and it flips at each of the three zeros: - on (0, 1), + on (-1, 0), - on (-\infty, -1). One test confirms it: at \displaystyle x = \frac{1}{2}, \displaystyle \frac{1}{32} + \frac{1}{2} = \frac{17}{32} is less than \displaystyle \frac{5}{2}, so \displaystyle x = \frac{1}{2} belongs. Including the zeros,

(-\infty, -1] \cup [0, 1].

Example 4

Day 1: rails on a toboggan run

Enkhjin inspects one section of a toboggan run. At a horizontal distance of x meters from the start of the section, the track is h(x) meters above the flat run-out below it, where

h(x) = \frac{1}{25}\left(-x^3 + 16x^2 - 65x + 100\right)

for 0 \le x \le 10. Sleds come into this section already moving fast, which is what carries them over the hump near the 8-meter mark.

Rails are required wherever the track is more than 2 meters above the run-out. Where are they required?

Solution

The condition is h(x) > 2. Multiply both sides by 25, a positive number, so the direction stays:

-x^3 + 16x^2 - 65x + 100 > 50.

Subtract 50, then multiply by -1, which reverses the inequality:

x^3 - 16x^2 + 65x - 50 < 0.

Try x = 1: 1 - 16 + 65 - 50 = 0. By the factor theorem, x - 1 is a factor, and synthetic division leaves x^2 - 15x + 50 = (x - 5)(x - 10). So the question is where

(x - 1)(x - 5)(x - 10) < 0.

With a positive leading coefficient and three simple zeros, the signs from the right are + on (10, \infty), - on (5, 10), + on (1, 5), - on (-\infty, 1). The product is negative for x < 1 and for 5 < x < 10. Only 0 \le x \le 10 is part of the track, so the answer is

[0, 1) \cup (5, 10).

Figure 2Example 4. The track is above the dashed line h = 2 on [0, 1) and on (5, 10).

The ends are decided by the context. The section starts at x = 0, where h(0) = 4, so 0 is in. At 1, 5 and 10 the track is exactly 2 meters high, which is not more than 2. Spot checks: \displaystyle h(3) = \frac{22}{25}, below the line, and \displaystyle h(7) = \frac{86}{25}, above it. Rails run along the first meter and from 5 to 10 meters, 6 meters of horizontal distance in all.

Example 5

Day 2: the step that multiplying skips

Solve \displaystyle \frac{x + 2}{x - 5} \ge 1.

Solution

It is tempting to multiply both sides by x - 5. That gives x + 2 \ge x - 5, or 2 \ge -5, which is true for every x. But x = 0 gives \displaystyle \frac{2}{-5} = -\frac{2}{5}, which is not at least 1. The step went wrong because x - 5 is negative whenever x < 5, and there the inequality should have flipped.

Instead, subtract 1 and write it over the common denominator x - 5:

\begin{aligned} \frac{x + 2}{x - 5} - \frac{x - 5}{x - 5} &= \frac{x + 2 - x + 5}{x - 5} \\ &= \frac{7}{x - 5} \ge 0. \end{aligned}

The numerator 7 is never zero, so the only critical number is 5, where the fraction is undefined. A positive number over x - 5 is positive exactly when x - 5 > 0, and it is never 0. So the answer is

(5, \infty).

Check: x = 6 gives \displaystyle \frac{8}{1} = 8 \ge 1, and x = 5 is out because \displaystyle \frac{7}{0} has no value.

Example 6

Day 2: zeros and breaks together

Solve \displaystyle \frac{x^2 - 9}{x^2 + 2x - 8} \le 0.

Solution

Factor the top and bottom:

\frac{x^2 - 9}{x^2 + 2x - 8} = \frac{(x + 3)(x - 3)}{(x + 4)(x - 2)}.

No factor cancels. The numerator is zero at -3 and 3; the denominator is zero at -4 and 2. The four critical numbers in order are -4, -3, 2, 3, and each comes from a factor with exponent 1. The leading coefficients of top and bottom are both 1, so the sign is + on (3, \infty) and flips at every critical number, giving +, -, +, -, + from left to right. Test numbers in the two short intervals confirm the minus signs:

\begin{aligned} x = -\tfrac{7}{2}&: \quad \frac{\tfrac{13}{4}}{-\tfrac{11}{4}} = -\frac{13}{11}, \\ x = \tfrac{5}{2}&: \quad \frac{-\tfrac{11}{4}}{\tfrac{13}{4}} = -\frac{11}{13}. \end{aligned}

The fraction is negative on (-4, -3) and (2, 3). The inequality allows 0, so the zeros -3 and 3 are in; -4 and 2 make the denominator zero, so they are out:

(-4, -3] \cup (2, 3].

Figure 3Example 6. Zeros of the numerator get filled dots, because the inequality allows 0; zeros of the denominator always get open circles.

Example 7

Day 2: two fractions and a numerator that is never zero

Solve \displaystyle \frac{x}{x - 4} \le \frac{3}{x + 2}.

Solution

Subtract the right side and use the common denominator (x - 4)(x + 2):

\frac{x(x + 2) - 3(x - 4)}{(x - 4)(x + 2)} = \frac{x^2 - x + 12}{(x - 4)(x + 2)} \le 0.

The numerator has discriminant (-1)^2 - 4 \cdot 1 \cdot 12 = -47, so it has no real zeros. Completing the square shows more: \displaystyle x^2 - x + 12 = \left(x - \frac{1}{2}\right)^2 + \frac{47}{4}, which is positive for every x. The fraction is therefore never 0, and its sign is the sign of the denominator (x - 4)(x + 2). The inequality needs that to be negative, which happens strictly between its zeros:

(-2, 4).

Check: at x = 0 the left side is 0 and the right side is \displaystyle \frac{3}{2}, so \displaystyle 0 \le \frac{3}{2} holds. At x = 5, \displaystyle 5 \le \frac{3}{7} fails, as it should.

Boundaries

The method needs the sign to change only at zeros and at breaks. That is true for polynomials, whose graphs are unbroken, and for rational functions, which break only where the denominator is zero. An inequality with a square root or an absolute value needs its own domain or cases first.
Every real zero has to be found. When a polynomial will not factor, the endpoints can only be approximated, and the answer is only as exact as they are.
Multiplying or dividing both sides by an expression in x is not allowed unless its sign is known on the whole set in question, as with (x - 1)^2 > 0 for x \ne 1.
A zero of the denominator is never part of the solution, even under \le or \ge. This includes a factor that cancels: \displaystyle \frac{(x - 3)(x + 2)}{x - 3} equals x + 2 only for x \ne 3.
A context can cut the domain, as the track’s 0 \le x \le 10 did, or allow only whole numbers.

Where this goes next

Sign charts answer every question of the form “where is this expression positive?”. They find the domain of \displaystyle \sqrt{\frac{x - 1}{x + 3}}, which is wherever the fraction under the root is at least 0. They locate where a polynomial’s graph lies above or below a line, which is how polynomial graphs are read. In calculus the same chart, built for a rate of change instead of the function itself, tells where a function rises and where it falls.

Exercises

Day 1: polynomial inequalities

(x - 4)(x + 1)(x + 5) > 0
x^2(x - 5) < 0
(2x - 1)(x + 4)^3 \le 0
(3 - x)(x + 2)(x - 6) > 0
x^3 - 2x^2 - 15x \ge 0
x^3 + 5x^2 < 4x + 20
x^4 - 17x^2 + 16 \le 0
x^4 + x^2 \ge 20
(x - 1)^2(x + 6)^2 \le 0
x^3 - 13x - 12 \le 0 (one zero is a small integer)
Another section of the toboggan run has height \displaystyle h(x) = \frac{1}{50}(x^3 - 15x^2 + 500) meters above the run-out at horizontal distance x meters, for 0 \le x \le 10. Moana finds that h(5) = 5. Where is the track at least 5 meters high?

Day 2: rational inequalities

\displaystyle \frac{x - 3}{x + 4} \le 0
\displaystyle \frac{x + 4}{(x - 1)^2} > 0
\displaystyle \frac{x^2 - 4x}{x^2 - 1} \ge 0
\displaystyle \frac{3x - 2}{x + 4} < 2
\displaystyle \frac{x + 6}{x} \ge x
\displaystyle \frac{4}{x - 3} \le \frac{2}{x + 1}
\displaystyle \frac{x^2 - x - 6}{x - 3} > 0
A season pass for the toboggan run costs 84 dollars, and with the pass each ride costs a further 2 dollars. A single ride without a pass costs 9 dollars. With the pass, the average cost per ride over n rides is \displaystyle \frac{84 + 2n}{n} dollars. For how many rides is that average less than 9 dollars?

Reasoning

Bayani solves \displaystyle \frac{x - 3}{x + 1} < 2 by multiplying both sides by x + 1, and gets x > -5. Find the true solution set, and say which numbers his answer gets wrong and why.
Explain why a polynomial changes sign at a zero of odd multiplicity but not at a zero of even multiplicity.
Write a polynomial inequality whose solution set is [-2, 1] \cup [4, \infty).
Enkhjin says: “In a rational inequality, the zeros of the denominator are never in the answer, and the zeros of the numerator are in the answer exactly when the sign is \le or \ge.” Is the second half always true? Use exercise 18 to decide.

Answers

(-5, -1) \cup (4, \infty). Critical numbers -5, -1, 4; the sign is + right of 4 and flips at each, giving -, +, -, + from the left. Strict, so the ends are out.
(-\infty, 0) \cup (0, 5). The zero 0 has even multiplicity, so the sign is - on both sides of it, and + only for x > 5. At x = 0 the value is 0, not less than 0.
\displaystyle \left[-4, \frac{1}{2}\right]. The zero -4 has odd multiplicity 3, so the sign flips there. The leading term is 2x^4, positive, giving +, -, + from the left.
(-\infty, -2) \cup (3, 6). Expanding the leading terms gives (-x)(x)(x) = -x^3, so the sign is - on (6, \infty), then + on (3, 6), - on (-2, 3), + on (-\infty, -2).
[-3, 0] \cup [5, \infty). x^3 - 2x^2 - 15x = x(x - 5)(x + 3); signs -, +, -, + across -3, 0, 5.
(-\infty, -5) \cup (-2, 2). Grouping, x^3 + 5x^2 - 4x - 20 = x^2(x + 5) - 4(x + 5), which is (x + 5)(x + 2)(x - 2). The signs -, +, -, + across -5, -2, 2.
[-4, -1] \cup [1, 4]. x^4 - 17x^2 + 16 = (x^2 - 1)(x^2 - 16), which is (x + 4)(x + 1)(x - 1)(x - 4). The signs +, -, +, -, + across -4, -1, 1, 4.
(-\infty, -2] \cup [2, \infty). x^4 + x^2 - 20 = (x^2 + 5)(x^2 - 4), and x^2 + 5 > 0, so the condition is x^2 - 4 \ge 0.
x = -6 \quad\text{or}\quad x = 1. The product of two squares is never negative, so it is \le 0 only where it equals 0.
(-\infty, -3] \cup [-1, 4]. x = -1 gives -1 + 13 - 12 = 0, and dividing by x + 1 leaves x^2 - x - 12 = (x - 4)(x + 3). Signs -, +, -, + across -3, -1, 4.
[0, 5]. h(x) \ge 5 means x^3 - 15x^2 + 500 \ge 250, or x^3 - 15x^2 + 250 \ge 0. Since h(5) = 5, x - 5 is a factor: x^3 - 15x^2 + 250 = (x - 5)(x^2 - 10x - 50). The quadratic has zeros \displaystyle 5 \pm 5\sqrt{3}, about -3.66 and 13.66, both outside [0, 10], and it is negative there (it is -50 at x = 0). So the product is \ge 0 exactly when x - 5 \le 0. The track is at least 5 meters high over the first 5 meters.
(-4, 3]. Signs +, -, + across -4 and 3; 3 makes it 0, -4 makes it undefined.
(-4, 1) \cup (1, \infty). (x - 1)^2 > 0 for x \ne 1, so the sign is that of x + 4, positive for x > -4; 1 is out because the fraction is undefined there.
(-\infty, -1) \cup [0, 1) \cup [4, \infty). \displaystyle \frac{x(x - 4)}{(x + 1)(x - 1)} has signs +, -, +, -, + across -1, 0, 1, 4; 0 and 4 are in, \pm 1 out.
(-4, 10). Subtracting 2 gives \displaystyle \frac{3x - 2 - 2x - 8}{x + 4}, which is \displaystyle \frac{x - 10}{x + 4} < 0.
(-\infty, -2] \cup (0, 3]. \displaystyle \frac{x + 6}{x} - x = \frac{-x^2 + x + 6}{x}, which is \displaystyle -\frac{(x - 3)(x + 2)}{x}, so the condition is \displaystyle \frac{(x - 3)(x + 2)}{x} \le 0. Signs -, +, -, + across -2, 0, 3.
(-\infty, -5] \cup (-1, 3). The difference is \displaystyle \frac{4(x + 1) - 2(x - 3)}{(x - 3)(x + 1)}, which is \displaystyle \frac{2(x + 5)}{(x - 3)(x + 1)} \le 0. The signs -, +, -, + across -5, -1, 3.
(-2, 3) \cup (3, \infty). \displaystyle \frac{(x - 3)(x + 2)}{x - 3} = x + 2 for x \ne 3, positive for x > -2; 3 is out because the original fraction is undefined there.
At least 13 rides. \displaystyle \frac{84 + 2n}{n} - 9 = \frac{84 - 7n}{n} < 0, true for n < 0 or n > 12. A number of rides is a positive whole number, so n \ge 13. At n = 12 the average is exactly \displaystyle \frac{108}{12} = 9 dollars.
(-\infty, -5) \cup (-1, \infty). \displaystyle \frac{x - 3}{x + 1} - 2 = \frac{-x - 5}{x + 1} < 0 is the same as \displaystyle \frac{x + 5}{x + 1} > 0. Multiplying by x + 1 keeps the direction only when x + 1 > 0. His answer wrongly includes -5 < x \le -1 (at x = -2 the left side is 5) and misses x < -5 (at x = -6 it is \displaystyle \frac{9}{5} < 2).
Near a zero r, every other factor keeps one sign. The factor (x - r)^k has the sign of (x - r) raised to the power k: an odd power keeps the sign of x - r, which flips at r; an even power is positive on both sides. So only odd k flips the product’s sign.
One answer: (x + 2)(x - 1)(x - 4) \ge 0. Its signs are -, +, -, + across -2, 1, 4, and \ge includes the zeros, so its solution set is [-2, 1] \cup [4, \infty).
No. In exercise 18, 3 is a zero of the numerator x^2 - x - 6, yet it is never in the answer, because it is also a zero of the denominator. A numerator zero belongs, under \le or \ge, only if the denominator is not also zero there. The first half of the claim is right.