Unit 2, lesson 34

Polynomial Inequalities

How to solve a polynomial inequality by finding its critical values, testing one point in each interval of a sign chart, and writing the solution set with the right endpoints.

  • Theory part 67
  • Practice part 68
  • Polynomial and Rational Inequalities

Throw a ball straight up and its height after tt seconds might be 20t−5t220t - 5t^2 meters. Asking when it is higher than 15 meters is not a question about one moment but about a whole stretch of time, and the answer is a set of numbers. Questions like this are polynomial inequalities. The method for solving them rests on one fact, that a polynomial can change sign only where it equals zero.

Key terms

Polynomial inequality

Definition A polynomial inequality in one variable is an inequality that can be written in one of the forms

p(x)>0,p(x)≥0,p(x)<0,p(x)≤0,\begin{gathered} p(x) > 0, \qquad p(x) \ge 0, \\[0.4em] p(x) < 0, \qquad p(x) \le 0, \end{gathered}

where pp is a polynomial with real coefficients. Its solution set is the set of all real numbers xx that make it true. An inequality between two polynomials can always be put in this form by subtracting one side from both sides, which does not change the solution set. Multiplying or dividing both sides by a positive constant keeps the direction of the inequality, and doing so by a negative constant reverses it. Multiplying or dividing by an expression that contains xx is safe only when that expression is positive for every real xx, as x2+1x^2 + 1 is, or negative for every real xx, in which case the inequality reverses. An expression such as xx is positive for some values, negative for others and zero at one, so dividing by it is not a valid step.

In plain words Move everything to one side so that a single polynomial is compared with zero. After that the whole question is about the sign of that polynomial.

Example The inequality x3>4xx^3 > 4x becomes x3−4x>0x^3 - 4x > 0. A tempting shortcut is to divide by xx and get x2>4x^2 > 4, but that would accept x=−3x = -3, where x3=−27x^3 = -27 and 4x=−124x = -12, and −27>−12-27 > -12 is false. Dividing by xx quietly assumed that xx is positive. The next sections solve x3−4x>0x^3 - 4x > 0 correctly.

Critical values

Definition The critical values of a polynomial inequality p(x)>0p(x) > 0, or of the same inequality with ≥\ge, << or ≤\le, are the real zeros of pp, that is, the real numbers cc with p(c)=0p(c) = 0. A polynomial is continuous on the whole real line, so it can change sign only at a critical value. On each stretch between two consecutive critical values, and to the left of the smallest one and to the right of the largest one, pp keeps a constant sign. If pp has no real zeros, it has one sign on the whole line. This meaning of critical value belongs to inequalities. Calculus uses the similar names critical point and critical number for something different, an interior point of the domain of a function where its derivative is zero or does not exist.

In plain words The zeros are the only places where the polynomial can switch between positive and negative. Find them first, usually by factoring.

Example Factoring the running example gives

x3−4x=x(x2−4)=x(x+2)(x−2),\begin{aligned} x^3 - 4x &= x(x^2 - 4) \\[0.4em] &= x(x + 2)(x - 2), \end{aligned}

so the critical values are −2-2, 00 and 22.

Number line

Definition Placing the critical values c1<c2<⋯<ckc_1 < c_2 < \cdots < c_k on the real number line splits it into the k+1k + 1 open intervals

(−∞,c1), (c1,c2), …,(ck−1,ck), (ck,∞).\begin{gathered} (-\infty, c_1), \ (c_1, c_2), \ \ldots, \\[0.4em] (c_{k-1}, c_k), \ (c_k, \infty). \end{gathered}

Each critical value is marked with a closed dot when the inequality is ≥\ge or ≤\le, because it makes p(x)=0p(x) = 0 and so satisfies the inequality. It is marked with an open dot when the inequality is strict, because then it does not.

In plain words The number line is the map of the problem. The critical values are the borders, and every open stretch between them is one region to check.

Example For x3−4x>0x^3 - 4x > 0 the critical values −2-2, 00 and 22 cut the line into the four intervals (−∞,−2)(-\infty, -2), (−2,0)(-2, 0), (0,2)(0, 2) and (2,∞)(2, \infty). The inequality is strict, so all three critical values get open dots.

Test points

Definition A test point for one of the open intervals is any single number chosen inside it. Since pp has constant sign on the interval, the sign of pp at the test point is the sign of pp at every point of that interval.

In plain words Pick an easy number in each region and plug it in. The sign you get holds for the whole region.

Example For p(x)=x3−4xp(x) = x^3 - 4x the test points −3-3, −1-1, 11 and 33 give the following.

IntervalTest pointValue of ppSign of pp
(−∞,−2)(-\infty, -2)−3-3−15-15negative
(−2,0)(-2, 0)−1-133positive
(0,2)(0, 2)11−3-3negative
(2,∞)(2, \infty)331515positive

The inequality asks where pp is positive, so the solution set of x3>4xx^3 > 4x is (−2,0)∪(2,∞)(-2, 0) \cup (2, \infty).

Sign chart

Definition Write pp as a nonzero constant times a product of factors, usually powers (x−c)m(x - c)^m of linear factors together with any quadratic factors that have no real zeros. A sign chart lists the intervals of the number line in order and records, for each interval, the sign of every factor and then the sign of the product. A product of nonzero numbers is positive when it has an even number of negative factors and negative when it has an odd number, where a negative constant counts as a negative factor. A factor (x−c)m(x - c)^m with mm a positive integer is negative exactly when x<cx < c and mm is odd. So the sign of pp changes at a critical value of odd multiplicity and stays the same at a critical value of even multiplicity.

In plain words Instead of plugging numbers into the whole polynomial, follow each factor on its own row and then combine the rows. An even power never turns negative, so it cannot flip the sign.

Example Take (x−1)2(x+3)≤0(x - 1)^2(x + 3) \le 0, with critical values −3-3 and 11.

Factor(−∞,−3)(-\infty, -3)(−3,1)(-3, 1)(1,∞)(1, \infty)
(x−1)2(x - 1)^2++++++
x+3x + 3−-++++
Product−-++++

The product is negative only on (−∞,−3)(-\infty, -3). Its sign does not change at 11, because 11 has multiplicity 22. A quick check with x=−4x = -4 gives 25⋅(−1)=−2525 \cdot (-1) = -25, which is negative as the chart says.

Factors with no real zeros

Definition A quadratic ax2+bx+cax^2 + bx + c with real coefficients and discriminant b2−4ac<0b^2 - 4ac < 0 has no real zeros, and its value has the same sign as aa for every real xx. Such a factor adds no critical values and does not affect where the product changes sign. Only its constant sign enters the sign chart.

In plain words Some factors never touch zero. They keep one sign everywhere, so you can note that sign once and leave them out of the hunt for critical values.

Example Solve (x2+x+1)(x−2)<0(x^2 + x + 1)(x - 2) < 0. The quadratic has discriminant 1−4=−3<01 - 4 = -3 < 0 and leading coefficient 11, so it is positive for every xx. The product is therefore negative exactly when x−2<0x - 2 < 0, and the solution set is (−∞,2)(-\infty, 2).

Writing the solution set

Definition The solution set of p(x)>0p(x) > 0 is the union of the open intervals where pp is positive, and the solution set of p(x)<0p(x) < 0 is the union of those where pp is negative. For p(x)≥0p(x) \ge 0 or p(x)≤0p(x) \le 0, add every critical value to the matching union. A critical value whose neighboring intervals both have the wrong sign is still a solution of a non-strict inequality, and it appears in the solution set as a single point.

In plain words Read the signs off the chart, then decide about the borders. With ≤\le or ≥\ge every zero belongs, even one that stands alone between two excluded regions.

Example From the sign chart above, (x−1)2(x+3)(x - 1)^2(x + 3) is negative on (−∞,−3)(-\infty, -3) and zero at −3-3 and 11. So

(x−1)2(x+3)≤0has solution set(−∞,−3]∪{1}.\begin{gathered} (x - 1)^2(x + 3) \le 0 \\[0.4em] \text{has solution set} \\[0.4em] (-\infty, -3] \cup \{1\}. \end{gathered}

The point 11 is easy to forget, since the product is positive on both sides of it. The strict version (x−1)2(x+3)<0(x - 1)^2(x + 3) < 0 has solution set (−∞,−3)(-\infty, -3) with no extra point.

Check yourself

  1. The ball from the start has height 20t−5t220t - 5t^2 meters after tt seconds. When is it higher than 15 meters?
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The inequality 20t−5t2>1520t - 5t^2 > 15 is the same as 5t2−20t+15<05t^2 - 20t + 15 < 0. Dividing by the positive number 55 gives

t2−4t+3=(t−1)(t−3)<0.t^2 - 4t + 3 = (t - 1)(t - 3) < 0.

The critical values are 11 and 33. The test points 00, 22 and 44 give 33, −1-1 and 33, so the product is negative only on (1,3)(1, 3). The ball is higher than 15 meters for 1<t<31 < t < 3, that is, between 1 and 3 seconds after the throw.

  1. Solve x4−5x2+4<0x^4 - 5x^2 + 4 < 0.
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Factoring gives

x4−5x2+4=(x2−1)(x2−4)=(x+2)(x+1)(x−1)(x−2),\begin{aligned} &x^4 - 5x^2 + 4 \\[0.4em] &\quad = (x^2 - 1)(x^2 - 4) \\[0.4em] &\quad = (x + 2)(x + 1)(x - 1)(x - 2), \end{aligned}

so the critical values are −2-2, −1-1, 11 and 22. The test points −3-3, −3/2-3/2, 00, 3/23/2 and 33 give 4040, −35/16-35/16, 44, −35/16-35/16 and 4040. The solution set is (−2,−1)∪(1,2)(-2, -1) \cup (1, 2).

  1. Solve x2(x−4)≥0x^2(x - 4) \ge 0.
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The factor x2x^2 is never negative and is zero only at 00, while x−4x - 4 is negative for x<4x < 4 and positive for x>4x > 4. So the product is negative on (−∞,0)(-\infty, 0) and on (0,4)(0, 4), positive on (4,∞)(4, \infty), and zero at 00 and 44. The solution set is {0}∪[4,∞)\{0\} \cup [4, \infty).

How this lesson connects

Where this leads in Calculus