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Topos Uranos
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Sign charts and inequalities

Summary

This lesson teaches how to determine the sign of a polynomial and of a quotient of polynomials from its factorization, organizing the analysis in a sign chart. It studies the role of the multiplicity of the roots and of the irreducible quadratic factors, distinguishes, among the zeros of the denominator, vertical asymptotes from removable points, with a first intuitive idea of limit, and applies the method to solving polynomial and rational inequalities, strict and non-strict, and to determining the domains of functions with quotients and roots, including nested roots. It is the first of three lessons devoted to the domain, the range and the graph of algebraic functions.

Learning objectives

  1. Construct the sign chart of a factored polynomial, taking into account the multiplicity of each root and the irreducible quadratic factors.
  2. Determine the sign of a quotient of polynomials and recognize, among the zeros of the denominator, the vertical asymptotes and the removable points.
  3. Solve polynomial and rational inequalities, strict and non-strict, without multiplying by expressions of unknown sign.
  4. Determine the domain of functions with quotients and roots, including nested roots, by means of sign charts.
  5. Translate applied problems into inequalities, such as the duration of a trip, the effect of a medication or the design of a container, and solve them with sign charts.

The sign of a polynomial

Many questions about a function come down to knowing where an expression is positive, negative or zero: the domain of a square root requires a non-negative radicand, an inequality asks where an expression is positive, and the graph of a function lies above the xx-axis exactly where the function is positive. The instrument that answers this question is the sign chart, and its raw material is the factorization studied in the lesson on factoring quadratic polynomials.

The sign of a linear factor

A linear factor ax+bax + b, with a≠0a \neq 0, vanishes only at x=−bax = -\frac{b}{a} and has constant sign on each side of that point. If a>0a > 0, it is negative to the left and positive to the right; if a<0a < 0, the other way round. For example, x−3x - 3 is positive exactly when x>3x > 3, and 2−x2 - x exactly when x<2x < 2:

x−3>0↔x>3,2−x>0↔x<2x - 3 > 0 \leftrightarrow x > 3, \qquad 2 - x > 0 \leftrightarrow x < 2
LemmaSign of a linear factor

Let ax+bax + b be a polynomial of degree 11, with root r=−bar = -\frac{b}{a}. If a>0a > 0, then ax+bax + b is positive exactly when x>rx > r and negative exactly when x<rx < r; if a<0a < 0, it is positive exactly when x<rx < r and negative exactly when x>rx > r.

The proof uses the rules for operating with inequalities, proved in the lesson on the real line, and the lemma on the sign of a product from the lesson on algebraic functions.

Proof

  1. ax+b=a(x+ba)=a (x−r)ax + b = a\left( x + \frac{b}{a} \right) = \dato{f}{a \, \resaltar{(x - r)}}

    We factor out aa: the linear factor is aa times the difference x−rx - r, which is positive exactly when x>rx > r, zero at rr and negative exactly when x<rx < r.

  2. a>0→(ax+b>0↔x>r)a > 0 \rightarrow \left( ax + b > 0 \leftrightarrow \resaltar{x > r} \right)

    By the lemma on the sign of a product, a(x−r)a(x - r) is positive exactly when aa and x−rx - r have the same sign. If a>0a > 0, this happens exactly when x>rx > r.

  3. a<0→(ax+b>0↔x<r)a < 0 \rightarrow \left( ax + b > 0 \leftrightarrow \resaltar{x < r} \right)

    If a<0a < 0, the product is positive exactly when x−rx - r is negative, that is, when x<rx < r; and in both cases the product is negative exactly when the signs are opposite.

The sign chart of a product

Suppose that a polynomial is written as a product of linear factors. The numbers at which some factor vanishes are called critical points; ordered from least to greatest, they divide the real line into open intervals. Now, each of these intervals lies entirely on one side of the root of each factor, so that, by the lemma on the sign of a linear factor, no factor changes sign within it; it follows that the product also has constant sign on each interval. That sign is given by the rule of signs, which extends to nn factors the lemma on the sign of a product of two factors, proved in the lesson on algebraic functions. The extension is proved by induction, as in the lesson on summations, progressions and induction.

LemmaRule of signs

A product of nn nonzero real factors is different from zero; it is negative if the number of its negative factors is odd, and positive if it is even.

Let c1,c2,…,cnc_1, c_2, \ldots, c_n be nonzero real numbers and Πn=c1c2⋯cn\Pi_n = c_1 c_2 \cdots c_n their product. We argue by induction on the number nn of factors. The figure shows the factors in a row and, below each one, the sign of the product accumulated up to it.

Proof

  1. c>0→(p⋅c>0↔p>0),c<0→(p⋅c>0↔p<0)\dato{pos}{c > 0 \rightarrow (p \cdot c > 0 \leftrightarrow p > 0)}, \quad \dato{neg}{c < 0 \rightarrow (p \cdot c > 0 \leftrightarrow p < 0)}

    By the lemma on the sign of a product of two factors, the product of a nonzero real pp by a nonzero real cc is not zero; if cc is positive, it has the sign of pp, and if cc is negative, the opposite sign.

  2. Π1=c1\Pi_1 = c_1

    Base case: if n=1n = 1, the product is the factor itself, which is negative exactly when there is one negative factor, and 11 is odd.

  3. Πn+1=Πn⋅cn+1\dato{h}{\Pi_{n + 1} = \Pi_n \cdot c_{n + 1}}

    Induction hypothesis: Πn\Pi_n is negative if the number knk_n of its negative factors is odd, and positive if it is even. We add one more factor.

  4. cn+1>0→kn+1=knc_{n + 1} > 0 \rightarrow \resaltar{k_{n + 1} = k_n}

    If cn+1c_{n + 1} is positive, the sign of the product does not change, and the number of negative factors does not change: the rule holds for n+1n + 1 factors.

  5. cn+1<0→kn+1=kn+1c_{n + 1} < 0 \rightarrow \resaltar{k_{n + 1} = k_n + 1}

    If cn+1c_{n + 1} is negative, the sign is reversed and the number of negative factors increases by one, which changes its parity. The rule holds as well, and by the principle of induction it is valid for every nn.

The sign chart organizes this reasoning. Its columns are the intervals determined by the critical points; it has one row for each factor and a last row for the product, which is obtained column by column with the rule of signs. At each critical point a 00 is written in the rows of the factors that vanish, and also in the row of the product.

ExampleThe sign chart of a product

Determine the sign of the polynomial P(x)=(x+2)(x−1)(x−3)P(x) = (x + 2)(x - 1)(x - 3) at each point of the real line.

Proof

  1. P(x)=0↔x=−2∨x=1∨x=3P(x) = 0 \leftrightarrow x = -2 \vee x = 1 \vee x = 3

    The critical points are the roots of the three factors. They divide the line into four intervals, which are the columns of the chart.

  2. x+2>0↔x>−2x + 2 > 0 \leftrightarrow x > -2

    The factor x+2x + 2 vanishes at −2-2; it is negative to the left of that point and positive to the right.

  3. x−1>0↔x>1x - 1 > 0 \leftrightarrow x > 1

    The factor x−1x - 1 changes sign at 11.

  4. x−3>0↔x>3x - 3 > 0 \leftrightarrow x > 3

    The factor x−3x - 3 changes sign at 33.

  5. P(x)<0↔x<−2∨1<x<3P(x) < 0 \leftrightarrow x < -2 \vee 1 < x < 3

    In each column we count the negative factors: three, two, one and none. The product is negative when their number is odd.

  6. P(x)>0↔−2<x<1∨x>3P(x) > 0 \leftrightarrow -2 < x < 1 \vee x > 3

    In the other columns the product is positive, and it vanishes at the three critical points. The graph confirms it: the curve lies above the axis exactly on those intervals, and crosses it at each critical point.

Multiplicity and irreducible factors

Not every polynomial splits into distinct linear factors. As seen in the lesson on the roots of higher-degree polynomials, rr is a root of multiplicity kk when the polynomial is (x−r)k(x - r)^k times another polynomial that does not vanish at rr, and this multiplicity is determined by repeated synthetic division; in the chart, the corresponding factor is (x−r)k(x - r)^k. On the other hand, by the theorem on the sign of the discriminant, proved in the lesson on factoring quadratic polynomials, a quadratic factor ax2+bx+cax^2 + bx + c with negative discriminant is irreducible and has no real roots. The following proposition states what role each of them plays in the chart.

PropositionSign and multiplicity

The sign of a factored polynomial changes on passing through a critical point if and only if the multiplicity of that root is odd. Irreducible quadratic factors have constant sign, that of their leading coefficient, and can be removed from an inequality when they are positive, because dividing by a positive number preserves the inequality.

Let rr be a root of multiplicity kk of the polynomial PP. The figure follows the proof with r=1r = 1 and the cofactor x+2x + 2, and at the end with the irreducible factor x2+x+1x^2 + x + 1.

Proof

  1. P(x)=(x−r)kC(x),C(r)≠0\dato{p}{P(x) = \resaltar{(x - r)^k} C(x), \quad C(r) \neq 0}

    We write PP separating the factor of rr. Between rr and the nearest critical point of CC, no factor of CC vanishes, and by the rule of signs CC has constant sign on both sides of rr.

  2. k=2j→(x−r)k=((x−r)j)2>0,x≠rk = 2j \rightarrow (x - r)^k = \resaltar{\left( (x - r)^j \right)^2} > 0, \quad x \neq r

    If kk is even, (x−r)k(x - r)^k is the square of a nonzero number on both sides of rr, and is positive: PP has there the sign of CC, which does not change.

  3. k=2j+1→(x−r)k=(x−r)((x−r)j)2k = 2j + 1 \rightarrow (x - r)^k = \resaltar{(x - r)} \left( (x - r)^j \right)^2

    If kk is odd, (x−r)k(x - r)^k is x−rx - r times a positive square, and has the sign of x−rx - r, which changes at rr: by the rule of signs, PP changes sign on passing through rr.

  4. ax2+bx+c=a((x+b2a)2+4ac−b24a2)ax^2 + bx + c = a \resaltar{\left( \left( x + \frac{b}{2a} \right)^2 + \frac{4ac - b^2}{4a^2} \right)}

    If b2−4ac<0b^2 - 4ac < 0, in the completed-square form the bracket is a square plus a positive number: the factor always has the sign of aa, contributes no critical points and, if a>0a > 0, dividing by it preserves the direction of an inequality.

ExampleA double root and an irreducible factor

Determine where the polynomial Q(x)=(x+1)2(2−x)(x2+x+1)Q(x) = (x + 1)^2 (2 - x)(x^2 + x + 1) is positive and where it is non-negative.

Proof

  1. Δ=12−4⋅1⋅1=−3<0\Delta = 1^2 - 4 \cdot 1 \cdot 1 = -3 < 0

    The critical points are −1-1, a double root, and 22. The quadratic factor contributes none, because its discriminant is negative.

  2. (x+1)2>0↔x≠−1(x + 1)^2 > 0 \leftrightarrow x \neq -1

    The factor (x+1)2(x + 1)^2 is a square: it vanishes at −1-1 and is positive on both sides.

  3. 2−x>0↔x<22 - x > 0 \leftrightarrow x < 2

    The factor 2−x2 - x has a negative leading coefficient: it is positive to the left of 22.

  4. x2+x+1=(x+12)2+34>0x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} > 0

    The irreducible factor is positive in every column, as the completed square shows.

  5. Q(x)≥0↔x≤2Q(x) \geq 0 \leftrightarrow x \leq 2

    With the rule of signs we obtain the row of QQ: the sign does not change at −1-1, a root of even multiplicity, and changes at 22. The non-strict inequality includes both zeros.

  6. Q(x)>0↔x<2∧x≠−1Q(x) > 0 \leftrightarrow x < 2 \wedge x \neq -1

    The strict inequality excludes the zeros of QQ, including the one at −1-1, even though the polynomial is positive on both sides of it.