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3.4 Real Zeros Of Polynomials Copyright © Cengage Learning. All rights reserved.

Bell Ringer Quiz 2. 3.

Objectives Rational Zeros of Polynomials

Real Zeros Of Polynomials The Factor Theorem tells us that finding the zeros of a polynomial is really the same thing as factoring it into linear factors. In this section we study some algebraic methods that help us to find the real zeros of a polynomial and thereby factor the polynomial. We begin with the rational zeros of a polynomial.

Rational Zeros of Polynomials

Rational Zeros of Polynomials To help us understand the next theorem, let’s consider the polynomial P (x) = (x – 2)(x – 3)(x + 4) = x3 – x2 – 14x + 24 From the factored form we see that the zeros of P are 2, 3, and – 4. When the polynomial is expanded, the constant 24 is obtained by multiplying (–2)  (–3)  4. This means that the zeros of the polynomial are all factors of the constant term. Factored form Expanded form

Rational Zeros of Polynomials The following generalizes this observation. We see from the Rational Zeros Theorem that if the leading coefficient is 1 or –1, then the rational zeros must be factors of the constant term.

Example 1 – Using the Rational Zeros Theorem Find the rational zeros of P (x) = x3 – 3x + 2. Solution: Since the leading coefficient is 1, any rational zero must be a divisor of the constant term 2. So the possible rational zeros are 1 and 2. We test each of these possibilities. P (1) = (1)3 – 3(1) + 2 = 0

Example 1 – Solution P (–1) = (–1)3 – 3(–1) + 2 cont’d P (–1) = (–1)3 – 3(–1) + 2 = 4 P (2) = (2)3 – 3(2) + 2 P (–2) = (–2)3 – 3(–2) + 2 = 0 The rational zeros of P are 1 and –2.

Rational Zeros of Polynomials The following box explains how we use the Rational Zeros Theorem with synthetic division to factor a polynomial.

Example 2 – Finding Rational Zeros Factor the polynomial P(x) = 2x3 + x2 – 13x + 6, and find all its zeros. Solution: By the Rational Zeros Theorem the rational zeros of P are of the form The constant term is 6 and the leading coefficient is 2, so

Example 2 – Solution cont’d The factors of 6 are 1, 2, 3, 6, and the factors of 2 are 1, 2. Thus, the possible rational zeros of P are Simplifying the fractions and eliminating duplicates, we get the following list of possible rational zeros:

Example 2 – Solution cont’d To check which of these possible zeros actually are zeros, we need to evaluate P at each of these numbers. An efficient way to do this is to use synthetic division.

Example 2 – Solution cont’d From the last synthetic division we see that 2 is a zero of P and that P factors as P(x) = 2x3 + x2 – 13x + 6 = (x – 2)(2x2 + 5x – 3) = (x – 2)(2x – 1)(x + 3) From the factored form we see that the zeros of P are 2, , and –3. Given polynomial From synthetic division Factor 2x2 + 5x – 3