Real Zeros of Polynomials through Factorization

1. 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.

2. Rational Zeros of Polynomials • To help us understand the next theorem, let’s consider the polynomialP(x) = (x – 2)(x – 3)(x – 4) = x3 – x2 – 14x + 24From 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

3. 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.

4. 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

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

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

7. 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

8. 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:

9. 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.

10. Example 2 – Solution • cont’d • From the last synthetic division we see that 2 is a zero of P and that P factors asP(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