Sullivan Algebra and Trigonometry: Section 5.2

1. Sullivan Algebra and Trigonometry: Section 5.2 • Objectives • Use the Remainder and Factor Theorems • Use Descartes’ Rule of Signs • Use the Rational Zeros Theorem • Find the Real Zeros of a Polynomial Function • Solve Polynomial Equations • Use the Theorem for Bounds on Zeros • Use the Intermediate Value Theorem

2. Division Algorithm for Polynomials If f(x) and g(x) denote polynomial functions and if g(x) is not the zero polynomial, then there are unique polynomial functions q(x) and r(x) such that dividend quotient divisor remainder

3. 3 2 f ( x ) = 3 x + 9 x - 18 x - 24 Find the remainder if is divided by x + 3. Remainder Theorem Let f be a polynomial function. If f (x) is divided by x - c, then the remainder is f (c). x + 3 = x - (-3)

4. Factor Theorem Let f be a polynomial function. Then x - c is a factor of f (x) if and only if f (c) = 0. In other words, if f(c) = 0, then the remainder found if f(x) is divided by x - c is zero. Hence, since x - c divided into f(x) evenly (remainder 0), x - c is a factor of f(x).

5. Use the Factor Theorem to determine whether the 3 2 f ( x ) = 3 x + 9 x - 18 x - 24 has the factor function (a) x + 3 (b) x + 4 a.) f(-3) = 30 Therefore, x + 3 does not divide into f. So, x + 3 is not a factor of f. b.) f(4) = 0 Therefore, x + 4 does divide into f. So, x + 4 is a factor of f.

6. Theorem: Number of Zeros A polynomial function cannot have more zeros than its degree. Theorem: Descartes’ Rule of Signs Let f denote of polynomial function. The number of positive real zeros of f either equals the number of variations in sign of the nonzero coefficients of f (x) or else equals that number less an even integer. The number of negative real zeros of f either equals the number of variations in sign of the nonzero coefficients of f (-x) or else equals that number less an even integer.

7. 3 2 f ( x ) = 2 x + 3 x - 23 x - 12 Discuss the real zeros of There are at most three zeros, since the function is a polynomial of degree three. Using Descartes’ Rule of Signs, f(x) has one sign change. So, there is one positive real zero. Using Descartes’ Rule of Signs, f (-x) has two sign changes. So, there are two or zero negative real zeros.

8. Rational Zeros Theorem Let f be a polynomial function of degree 1 or higher of the form where each coefficient is an integer. If p/q, in lowest terms, is a rational zero of f, then p must be a factor of a0 and q must be a factor of an.

9. 3 2 f ( x ) = 2 x + 3 x - 23 x - 12 Factors of p: Factors of q: Potential Rational Zeros List the potential rational zeros of According the the theorem, the numerator of potential rational zeros will be factors of p = - 12 and the denominator will be factors of q = 2

10. 5 4 3 2 f ( x ) = x + x - 9 x - x + 20 x - 12 Find the real zeros of Factor f over the reals. First, determine the nature of the zeros. Since the polynomial is degree 5, there are at most five zeros. Using Descartes’ Rule of Signs, there are three or one positive real zero(s). Using Descartes’ Rule of Signs again, there are two or no negative real zeros.

11. 5 4 3 2 f ( x ) = x + x - 9 x - x + 20 x - 12 Now, list all possible rational zeros p/q by factoring the first and last coefficients of the function. Now, begin testing each potential zero using synthetic division. If a potential zero k is in fact a zero, then x - k divides into f (remainder will be zero) and is a factor of f.

12. 5 4 3 2 f ( x ) = x + x - 9 x - x + 20 x - 12 Test k = -3 Test k = -2 Thus, -3 is a zero of f and x + 3 is a factor of f. Thus, -2 is a zero of f and x + 2 is a factor of f.

13. Test k = 1 We know from Descartes’ Rule of Signs that there are no more negative real zeros. Thus, 1 is a zero of f and x - 1 is a factor of f.

14. Theorem: Bounds on Zeros Let f denote a polynomial function whose leading coefficient is 1. A bound M on the zeros of f is the smaller of the two numbers: where Max { } means “choose the largest entry in { }

15. Find a bound on the zeros of Every zero of f lies between -21 and 21.

16. Intermediate Value Theorem Let f denote a continuous function. If a < b and if f(a) and f(b) are of opposite sign, then the graph of f has at least one x-intercept between a and b. y f(b) x-intercept a x b f(a)

17. 3 2 f ( x ) = x + 0 . 2 x - 5 x - 1 Use the intermediate value theorem to show that the graph of the function has an x - intercept in the interval [-3, -2] f (- 3) = -11.2 < 0 f (-2) = 1.8 > 0 Therefore, since f (-3) < 0 and f (-2) > 0, there exists a number between - 3 and -2 where f (x) = 0. So, the function has an x - intercept in the interval [-3,-2]