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3.5: More on Zeros of Polynomial Functions Upper and Lower Bounds for Roots The Upper and Lower Bound Theorem helps us rule out many of a polynomial equation's possible rational roots. The Upper and Lower Bound Theorem Let f(x) be a polynomial with real coefficients and a positive leading coefficient, and let a and b be nonzero real numbers. 1. Divide f(x) by x - b (where b > 0) using synthetic division. If the last row containing the quotient and remainder has no negative numbers, then b is an upper bound for the real roots of f(x) = 0. 2. Divide f(x) by x - a (where a < 0) using synthetic division. If the last row containing the quotient and remainder has numbers that alternate in sign (zero entries count as positive or negative), then a is a lower bound for the real roots of f(x) = 0. 3.5: More on Zeros of Polynomial Functions EXAMPLE: Finding Bounds for the Roots Show that all the real roots of the equation 8x3 + 10x2 - 39x + 9 = 0 lie between –3 and 2. Solution We begin by showing that 2 is an upper bound. Divide the polynomial by x - 2. If all the numbers in the bottom row of the synthetic division are nonnegative, then 2 is an upper bound . 2 8 10 -39 9 16 52 26 8 26 13 35 All numbers in this row are nonnegative. more 3.5: More on Zeros of Polynomial Functions EXAMPLE: Finding Bounds for the Roots Show that all the real roots of the equation 8x3 + 10x2 - 39x + 9 = 0 lie between –3 and 2. Solution The nonnegative entries in the last row verify that 2 is an upper bound. Next, we show that -3 is a lower bound. Divide the polynomial by x (-3), or x + 3. If the numbers in the bottom row of the synthetic division alternate in sign, then -3 is a lower bound. Remember that the number zero can be considered positive or negative. -3 8 10 -39 -24 9 42 -9 Counting zero as negative, the signs alternate: +, -, +, -. 8 26 13 35 By the Upper and Lower Bound Theorem, the alternating signs in the last row indicate that -3 is a lower bound for the roots. (The zero remainder indicates that -3 is also a root.) 3.5: More on Zeros of Polynomial Functions The Intermediate Value Theorem The Intermediate Value Theorem for Polynomials Let f(x) be a polynomial function with real coefficients. If f(a) and f(b) have opposite signs, then there is at least one value of c between a and b for which f(c) = 0. Equivalently, the equation f(x) = 0 has at least one real root between a and b. 3.5: More on Zeros of Polynomial Functions EXAMPLE: Approximating a Real Zero a. Show that the polynomial function f(x) = x3 - 2x - 5 has a real zero between 2 and 3. b. Use the Intermediate Value Theorem to find an approximation for this real zero to the nearest tenth Solution a. Let us evaluate f(x) at 2 and 3. If f(2) and f(3) have opposite signs, then there is a real zero between 2 and 3. Using f(x) = x3 - 2x - 5, we obtain f(2) = 23 - 2 2 - 5 = 8 - 4 - 5 = -1 and f (2) is negative. f (3) is positive. f(3) = 33 - 2 3 - 5 = 27 - 6 - 5 = 16. This sign change shows that the polynomial function has a real zero between 2 and 3. 3.5: More on Zeros of Polynomial Functions EXAMPLE: Approximating a Real Zero a. Show that the polynomial function f(x) = x3 - 2x - 5 has a real zero between 2 and 3. b. Use the Intermediate Value Theorem to find an approximation for this real zero to the nearest tenth Solution b. A numerical approach is to evaluate f at successive tenths between 2 and 3, looking for a sign change. This sign change will place the real zero between a pair of successive tenths. x f(x) = x3 - 2x - 5 2 f(2) = 23 - 2(2) - 5 2.1 = -1 f(2.1) = (2.1)3 - 2(2.1) - 5 = 0.061 Sign change Sign change The sign change indicates that f has a real zero between 2 and 2.1. more 3.5: More on Zeros of Polynomial Functions EXAMPLE: Approximating a Real Zero a. Show that the polynomial function f(x) = x3 - 2x - 5 has a real zero between 2 and 3. b. Use the Intermediate Value Theorem to find an approximation for this real zero to the nearest tenth Solution b. We now follow a similar procedure to locate the real zero between successive hundredths. We divide the interval [2, 2.1] into ten equal subintervals. Then we evaluate f at each endpoint and look for a sign change. f (2.00) = -1 f (2.04) = -0.590336 f (2.08) = -0.161088 f (2.01) = -0.899399 f (2.05) = -0.484875 f (2.09) = -0.050671 f (2.02) = -0.797592 f (2.06) = -0.378184 f (2.1) = 0.061 f (2.03) = -0.694573 f (2.07) = -0.270257 Sign change The sign change indicates that f has a real zero between 2.09 and 2.1. Correct to the nearest tenth, the zero is 2.1. 3.5: More on Zeros of Polynomial Functions The Fundamental Theorem of Algebra We have seen that if a polynomial equation is of degree n, then counting multiple roots separately, the equation has n roots. This result is called the Fundamental Theorem of Algebra. The Fundamental Theorem of Algebra If f(x) is a polynomial of degree n, where n 1, then the equation f(x) = 0 has at least one complex root. 3.5: More on Zeros of Polynomial Functions The Linear Factorization Theorem Just as an nth-degree polynomial equation has n roots, an nth-degree polynomial has n linear factors. This is formally stated as the Linear Factorization Theorem. The Linear Factorization Theorem If f(x) = anxn + an-1xn-1 + … + a1x + a0 b, where n 1 and an 0 , then f (x) = an (x - c1) (x - c2) … (x - cn) where c1, c2,…, cn are complex numbers (possibly real and not necessarily distinct). In words: An nth-degree polynomial can be expressed as the product of n linear factors. 3.5: More on Zeros of Polynomial Functions EXAMPLE: Finding a Polynomial Function with Given Zeros Find a fourth-degree polynomial function f(x) with real coefficients that has -2, 2, and i as zeros and such that f(3) = -150. Solution Because i is a zero and the polynomial has real coefficients, the conjugate must also be a zero. We can now use the Linear Factorization Theorem. f(x) = an(x - c1)(x - c2)(x - c3)(x - c4) This is the linear factorization for a fourthdegree polynomial. = an(x + 2)(x -2)(x - i)(x + i) Use the given zeros: c1 = -2, c2 = 2, c3 = i, and, from above, c4 = -i. = an(x2 - 4)(x2 + i) Multiply f(x) = an(x4 - 3x2 - 4) Complete the multiplication more 3.5: More on Zeros of Polynomial Functions EXAMPLE: Finding a Polynomial Function with Given Zeros Find a fourth-degree polynomial function f(x) with real coefficients that has -2, 2, and i as zeros and such that f(3) = -150. Solution f (3) = an(34 - 3 32 - 4) = -150 To find an, use the fact that f (3) = -150. an(81 - 27 - 4) = -150 Solve for an. 50an = -150 an = -3 Substituting -3 for an in the formula for f(x), we obtain f(x) = -3(x4 - 3x2 - 4). Equivalently, f(x) = -3x4 + 9x2 + 12.