1. 4.5: More on Zeros of Polynomial Functions 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. 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. 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. 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. 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. 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. Upper and Lower Bounds for Roots

2. EXAMPLE: Finding Bounds for the Roots Show that all the real roots of the equation 8x 3  10x 2  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. 2810  39 9 165226 8 1335 All numbers in this row are nonnegative. 4.5: More on Zeros of Polynomial Functions more

3. EXAMPLE: Finding Bounds for the Roots Show that all the real roots of the equation 8x 3  10x 2  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3 810  39 9  24 42 99 8-1430 Counting zero as negative, the signs alternate: , , , . 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.) 4.5: More on Zeros of Polynomial Functions

4. EXAMPLE: Finding Bounds for the Roots Show that all the real roots of the equation 8x 3  10x 2  39x + 9  0 lie between –3 and 2. 2 nd Method For Finding Lower Bound: Similar to Des Carte’s Rule of signs, we can find f(-x). Divide the polynomial by x – 3 using synthetic division (note 3 is now positive). If all of the sings in the bottom row are negative in value, then  3 is a lower bound. Remember that the number zero can be considered positive or negative. 3-810399  24 -42 99 -8-14-30 Counting zero as negative, the signs alternate: , , , . By the Upper and Lower Bound Theorem, the negative 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.) A lower/upper bound may not be a root/zero. 4.5: More on Zeros of Polynomial Functions

5. EXAMPLE 2: Finding Bounds for the Roots Use the Upper Bound Theorem to find an integral upper bound and the Lower Bound to find an integral lower bound of the zeros of f(x) =x 3  5x 2  3x - 20 1.Determine the number of complex roots 1.List the possible rational zeros. 1.Find Possible positive and negative real zeros 1.Determine interval where the zeros are located 1.Determine Upper and Lower Bounds 1.Determine zeros to the nearest tenth. 4.5: More on Zeros of Polynomial Functions

6. 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. 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

7. EXAMPLE:Approximating a Real Zero a. Show that the polynomial function f (x)  x 3  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 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)  x 3  2x  5, we obtain Solution This sign change shows that the polynomial function has a real zero between 2 and 3. and f (3)  3 3  2  3  5  27  6  5  16. f (3) is positive. f (2)  2 3  2  2  5  8  4  5  1 f (2) is negative. 3.5: More on Zeros of Polynomial Functions

8. EXAMPLE:Approximating a Real Zero 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. Solution a. Show that the polynomial function f (x)  x 3  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 The sign change indicates that f has a real zero between 2 and 2.1. x f (x)  x 3  2x  5 2 f (2)  2 3  2(2)  5   1 2.1 f (2.1)  (2.1) 3  2(2.1)  5  0.061 Sign change 3.5: More on Zeros of Polynomial Functions more

9. f (2.00)  1f (2.04)  0.590336f (2.08)  0.161088 f (2.01)  0.899399f (2.05)  0.484875f (2.09)  0.050671 f (2.02)  0.797592f (2.06)  0.378184f (2.1)  0.061 f (2.03)  0.694573f (2.07)  0.270257 EXAMPLE:Approximating a Real Zero 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 sub- intervals. Then we evaluate f at each endpoint and look for a sign change. Solution a. Show that the polynomial function f (x)  x 3  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 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. Sign change 3.5: More on Zeros of Polynomial Functions

10. 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. 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. The Fundamental Theorem of Algebra

11. The Linear Factorization Theorem If f (x)  a n x n  a n  1 x n  1  …  a 1 x  a 0 b, where n  1 and a n  0, then f (x)  a n (x  c 1 ) (x  c 2 ) … (x  c n ) where c 1, c 2,…, c n 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. The Linear Factorization Theorem If f (x)  a n x n  a n  1 x n  1  …  a 1 x  a 0 b, where n  1 and a n  0, then f (x)  a n (x  c 1 ) (x  c 2 ) … (x  c n ) where c 1, c 2,…, c n 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. 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. 3.5: More on Zeros of Polynomial Functions

12. EXAMPLE:Finding a Polynomial Function with Given Zeros Find a fourth-degree polynomial function f (x) with real coefficients that has  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.  a n (x  2)(x  2)(x  i)(x  i) Use the given zeros: c 1  2, c 2  2, c 3  i, and, from above, c 4  i. f (x)  a n (x  c 1 )(x  c 2 )(x  c 3 )(x  c 4 ) This is the linear factorization for a fourth- degree polynomial.  a n (x 2  4)(x 2  i) Multiply f (x)  a n (x 4  3x 2  4) Complete the multiplication more

13. 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, and i as zeros and such that f (3)  150. Substituting  3 for a n in the formula for f (x), we obtain f (x)  3(x 4  3x 2  4). Equivalently, f (x)  3x 4  9x 2  12. Solution f (3)  a n (3 4  3  3 2  4)  150 To find a n, use the fact that f (3)  150. a n (81  27  4)  150 Solve for a n. 50a n  150 a n  3