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Upper and Lower Bounds for Roots 3.5: 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. 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.
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EXAMPLE: Finding Bounds for the Roots Show that all the real roots of the equation 8 x 3 10 x 2 39 x + 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 . All numbers in this row are nonnegative. 3.5: More on Zeros of Polynomial Functions 35 13 26 8 26 52 16 9 39 10 8 2 more more
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EXAMPLE: Finding Bounds for the Roots Show that all the real roots of the equation 8 x 3 10 x 2 39 x + 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. 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.) 3.5: More on Zeros of Polynomial Functions 35 13 26 8 9 42 24 9 39 10 8 3
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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
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EXAMPLE : Approximating a Real Zero a. Show that the polynomial function f ( x ) x 3 2 x 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 3.5: More on Zeros of Polynomial Functions 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 2 x 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.
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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 2 x 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. Sign change Sign change 3.5: More on Zeros of Polynomial Functions f (2.1) (2.1) 3 2(2.1) 5 0.061 2.1 f (2) 2 3 2(2) 5 1 2 f ( x ) x 3 2 x 5 x more more
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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 2 x 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 f (2.07) 0.270257 f (2.03) 0.694573 f (2.1) 0.061 f (2.06) 0.378184 f (2.02) 0.797592 f (2.09) 0.050671 f (2.05) 0.484875 f (2.01) 0.899399 f (2.08) 0.161088 f (2.04) 0.590336 f (2.00) 1
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The Fundamental Theorem of Algebra 3.5: More on Zeros of Polynomial Functions 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.
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The Linear Factorization Theorem 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 n th-degree polynomial can be expressed as the product of n linear factors. Just as an n th-degree polynomial equation has n roots, an n th-degree polynomial has n linear factors. This is formally stated as the Linear Factorization Theorem. 3.5: More on Zeros of Polynomial Functions
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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. 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 3 x 2 4) Complete the multiplication more more
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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 3 x 2 4) . Equivalently, f ( x ) 3 x 4 9 x 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 . 50 a n 150 a n 3
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