Zeros of Polynomial Functions


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1 Zeros of Polynomial Functions The Rational Zero Theorem If f (x) = a n x n + a n1 x n a 1 x + a 0 has integer coefficients and p/q (where p/q is reduced) is a rational zero, then p is a factor of the constant term a 0 and q is a factor of the leading coefficient a n. Example Find all of the possible real, rational roots of f(x) = 2x 33x 2 +5 : p is a factor of 5 = 1, 5 q is a factor of 2 = 1, 2 p/q = 1, 1/2, 5, 5/2 1
2 Properties of Polynomial Equations If a polynomial equation is of degree n, then counting multiple roots separately, the equation has n roots If a+bi is a root of the equation, then abi is also a root. Example Find all zeros of f(x) = x 3 +12x 2 +21x+10 p/q = 1, 2, 5, 10 f(1) = 44 f(1) = 0 Divide out 1 to get x 2 +11x10 Use the quadratic formula to find the last 2 zeros. x= and.844 The solutions are 1, , and.844 Text Example Solve: x 4 6x 2 8x + 24 = 0. The graph of f(x) = x 4 6x 2 8x + 24 is shown the figure below. Because the xintercept is 2, we will test 2 by synthetic division and show that it is a root of the given equation. x intercept: The zero remainder indicates that 2 is a root of x 4 6x 2 8x + 24 = 0. 2
3 Text Example cont. Solve: x 4 6x 2 8x + 24 = 0. Now we can rewrite the given equation in factored form. x 4 6x 2 + 8x + 24 = 0 This is the given equation. (x 2)(x 3 + 2x 2 2x 12) = 0 This is the result obtained from the synthetic division. x 2 = 0 or x 3 + 2x 2 2x 12 = 0 Set each factor equal to zero. Text Example cont. Solve: x 4 6x 2 8x + 24 = 0. We can use the same approach to look for rational roots of the polynomial equation x 3 + 2x 2 2x 12 = 0, listing all possible rational roots. However, take a second look at the figure of the graph of x 4 6x 2 8x + 24 = 0. Because the graph turns around at 2, this means that 2 is a root of even multiplicity. Thus, 2 must also be a root of x 3 + 2x 2 2x 12 = 0, confirmed by the following synthetic division. These are the coefficients of x 3 + 2x 2 2x 12 = 0. x intercept: The zero remainder indicates that 2 is a root of x 3 + 2x 2 2x 12 = 0. Text Example cont. Solve: x 4 6x 2 8x + 24 = 0. Now we can solve the original equation as follows. x 4 6x 2 + 8x + 24 = 0 This is the given equation. (x 2)(x 3 + 2x 2 2x 12) = 0 This was obtained from the first synthetic division. (x 2)(x 2)(x 2 + 4x + 6) = 0 This was obtained from the second synthetic division. x 2 = 0 or x 2 = 0 or x 2 + 4x + 6 = 0 Set each factor equal to zero. x = 2 x = 2 x 2 + 4x + 6 = 0 Solve. 3
4 Text Example cont. Solve: x 4 6x 2 8x + 24 = 0. We can use the quadratic formula to solve x 2 + 4x + 6 = 0. The solution set of the original equation is: {2, 2  i 2, 2+i 2} Descartes s Rule of Signs If f (x) = a n x n + a n 1 x n a 2 x 2 + a 1 x + a 0 be a polynomial with real coefficients. 1. The number of positive real zeros of f is either equal to the number of sign changes of f (x) or is less than that number by an even integer. If there is only one variation in sign, there is exactly one positive real zero. 2. The number of negative real zeros of f is either equal to the number of sign changes of f ( x) or is less than that number by an even integer. If f ( x) has only one variation in sign, then f has exactly one negative real zero. Determine the possible number of positive and negative real zeros of f(x) = x 3 + 2x 2 + 5x To find possibilities for positive real zeros, count the number of sign changes in the equation for f(x). Because all the terms are positive, there are no variations in sign. Thus, there are no positive real zeros. 2. To find possibilities for negative real zeros, count the number of sign changes in the equation for f( x). We obtain this equation by replacing x with x in the given function. f(x) = x 3 + 2x 2 + 5x + 4 This is the given polynomial function. Replace x with x. f( x) = ( x) 3 + 2( x) 2 + 5( x) + 4 = x 3 + 2x 2 5x + 4 Text Example 4
5 Determine the possible number of positive and negative real zeros of f(x) = x 3 + 2x 2 + 5x + 4. Now count the sign changes. Text Example cont. f( x) = x 3 + 2x 2 5x There are three variations in sign. The number of negative real zeros of f is either equal to the number of sign changes, 3, or is less than this number by an even integer. This means that there are either 3 negative real zeros or 3 2 = 1 negative real zero. Zeros of Polynomial Functions More on Zeros of Polynomial Functions 5
6 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) = 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. Text Example Show that all the real roots of the equation 8x x 2 39x + 9 = 0 lie between 3 and 2. 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. Show that all the real roots of the equation 8x x 2 39x + 9 = 0 lie between 3 and 2. 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 Text Example cont 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.) 6
7 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. 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 and Text Example f(2) = = = 1 f(3) = = = 16. f (2) is negative. f (3) is positive. This sign change shows that the polynomial function has a real zero between 2 and 3. Text Example cont. 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 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) = x 3 2x 5 f(2) = 2 3 2(2) 5 = 1 f(2.1) = (2.1) 3 2(2.1) 5 = Sign change Sign change The sign change indicates that f has a real zero between 2 and
8 Text Example cont. 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 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) = f (2.08) = f (2.01) = f (2.02) = f (2.05) = f (2.06) = f (2.09) = f (2.1) = Sign change f (2.03) = f (2.07) = 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. The Fundamental Theorem of Algebra If f (x) is a polynomial of degree n, where n I, then the equation f (x) = 0 has at least one complex root. The Linear Factorization Theorem If f (x) = a n x n + a n 1 x n a 1 x + a 0 b, where n I 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 nthdegree polynomial can be expressed as the product of n linear factors. 8
9 Find a fourthdegree polynomial function f (x) with real coefficients that has 2, 2, and i as zeros and such that f (3) = 150. 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) = a n (x c 1 )(x c 2 )(x c 3 )(x c 4 ) = a n (x + 2)(x 2)(x i)(x + i) = a n (x 2 4)(x 2 + i) Multiply f(x) = a n (x 4 3x 2 4) Text Example This is the linear factorization for a fourthdegree polynomial. Use the given zeros: c 1 = 2, c 2 = 2, c 3 = i, and, from above, c 4 = i. Complete the multiplication Find a fourthdegree polynomial function f (x) with real coefficients that has 2, 2, and i as zeros and such that f (3) = 150. a n ( ) = 150 Solve for a n. 50a n = 150 a n = 3 Substituting 3 for a n in the formula for f(x), we obtain Equivalently, Text Example cont. f (3) = a n (3 4 3*3 2 4) = 150 To find a n, use the fact that f (3) = 150. f(x) = 3(x 4 3x 2 4). f(x) = 3x 4 + 9x Example Use the roots to find the linear factorization of the polynomial equation x 37x 2 +16x10 : The solutions are 3+i, 3i, and 1 Therefore, x=3+i, x=3i, and x=1 Getting zero on one side we have the factors (x (3+i)), (x(3i)), and (x1) The linear factorization is: (x3i)(x3+i)(x1)= 0 9
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