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Zeros of Polynomial Functions The Rational Zero Theorem • If f (x) = anxn + an-1xn-1 +…+ a1x + a0 has integer coefficients and p/q (where p/q is reduced) is a rational zero, then p is a factor of the constant term a0 and q is a factor of the leading coefficient an. Example Find all of the possible real, rational roots of f(x) = 2x3-3x2+5 Solution: 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 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 a-bi is also a root. Example Find all zeros of f(x) = x3+12x2+21x+10 p/q = 1, 2, 5, 10 f(1) = 44 f(-1) = 0 Divide out -1 to get x2+11x-10 Use the quadratic formula to find the last 2 zeros. x=-11.844 and .844 The solutions are -1, -11.844, and .844 Text Example Solve: x − 6x2 − 8x + 24 = 0. 4 Solution The graph of f (x) = x4 − 6x2 − 8x + 24 is shown the figure below. Because the x-intercept is 2, we will test 2 by synthetic division and show that it is a root of the given equation. 2 • 1 xintercept: 2 0 −6 −8 24 2 4 −4 −24 2 −2 −12 0 The zero remainder indicates that 2 is a root of x4 − 6x2 − 8x + 24 = 0. 2 Text Example cont. Solve: Solution x4 − 6x2 − 8x + 24 = 0. Now we can rewrite the given equation in factored form. x4 − 6x2 + 8x + 24 = 0 (x – 2)(x3 + 2x2 − 2x − 12) = 0 x–2=0 or x + 2x − 2x − 12 = 0 3 2 This is the given equation. This is the result obtained from the synthetic division. Set each factor equal to zero. Text Example cont. Solve: x4 − 6x2 − 8x + 24 = 0. Solution We can use the same approach to look for rational roots of the polynomial equation x3 + 2x2 − 2x − 12 = 0, listing all possible rational roots. However, take a second look at the figure of the graph of x4 − 6x2 − 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 x3 + 2x2 − 2x − 12 = 0, confirmed by the following synthetic division. These are the coefficients of x3 + 2x2 − 2x − 12 = 0. • 1 1 2 −2 −12 2 8 12 4 6 0 The zero remainder indicates that 2 is a root of x3 + 2x2 − 2x − 12 = 0. xintercept: 2 Text Example cont. Solve: Solution x4 − 6x2 − 8x + 24 = 0. Now we can solve the original equation as follows. x4 − 6x2 + 8x + 24 = 0 This was obtained from the first synthetic division. (x – 2)(x – 2)(x + 4x + 6) = 0 This was obtained from the second synthetic division. 2 x – 2 = 0 or x – 2 = 0 or x + 4x + 6 = 0 2 x=2 This is the given equation. (x – 2)(x3 + 2x2 − 2x − 12) = 0 x=2 x2 + 4x + 6 = 0 Set each factor equal to zero. Solve. 3 Text Example cont. Solve: Solution x4 − 6x2 − 8x + 24 = 0. We can use the quadratic formula to solve x2 + 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) = anxn + an−−1xn−−1 + … + a2x2 + a1x + a0 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. Text Example Determine the possible number of positive and negative real zeros of f (x) = x3 + 2x2 + 5x + 4. Solution 1. 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) = x3 + 2x2 + 5x + 4 This is the given polynomial function. Replace x with −x. f (−x) = (−x)3 + 2(−x)2 + 5(−x) + 4 = −x3 + 2x2 − 5x + 4 4 •Text Example cont. Determine the possible number of positive and negative real zeros of f (x) = x3 + 2x2 + 5x + 4. Solution Now count the sign changes. f (−x) = −x3 + 2x2 − 5x + 4 1 3 2 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 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. Text Example 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. Text Example cont. 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.) 6 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. Text Example 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 2 − 5 = 8 − 4 − 5 = −1 and f (2) is negative. f (3) is positive. f (3) = 3 − 2 2 3 − 5 = 27 − 6 − 5 = 16. 3 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) = 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 2 2.1 f(x) = x3 − 2x − 5 f (2) = 23 − 2(2) − 5 = −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. 7 Text Example cont. 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. 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) = anxn + an−−1xn−−1 + … + a1x + a0 b, where n ∈ I 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. 8 Text Example 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) = an(x + 2)(x −2)(x − i)(x + i) = an(x − 4)(x + i) 2 2 This is the linear factorization for a fourthdegree polynomial. Use the given zeros: c1 = −2, c2 = 2, c3 = i, and, from above, c4 = −i. Multiply f (x) = an(x4 − 3x2 − 4) Complete the multiplication Text Example cont. 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. Example Use the roots to find the linear factorization of the polynomial equation x3-7x2+16x-10 Solution: The solutions are 3+i, 3-i, and 1 Therefore, x=3+i, x=3-i, and x=1 Getting zero on one side we have the factors (x(3+i)), (x-(3-i)), and (x-1) The linear factorization is: (x-3-i)(x-3+i)(x-1)= 0 9 More on Zeros of Polynomial Functions 10