The Real Zeros of a Polynomial Function 1 Learning Objectives 1. Use the Remainder and Factor Theorems 2. Use the Rational Zeros Theorem to list the potential rational zeros of a polynomial function 3. Find the real zeros of a polynomial function 4. Solve polynomial equations 5. Use the Theorem for Bounds on Zeros 6. Use the Intermediate Value Theorem 2 Zeroes There are two types of zeroes (roots) Real x = a Complex x = a + bi It is important to understand that complex roots always appear in conjugate pairs. That is if a + bi is one root then a - bi is another (a + bi)(a bi) = a 2 + b 2 1
Number and Types of Zeroes Degree of polynomial Real Zeros Complex Zeros Total Zeros 2 0,2 0,2 2 3 1,3 0,2 3 4 0,2,4 0,2,4 4 5 1,3,5 0,2,4 5 6 0,2,4,6 0,2,4,6 6 7 1,3,5,7 0,2,4,6 7 Intermediate Value Theorem IVT The intermediate value theorem states if y = f (x ) is continuous on [ a, b ], and N is a number between f (a ) and f (b ), then there is at least one c where a < c < b such that f (c ) = N. f (b ) y above N f (a ) below a b x IVT and Zeroes If f (x) > 0 (above the x-axis) at one point and f (x) < 0 (below the x-axis) at another point, then f (x) = 0 (on the x-axis) at some point between + - 0 This is useful when finding real zeros 2
Show there is a real zero on [0,1] f x x 6x 9x 3 3 2 f 0 f 1 3 1 0 Hence there is a zero on 0,1 by the IVT Remarks The IVT only works for continuous functions. Intuitively, a function is continuous if you can draw it without lifting your pen from the paper. The function on the left is continuous throughout, but the function on the right is not. It is "discontinuous" at x = c continuous not continuous The Rational Zero Theorem If f (x) = a n x n + a n-1 x n-1 + + a 1 x + a 0 has p integer coefficients and (where p is reduced) q q 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 3
List all possible rational zeros of The constant term is 2 and the leading coefficient is 15 Factors of the constant term, 2 1, 2 Factors of the leading coefficient, 15 1, 3, 5, 15 Factors of the constant term Possible rational zeros Factors of the leading coefficient 1, 2 1, 3, 5, 15 1, 2, 1 2 1 2 1 2 3, 3, 5, 5, 15, 15 Possible rational zeros 1, 2, 1 2 1 2 1 2 3, 3, 5, 5, 15, 15 List all possible rational zeros of We enter f(x) We check the table for the zeroes Zeros 1, 1 2 3, 5 List all possible rational zeros of We compare our results with the graph x = -1 x = -1/3 x = 2/5 4
List all possible rational zeros of We compare with results from plysmt2 x = -1 x = -1/3 x = 2/5 Test for -1 and 1 If the sum of the coefficients is zero, then x=1 is a zero and (x-1) is a factor For example f(x) = 15x 3-14x 2-3x + 2 15-14-3+2=0 so x=1 is a zero If the sum of the coefficients is zero after changing the sign of those with odd degree, then x=-1 is a zero and (x+1) is a factor For example f(x) = 7x 3 + 9x 2-3x - 5-7+9+3-5=0 so x=-1 is a zero Descartes' Rule of Signs Descartes' Rule of Signs will not tell us where the polynomial's zeroes are and how many roots we can expect to find We will then need to use the Rational Roots Test and synthetic division or our graphing calculator to find the zeros 5
Descartes Rule of Signs Let P(x) be a polynomial with real coefficients The number of positive zeros of P is either equal to the number of variations in sign of P(x) or less than this by an even number The number of negative real zeros of P is either equal to the number of variations in sign of P( x) or less than this by an even number 4 2 P x x 2x x 3 4 2 P x x 2x x 3 4 2 P x x 2x x 3 Since P(x) has one variation in sign, the polynomial has one positive real zero There is one variation in sign in P( x) so the polynomial has one negative real zero Since 0 is not a zero for the polynomial and the degree of P(x) is 4, the remaining 2 zeros are complex numbers Finding Zeros and Factors of a Polynomial List the possible rational zeros using the Rational Zero Test Apply the Descartes Rule of Signs to determine the number of possible positive and negative zeros Check the candidates for possible rational zeros, substituting the values from the smallest in magnitude to the largest 6
Apply Descartes' rule of signs to p(x) = x 4 5x 3 + 5x 2 + 5x 6 We have a degree four polynomial, so we know there are four zeros Possible rational zeros are 1 2 3 6,,, 1 1 1 1 Or 6, 3, 2, 1,1, 2,3, 6 Apply Descartes' rule of signs to p(x) = x 4 5x 3 + 5x 2 + 5x 6 x x 4 5x 3 + 5x 2 + 5x 6 + + + We have three sign changes x x 4 5x 3 + 5x 2 + 5x 6 + + + - - We have one sign changes We now believe we have one negative root and three positive roots from the list 6, 3, 2, 1,1, 2,3, 6 Apply Descartes' rule of signs to p(x) = x 4 5x 3 + 5x 2 + 5x 6 We find the zeroes are -1, 1, 2, and 3 7
Upper Bounds Theorem If we divide a polynomial function f (x) by (x - c), where c > 0, using synthetic division and this yields all positive numbers, then c is an upper bound to the real roots of the equation f (x) = 0 Two things must occur for c to be an upper bound. One is c > 0 or positive The other is that all the coefficients of the quotient as well as the remainder are positive Lower Bounds Theorem If you divide a polynomial function f (x) by (x - c), where c < 0, using synthetic division and this yields alternating signs, then c is a lower bound to the real roots of the equation f (x) = 0 Zeros can be counted as either positive or negative Two things must occur for c to be a lower bound One is c < 0 or negative The other is that successive coefficients of the quotient and the remainder have alternating signs Show roots of 5 3 2 P x x 5x 10x 12x 20 are bounded by 4 and 4 4 1 0 5 10 12 20 4 16 44 216 912 1 4 11 54 228 932 We have -4 < 0 AND the successive signs in the bottom row of our synthetic division alternate - 4 is a lower bound for the real roots of this equation 8
Show roots of 5 3 2 P x x 5x 10x 12x 20 are bounded by 4 and 4 4 1 0 5 10 12 20 4 16 44 136 592 1 4 11 34 148 572 We have 4 > 0 AND the all of the signs in the bottom row of our synthetic division are positive 4 is an upper bound for the real roots of this equation Show roots of 5 3 2 P x x 5x 10x 12x 20 are bounded by 4 and 4 Since - 4 is a lower bound and 4 is an upper bound for the real roots of the equation, then that means all real roots of the equation lie between - 4 and 4 We confirm our results Find zeros of Using TI-84 and Poly Root Finder This App is available for the TI-83+, TI-83+ Silver, TI-84 Plus, and TI-84+ Silver at www.education.ti.com 9
Find zeros of Using TI-89 10