Given the following polynomial,p(x)=7x³10x² +8x-5According to the rational zero theorem, the possible zeros for p(x) are ±a =It turns out x= 1 is a true zero of p(x). You can perform synthetic division to rewritep(x) in the factored form given below. Find the values for a,b,c.p(x) = (x - 1)(ax²+bx+c)b =C =N1,5A/~/~

The given polynomial function is and one zero is

Given the following polynomial, p(x)=7x³10x² +8x-5 According to the rational zero theorem, the possible zeros for p(x) are + 1,5 1,7 It turns out x= 1 is a true zero of p(x). You can perform synthetic division to rewrite p(x) in the factored form given below. Find the values for a,b,c. p(x) = (x - 1)(ax²+bx+c) a = b = C = 1 신 T 1

Given the following polynomial, p(x)=7x³10x² +8x-5 According to the rational zero theorem, the possible zeros for p(x) are + 1,5 1,7 It turns out x= 1 is a true zero of p(x). You can perform synthetic division to rewrite p(x) in the factored form given below. Find the values for a,b,c. p(x) = (x - 1)(ax²+bx+c) a = b = C = 1 신 T 1

College Algebra (MindTap Course List)

12th Edition

ISBN:9781305652231

Author:R. David Gustafson, Jeff Hughes

Publisher:R. David Gustafson, Jeff Hughes

Chapter4: Polynomial And Rational Functions

Section4.CR: Chapter Review

Problem 61E

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