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Archived Discussion => 2009 => End-of-year exams => Exam Discussion => Victoria => Mathematical Methods (and CAS) => Topic started by: tolga on December 24, 2009, 08:20:35 am

Title: help with polynomials
Post by: tolga on December 24, 2009, 08:20:35 am: 2)The functions y=x^3-2x^2+ax+10 and y=6+(a+b)x-4x^2-x^3 both have (-1,0) as an x-intercept. Find the values of a and b
i know that we are suposed to do simultaneous questions but i fon't know what to do when i form the equations
Title: Re: help with polynomials
Post by: GerrySly on December 24, 2009, 10:11:09 am: $y=x^3-2x^2+ax+10$ and $y=6+(a+b)x-4x^2-x^3$

Sub in $(-1, 0)$ into both equations.

$0=(-1)^3-2(-1)^2+(-1)a+10$ and $0=6+(a+b)(-1)-4(-1)^2-(-1)^3$

Leaving you with the two equations... $0=7-a\implies a=7$

Sub $a=7$ into the second equation to get b, $0=3-a-b\implies0=-4-b\implies b=-4$

$\therefore a=7 , b=-4$