the polynomial P(x)=x^3+ax^2+3x+b has a reminder of -8 when divided by x-3. One of the factors of p(x) is x+1. Find a and b.
Do you think for this question need to do system of equation?
i'm not sure if this is right: a=146/8 and b=98/8
So this is basically long division twice, which results in some algebraic equations:
x+1 is one factor mean x=-1 is root os equation
I used substitution since i found that easiest Check my messy work here : http://www.twiddla.com/Redirector.aspx?404;http://www.twiddla.com:80/1581717
root will satisfy the equation and p(-1)=0 p(-1)=0=(-1)^3+a*(-1)^2+3*(-1)+b
You can probably use synthetic division which is neater
0=-1+a-3+b 0=-4+a+b a+b=4
So that gives us one equation for a, b... then the next one is similar we divide x+1 into P(x) get an expression for the remainder and set it equal to 0 in this case
now on dividing with x-3 it gives remainder =-8
now if we add 8 in the equation than it will give remainder =0
if remainder is aero than x-3 will be its factor
that means x=3 is a factor of the equation
x^3+ax^2+3x+b +8 x=3 is a factor of this equation
put x=3 which should give result =0
3^3+a3^2+3*3+b+8=0
27+9a+9+b+8=0 9a+b=-44
9a+b=-44 a+b=4 means b=4-a putting it in above equation 9a+4-a=-44 8a+4=-44 8a=-44+4=-40 a=-40/8=-5 and b=9
so a=-5 and b=9
a = -6 @gorv
and b = 10
mine is comming -5
and 9
Well, mine came out as a= -6 and b= 10... Unless I have the wrong P(x) or factors But given the problem statement, I'm confident in this answer now
I really doubt it unless I don't have the right P(x) or factors or maybe the wrong remainder
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