Question

If t^2 - 1 is a factor of at^3+t^2 - 2t + b, find the values of a and b

Answer 1

`t^2-1=t^2-1^2=(t-1)(t+1)`

If t²-1 is a factor, then both t-1 and t+1 are factors.

According to the remainder theorem, when a binomial x-a is a factor of a polynomial p(x), then p(a)=0.

If t-1 and t+1 are factors of p(t)=at³+t²-2t+b, then p(1)=0 and p(-1)=0.


`p(1)=a * 1^3 + 1^2 -2 * 1+b=a+1-2+b=a+b-1 \\ p(-1)=a * (-1)^3+ (-1)^2-2 * (-1)+b=-a+1+2+b= \\ =-a+b+3`


`p(1)=0 \\ p(-1)=0 \\ \\ a+b-1=0 \\ \underline{-a+b+3=0} \\ 2b+2=0 \\ 2b=-2 \\ b=(-2)/(2) \\ b=-1 \\ \\ a+b-1=0 \\ a-1-1=0 \\ a-2=0 \\ a=2 \\ \\ \boxed{a=2} \\ \boxed{b=-1}`