[tex]t^2-1=t^2-1^2=(t-1)(t+1)[/tex]
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.
[tex]p(1)=a \times 1^3 + 1^2 -2 \times 1+b=a+1-2+b=a+b-1 \\
p(-1)=a \times (-1)^3+ (-1)^2-2 \times (-1)+b=-a+1+2+b= \\
=-a+b+3[/tex]
[tex]p(1)=0 \\
p(-1)=0 \\ \\
a+b-1=0 \\
\underline{-a+b+3=0} \\
2b+2=0 \\
2b=-2 \\
b=\frac{-2}{2} \\
b=-1 \\ \\
a+b-1=0 \\
a-1-1=0 \\
a-2=0 \\
a=2 \\ \\
\boxed{a=2} \\ \boxed{b=-1}[/tex]
