Functions Alright

Write $p(x)=q(x)(x^2-1)+r(x)=q(x)(x^2-1)+ax+b$ .

We are told that $p(1)=a+b=-1$ and $p(-1)=-a+b=5$ so $a=-3$ and $b=2$ .

The remainder is $r(x)=\boxed{-3x+2}$

let $p(x) = (a_{0})(x^{n}) + (a_{1})(x^{n-1}) + ... + (a_{n})$

$p(1) = -1 = a_{0} + a_{1} + ... + a_{n}$ (1)

$p(-1) = 5 = -a_{0} + a_{1} - ... + a_{n}$ (2)

or

$p(-1) = 5 = a_{0} - a_{1} + ... + a_{n}$ (3)

by synthetic division,

$[p(x)] mod (x^2 - 1) = (a_{n-1} + a_{n-3} + ... + a_{1})x + (a_{n} + a_{n-2} + a_{n-4} + ... + a_{0})$ )

$(3) + (1)$

$4 = 2(a_{0} + a_{2} + ... + a_{n})$

$2 = a_{0} + a_{2} + ... + a_{n}$

$(1) - (3)$

$-6 = 2(a_{1} + a_{3} + a_{5} + ... + a_{n-1})$

$-3 = a_{1} + a_{3} + a_{5} + ... + a_{n-1}$

$therefore$ ,

$[p(x)] mod (x^2 - 1) = (a_{n-1} + a_{n-3} + ... + a_{1})x + (a_{n} + a_{n-2} + a_{n-4} + ... + a_{0}) = \boxed{-3x + 2}$