Let $f(x) = x^{10}+5x^9-8x^8+7x^7-x^6-12x^5+4x^4-8x^3+12x^2-5x-5. $
Without using long division (which would be horribly nasty!), find the remainder when $f(x)$ is divided by $x^2-1$.
I'm not sure how to do this, as the only way I know of dividing polynomials other than long division is synthetic division, which only works with linear divisors. I thought about doing $f(x)=g(x)(x+1)(x-1)+r(x)$, but I'm not sure how to continue. Thanks for the help in advance.
Best Answer
Plug in $1$ and $-1$ to get two values of $r(x)$ which is linear. From there you can get what $a,b$ are in $ax+b.$
Since $$f(x)=g(x)(x+1)(x-1)+r(x)$$
we have
$$ f(1)=g(1)(1+1)(1-1)+r(1)=r(1)=-10$$ $$ f(-1)=g(1)(-1+1)(-1-1)+r(-1)=r(-1)=16$$
We know the remainder is of degree $1$, so
$r(x)=ax+b$
and now we know, $$r(1)=ax+b=a+b=-10$$ $$r(-1)=ax+b=-a+b=16$$
so, solve
$$a+b=-10$$ $$-a+b=16$$
which yields, $a=-13$ $b=3$, so
$$r(x)=-13x+3$$