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
