Find the value of k such that `(x^3-k*x^2+2)` / `(x-1)` has a remainder of 8.

Hello!

There is the Polynomial remainder theorem (its prove is simple provided we know that the remainder is a number (a monomial of degree 0)):

if a polynomial `P(x)` is divided by a binomial `(x-c)` then the remainder is equal to `P(c).`

Here `P(x)=x^3-k*x^2+2` and `c=1,` therefore the remainder is equal to

`P(1)=1-k+2=3-k.`

And it is equal to 8 by the condition of the problem,

`3-k=8.`

Therefore k=3-8=-5. This is the (unique) answer, k=-5.

Check the answer by the direct division:

`(x^3+5x^2+2)/(x-1)=[x^2(x-1)+x^2+5x^2+2]/(x-1)=`

`=x^2+[6x^2+2]/(x-1)=x^2+[6x(x-1)+6x+2]/(x-1)=`

`=x^2+6x+[6x+2]/(x-1)=x^2+6x+[6(x-1)+6+2]/(x-1)=`

`=(x^2+6x+6)+8/(x-1),`

or

`(x^3+5x^2+2)=(x^2+6x+6)(x-1)+8.`