Question

What should be added to $x^5-1$ to be completely divisible by $x^2+3x-1$?

Answer 1

Steps:

Step 1) Multiply $x^2+3x-1$ with $x^3$, to get $x^5$,Remainder $=-3x^4+x^3-1$

Continue it to further two steps

Step 4)Multiply $x^2+3x-1$ with $-33$, to get, $-33x^2+10x-1$, Remainder $=109x-34$

$\therefore x^5-1$ to be completely divisible by $x^2+3x-1,34-109x$ needs to be added.