Question

Find the remainder when $3x^5+11x^4+90x^2-19x+53$ is divided by $x+5$.

Answer 1

Given, polynomial $3x^5+11x^4+90x^2-19x+53$ divided by $(x+5)$.

Then, $f\left(x\right)=$ $3x^5+11x^4+90x^2-19x+53$.

The polynomial is divided by $(x+5)$ .

Then put$(x+5)=0⟹$ $x=-5$, we get,

$f(-5)=$ $3(-5{)}^5+11(-5{)}^4+90(-5{)}^2-19(-5)+53$

$=(3\times -3125)+11\times 625+90\times 25+95+53$

$=-9375+6875+2250+95+53$ $=-102$.

So, when $f\left(x\right)=x^3+1$ is divided by $x+1$, the remainder obtained is $-102$.