Question

If remainder is same when polynomial$p\left(x\right)=x^3+8x^2+17x+ax$ is divided by ( x + 2 ) and ( x + 1 ).Find the value of a.

Answer 1

According to the Remainder theorem, when a polynomial p(x) is divided by (x−a), then the remainder is the value of p(x) at x=a

$\text{Here}p\left(x\right)=x^3+8x^2+17x+ax$

When it is divided by (x+2), Remainder is = p(−2)

$\therefore p(-2)=(-2{)}^3+8(-2{)}^2+17(-2)+a(-2)$

$=-8+32-34-2a$

$=-10-2a$

When it is divided by ( x + 1 ), Remainder is = p ( −1 )

$\therefore p(-1)=(-1{)}^3+8(-1{)}^2+17(-1)+a(-1)$

$=-1+8–17–a$

$=-10-a$

Now, both remainders are equal.

$-10-2a=-10-a$

$-2a+a=0$

$-a=0$

$a=0$