Question

The polynomial $p\left(x\right)=a{x}^3+4x^2+3x-4$ and $q\left(x\right)=x^3-4x+a$ leave same remainder when divided by $(x-3)$. Find $a$ and hence find the remainder when $p\left(x\right)$ is divided by $(x-2).$

Answer 1

A] given $a{x}^3+4x^2+3x-4=0$ &

$x^3-4x+a=0$ leave same

remainder when divided by x-3

P(x) = $a{x}^3+4x^2+3x-4$

q(x)= $x^3-4x+a$

Remainder theorem,

$P\left(3\right)=q\left(3\right)$

$a(3{)}^3+4(3{)}^2+3\left(3\right)-4=3^3-4\left(3\right)+a$

$27a+36+9-4=27-12+a$

$26a=15-41$

$26a=-26$

$\therefore a=-1$

$P\left(x\right)=-x^3+4x^2+3x-4$

when divide by (x-2)

$P\left(2\right)=-(2{)}^3+4(2{)}^2+3\left(2\right)-4$

$=-8+16+6-4$

$=8+2$

$=10$