Question

The polynomial $p\left(x\right)=x^4-2x^3-ax+3a-7$ when divided by $x+1$ leaves a remainder $19$. Fnd the remainder when $p\left(x\right)$ is divided by $x+2.$

Answer 1

Given

$p\left(x\right)=x^4-2x^3+3x^3-ax+3a-7$

By remainder theorem,

If $p\left(x\right)$ is divided by $x+a,$ the remainder $r\left(x\right)=p(-a).$

Here,

$p\left(x\right)$ is divided by $x+1,$

$\therefore$ Remainder $r\left(x\right)=p(-1)=19$

$(-1{)}^4-2(-1{)}^3+3(-1{)}^2-a(-1)+3a-7=19$

$\Rightarrow 1-2(-1)+3\left(1\right)+a+3a-7=19$

$\Rightarrow 1+2+3+4a-7=19$

$\Rightarrow -1+4a=19$

$\Rightarrow 4a=20$

$\Rightarrow a=5$

Again, when $p\left(x\right)$ is divided by $x+2,$

Remainder $r\left(x\right)=p(-2)$

$=(-2{)}^4-2(-2{)}^3+3(-2{)}^2-a(-2)+3a-7$

$=16+16+12+2a+3a-7$

$=37+5a$

$=37+5\left(5\right)$

$=37+25=62$