Question

On dividing $x^4-x^3-3x^2+3x+2$ by polynomial $g\left(x\right)$, the quotient and the remainder are $x^2-x-2$ and $2x$ respectively. Find $g\left(x\right)=x^2-m$.Find $m$

Answer 1

We are given the following:

Dividend $=x^4-x^3-3x^2+3x+2$

Divisor$=x^2-m$

Quotient $=x^2-x-2$

Remainder$=2x$

We know that Dividend=Divisor$\times$Quotient$+$ Remainder

Now, substituting the respective values, we get:

$x^4-x^3-3x^2+3x+2=(x^2-m)(x^2-x-2)+2x\Rightarrow x^4-x^3-3x^2+3x+2=\left[x^2\right(x^2-x-2)-m(x^2-x-2\left)\right]+2x\Rightarrow x^4-x^3-3x^2+3x+2=x^4-x^3-2x^2-m{x}^2+mx+2m+2x\Rightarrow x^4-x^3-3x^2+3x+2=x^4-x^3-(2+m)x^2+(2+m)x+2m$

Now, comparing the coefficients, we get

$2+m=3$ or $2m=2$

$⟹m=1$

Hence, $m=1$.