Question

On dividing${\text{x}}^3-3{\text{x}}^2\text{+x+2}$ by a polynomial$\text{g}\left(x\right)$, the quotient and remainder were respectively$\text{x-2 and -2x+4}$respectively. Find$\text{g}\left(x\right)$.

Answer 1

Finding the polynomial $\text{g}\left(x\right)$:

By division algorithm, we know that -

$\text{Dividend =}\left(\text{divisor×quotient}\right)\text{+ remainder}$

According to the question -

$$\begin{gathered} {\text{x}}^3-3{\text{x}}^2\text{+x+2 = g}\left(\text{x}\right)\text{×(x-2) + (-2x+4)} \\ {x}^3-3x^2+x+2-(-2x+4)=g\left(x\right)\times (x-2) \\ {x}^3-3x^2+x+2+2x-4=g\left(x\right)\times (x-2) \\ {x}^3-3x^2+3x-2=g\left(x\right)\times (x-2) \\ g\left(x\right)=\frac{x^3-3x^2+3x-2}{x-2} \\ g\left(x\right)=\frac{x^3-2x^2-x^2+2x+x-2}{x-2} \\ g\left(x\right)=\frac{x^2\left(x-2\right)-x\left(x-2\right)+x-2}{x-2} \\ g\left(x\right)=\frac{(x-2)(x^2-x+1)}{x-2} \\ g\left(x\right)=x^2-x+1 \end{gathered}$$

Hence, the value of$\text{g}\left(x\right)$ is $x^2-x+1.$