NTSE Stage 2 SAT 2018 Exam Question Paper

The remainder when the polynomial
${\mathit{x}}^4-6{\mathit{x}}^3+16{\mathit{x}}^2-25\mathit{x}+10$ is divided by ${\mathit{x}}^2-2\mathit{x}+\mathit{k}$ is $\left(2\mathit{k}-9\right)\mathit{x}+\left({\mathit{k}}^2-8\mathit{k}+10\right)$
But, the remainder is $\mathit{x}+\mathit{a}$
Then,
$\left(2\mathit{k}-9\right)\mathit{x}+\left({\mathit{k}}^2-8\mathit{k}+10\right)=\mathit{x}+\mathit{a}$
Compare the coefficients of x and constant term on both the sides,
Then $2\mathit{k}-9=1$ and ${\mathit{k}}^2-8\mathit{k}+10=\mathit{a}$
$2\mathit{k}=10$
$\mathit{k}=5$
Now
$\mathit{a}={\mathit{k}}^2-8\mathit{k}+10$
$=25-40+10=35-40=-5$
Hence, the required value of a is – 5.