Question

If the polynomial $f\left(x\right)=x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, the remainder comes out to be $(x+a)$, then values of $k$ and $a$ are

• A. $k=-2$ & $a=4$
• B. $k=5$ & $a=-5$
• C. $k=-3$ & $a=-7$
• D. None of these

Answer 1

The correct option is B $k=5$ & $a=-5$

$x^2-2x+k\left)x^4-6x^3+16x^2-25x+10\right(x^2-4x+(8-k)$
$\underset{–}{\underset{-}{x^4}\underset{+}{-2x^3}\underset{-}{+}k{x}^2}$
$-4x^3+(16-k)x^2-25x+10$
$\underset{–}{\underset{-}{-4x^3}\underset{-}{+8x^2}\underset{+}{-}4kx}$
$(8-k)x^2+(4k-25)x+10$
$\underset{–}{\underset{-}{(}8-k)x^2+\underset{-}{(}2k-16)x\underset{-}{+}(8k-k^2)}$
$(2k-9)x+(k^2-8k+10)$
But remainder is given $x+a$
$\therefore x+a=(2k-9)x+(k^2-8k+10)$
On equating coefficient, we get
$2k-9=1\Rightarrow k=5$
and $a=k^2-8k+10\Rightarrow a=25-40+10=-5$
Hence, $k=5,a=-5$