Question

If the polynomial $x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, the remainder comes out to be x+a, find k and a.

Answer 1

We know that,

Dividend$=$Divisor$\times$Quotient$+$Remainder

$\Rightarrow$Dividend$-$Remainder$=$Divisor$\times$Quotient

$\Rightarrow$Dividend$-$Remainder is always divisible by the divisor.

Now, it is given that the polynomial $f\left(x\right)=x^4-6x^3+16x^2-25x+10$ when divided by another polynomial $x^2-2x+k$, leaves $x+a$ as remainder.

Therefore, we have

$f\left(x\right)-(x+a)=x^4-6x^3+16x^2-25x+10-(x+a)\Rightarrow f\left(x\right)-(x+a)=x^4-6x^3+16x^2-26x+10-a$

Now, we divide $x^4-6x^3+16x^2-26x+10-a$ by $x^2-2x+k$ as shown in the above image:

So, for $f\left(x\right)$ to be completely divisible by $x^2-2x+k$, remainder must be equal to zero. Thus, we have,

$(-10+2k)x+(10-a-8k+k^2)=0\Rightarrow -10+2k=0,10-a-8k+k^2=0\Rightarrow 2k=10,10-a-8k+k^2=0\Rightarrow k=5,10-a-(8\times 5)+5^2=0\Rightarrow k=5,10-a-40+25=0\Rightarrow k=5,-a-5=0\Rightarrow k=5,a=-5$

Hence, $k=5$ and $a=-5$.