Question

Find the values of a and b so that $x^4+x^3+8x^2+ax+b$ is divisible by $x^2+1$.

Answer 1

Let us first divide the given polynomial $x^4+x^3+8x^2+ax+b$ by $(x^2+1)$ as shown in the above image:

From the division, we observe that the quotient is $x^2+x+7$ and the remainder is $(a-1)x+(b-7)$.

Since it is given that $x^4+x^3+8x^2+ax+b$ is exactly divisible by $x^2+1$, therefore, the remainder must be equal to $0$ that is:

$(a-1)x+(b-7)=0\Rightarrow (a-1)x+(b-7)=0\cdot x+0\Rightarrow (a-1)=0,(b-7)=0\left(Bycomparingcoefficients\right)\Rightarrow a=1,b=7$

Hence, $a=1$ and $b=7$.