Question

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

Answer 1

Hint: In this question, we are given two polynomials of x and we have to find the values of a and b so that ${{x}^{4}}+{{x}^{3}}+8{{x}^{2}}+ax+b$ becomes divisible by ${{x}^{2}}+1$ therefore use long division method to divide one polynomial by another polynomial.

Complete step-by-step answer:

In this question, we have to find the values of a and b such that one ${{x}^{4}}+{{x}^{3}}+8{{x}^{2}}+ax+b$ leaves remainder 0 when divided by ${{x}^{2}}+1$ . Therefore, we can directly divide the polynomials using the long division method to obtain
\[{{x}^{2}}+1\overset{{{x}^{2}}+x+7}{\overline{\left){\begin{align}
  & {{x}^{4}}+{{x}^{3}}+8{{x}^{2}}+ax+b \\
 & -\left( {{x}^{4}}+{{x}^{2}} \right) \\
 & 0+{{x}^{3}}+7{{x}^{2}}+ax+b \\
 & -\left( {{x}^{3}}+x \right) \\
 & 7{{x}^{2}}+\left( a-1 \right)x+b \\
 & -\left( 7{{x}^{2}}+7 \right) \\
 & \left( a-1 \right)x+\left( b-7 \right) \\
\end{align}}\right.}}\]
Thus, by long division, the remainder is equal to \[\left( a-1 \right)x+\left( b-7 \right)\] . Now, if ${{x}^{4}}+{{x}^{3}}+8{{x}^{2}}+ax+b$ is divisible by ${{x}^{2}}+1$ , then the remainder should be equal to zero.
Therefore, we should have
\[\left( a-1 \right)x+\left( b-7 \right)=0=0x+0.................(1.1)\]
Now, we know that in an equation involving variables, the coefficient of each power of x should be separately equal to zero. Therefore, from (1.1), we obtain two separate equations as
$a-1=0\Rightarrow a=1$ and
$b-7=0\Rightarrow b=7$
Thus, we find that the values of a and b so that ${{x}^{4}}+{{x}^{3}}+8{{x}^{2}}+ax+b$ becomes divisible by ${{x}^{2}}+1$ is a=1 and b=7 which is the required answer to the given question.

Note: We should note that even though the expression has a term with fourth power of x, therefore, we should not think that we will get four values of a and b satisfying the equation. The statement that the number of solutions of an nth degree polynomial is n holds if we are trying to find x such that $f(x)=0$ where f is an nth degree polynomial. Here, we are asked to a and b and not x, so the values of a and b are unique as obtained in the solution.