Question

The product of two expressions is \[\left( {{x}^{5}}+\text{ }{{x}^{3}}+\text{ }x \right)\] . If one of the two expressions is \[\left( {{x}^{2}}+\text{ }x\text{ }+\text{ }1 \right)\] ,then find the other.

Answer 1

Hint:-We will take the other expression to be (a).
As \[\left( {{x}^{2}}+\text{ }x\text{ }+\text{ }1 \right)\times ~\left( a \right)\text{ }=\left( {{x}^{5}}+\text{ }{{x}^{3}}+\text{ }x \right)\]

Complete step-by-step answer:
Therefore, to find (a), we will have to divide \[\left( {{x}^{5}}+\text{ }{{x}^{3}}+\text{ }x \right)\text{ }by\text{ }\left( {{x}^{2}}+\text{ }x\text{ }+\text{ }1 \right)\] .
So, let us now divide \[\left( {{x}^{5}}+\text{ }{{x}^{3}}+\text{ }x \right)\text{ }by\text{ }\left( {{x}^{2}}+\text{ }x\text{ }+\text{ }1 \right)\] .
\[\begin{align}
  & x_{{}}^{2}+x+1\overset{x_{{}}^{3}-x_{{}}^{2}+x}{\overline{\left){\begin{align}
  & +x_{{}}^{5}+x_{{}}^{3}+x \\
 & \underline{\begin{align}
  & +x_{{}}^{5}+x_{{}}^{4}+x_{{}}^{3} \\
 & \left( - \right)\ \left( - \right)\ \ \left( - \right) \\
\end{align}} \\
\end{align}}\right.}} \\
 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -x_{{}}^{4}+x \\
 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{\begin{align}
  & -x_{{}}^{4}-x_{{}}^{3}-x_{{}}^{2} \\
 & \left( + \right)\ \left( + \right)\ \ \left( + \right) \\
\end{align}} \\
 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +x_{{}}^{3}+x_{{}}^{2}+x \\
 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{\begin{align}
  & +x_{{}}^{3}+x_{{}}^{2}+x \\
 & \left( - \right)\ \left( - \right)\ \ \left( - \right) \\
\end{align}} \\
 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\,\ \ \ 00 \\
\end{align}\]
We can see that the quotient of this question is \[\left( {{x}^{3}}-\text{ }{{x}^{2}}+\text{ }x \right)\] .
Therefore, the other number is \[\left( {{x}^{3}}-\text{ }{{x}^{2}}+\text{ }x \right)\] .
Hence, the answer of this question is \[\left( {{x}^{3}}-\text{ }{{x}^{2}}+\text{ }x \right)\] .
Let us now verify this question.
Verification:-
\[\begin{array}{*{35}{l}}
   \left( {{x}^{3}}-\text{ }{{x}^{2}}+\text{ }x \right)\text{ }\left( {{x}^{2}}+\text{ }x\text{ }+\text{ }1 \right)\text{ }=\text{ }{{x}^{5}}+\text{ }{{x}^{3}}+\text{ }x \\
\Rightarrow {{x}^{3}}\left( {{x}^{2}}+\text{ }x\text{ }+\text{ }1 \right)-\text{ }{{x}^{2}}\left( {{x}^{2}}+\text{ }x\text{ }+\text{ }1 \right)\text{ }+\text{ }x\text{ }\left( {{x}^{2}}+\text{ }x\text{ }+\text{ }1 \right)\text{ }=\text{ }{{x}^{5}}+\text{ }{{x}^{3}}+\text{ }x \\
\Rightarrow {{x}^{5}}+\text{ }{{x}^{4}}+\text{ }{{x}^{3}}-\text{ }{{x}^{4}}-\text{ }{{x}^{3}}-\text{ }{{x}^{2}}+\text{ }{{x}^{3}}+\text{ }{{x}^{2}}+\text{ }x\text{ }=\text{ }{{x}^{5}}+\text{ }{{x}^{3}}+\text{ }x \\
\Rightarrow {{x}^{5}}+\text{ }{{x}^{3}}+\text{ }x\text{ }=\text{ }{{x}^{5}}+\text{ }{{x}^{3}}+\text{ }x \\
\end{array}\]
Hence, proved

Note:-One must do all the calculations in this question very carefully.
Also not only in this question, the students must be very careful while solving any such questions as if there is any mistake in the calculus, then the answer can come out to be wrong.