Question

What must be added to the polynomial $p\left( x \right)={{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1$ so that the resulting polynomial is exactly divisible by ${{x}^{2}}+2x-3$.

Answer 1

Hint: We will divide the polynomial $p\left( x \right)={{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1$ with ${{x}^{2}}+2x-3$ and then we will find the remainder. We will use the concept of division that \[divisor=dividend\times quotient+remainder\]. The remainder of the division will be our required polynomial.

Complete step-by-step answer:
We have been asked in the question that what must be added to the polynomial $p\left( x \right)={{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1$ so that the resulting polynomial is exactly divisible by ${{x}^{2}}+2x-3$.
We know that in any division, \[divisor=dividend\times quotient+remainder\], so from this we can write, \[divisor-remainder=dividend\times quotient.........\left( i \right)\]
Here, the divisor is $p\left( x \right)={{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1$ and the dividend is ${{x}^{2}}+2x-3$.
We will find the remainder when we divide the polynomial $p\left( x \right)={{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1$ with ${{x}^{2}}+2x-3$ and then this remainder will be our required polynomial.
${{x}^{2}}+2x-3\overset{{{x}^{2}}+1}{\overline{\left){\begin{align}
  & {{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1 \\
 & \underline{{{x}^{4}}+2{{x}^{3}}-3{{x}^{2}}} \\
 & 0+0+{{x}^{2}}+x-1 \\
 & \underline{0+0+{{x}^{2}}+2x-3} \\
 & -x+2 \\
\end{align}}\right.}}$
So, we get the remainder as (-x + 2).
Therefore, we should subtract (-x + 2) from the divisor, $p\left( x \right)={{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1$ to be divisible by ${{x}^{2}}+2x-3$ as per equation (i). Or we can also say that we have to add (x-2) to $p\left( x \right)={{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+x-1$, to be divisible by ${{x}^{2}}+2x-3$ as per equation (i). So, we will get, $p\left( x \right)={{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+2x-3$. And this polynomial is exactly divisible by ${{x}^{2}}+2x-3$.
${{x}^{2}}+2x-3\overset{{{x}^{2}}+1}{\overline{\left){\begin{align}
  & {{x}^{4}}+2{{x}^{3}}-2{{x}^{2}}+2x-3 \\
 & \underline{{{x}^{4}}+2{{x}^{3}}-3{{x}^{2}}} \\
 & 0+0+{{x}^{2}}+2x-3 \\
 & \underline{0+0+{{x}^{2}}+2x-3} \\
 & 0 \\
\end{align}}\right.}}$
Therefore, we have got the required polynomial as (x-2).

Note: Most of the students make mistakes while taking the sign in the long division and make silly calculation mistakes. They may get the remainder as (3x -4), which is not correct. Therefore it is strongly recommended that the students do the calculations step by step and very carefully in order to avoid making any small mistakes.