Question

Divide the given polynomial function $p\left( x \right)={{x}^{4}}-3{{x}^{2}}+4x+5$ by a polynomial function $g\left( x \right)={{x}^{2}}+1-x$ .
a) $Q={{x}^{2}}+x-5,R=0$
b) $Q={{x}^{2}}+x-3,R=8$
c) $Q={{x}^{2}}+x-3,R=0$
d) $Q={{x}^{2}}+x-5,R=8$

Answer 1

Hint: Here, in this question, we can use the fact that the dividend is equal to the product of the divisor and the quotient when added with the remainder, that is,
$p\left( x \right)=g\left( x \right)\times q\left( x \right)+r\left( x \right)$ where p$\left( x \right)$, g$\left( x \right)$, q$\left( x \right)$ and r$\left( x \right)$ are the dividend, divisor, quotient and remainder respectively and do the long division method to get the quotient and remainder.

Complete step-by-step solution -
In this given question, we are asked to divide $p\left( x \right)={{x}^{4}}-3{{x}^{2}}+4x+5$ by $g\left( x \right)={{x}^{2}}+1-x$ and find the quotient and the remainder.
We are going to use the fact that the dividend is equal to the product of the divisor and the quotient when added with the remainder, that is,
$p\left( x \right)=g\left( x \right)\times q\left( x \right)+r\left( x \right)$ where p$\left( x \right)$, g$\left( x \right)$, q$\left( x \right)$ and r$\left( x \right)$ are the dividend, divisor, quotient and remainder respectively.
Here, we are, going to perform the long division method to divide $p\left( x \right)={{x}^{4}}-3{{x}^{2}}+4x+5$ by $g\left( x \right)={{x}^{2}}+1-x$ to find our answers as the quotient and remainder. In the long division method, the first term of the quotient is chosen such that after multiplying with it the divisor (the polynomial by which the other polynomial is being divided) and then subtracting from the dividend (the polynomial to be divided), the highest power of x from the dividend gets cancelled out. We continue in this way with the resulting remainder as the new dividend until the power of the remainder becomes less than the power of the divisor.
\[{{x}^{2}}+1-x\overset{{{x}^{2}}+x-3}{\overline{\left){\begin{align}
  & {{x}^{4}}-3{{x}^{2}}+4x+5 \\
 & -\left( {{x}^{4}}-{{x}^{3}}+{{x}^{2}} \right) \\
 & {\overline {{{x}^{3}}-4{{x}^{2}}+4x+5}} \\
 & -\left( {{x}^{3}}-{{x}^{2}}+x \right) \\
 &{\overline {-3{{x}^{2}}+3x+5} }\\
 & {\underline{-\left( -3{{x}^{2}}+3x-3 \right)}} \\
 & 8 \\
\end{align}}\right.}}\]
From the above, we get the quotient as \[{{x}^{2}}+x-3\] and the remainder as 8.
So, \[{{x}^{4}}-3{{x}^{2}}+4x+5=\left( {{x}^{2}}+1-x \right)\times \left( {{x}^{2}}+x-3 \right)+8\].
Therefore, the correct answer to this question is option (b)$Q={{x}^{2}}+x-3,R=8$.

Note: In this sort of question, we must be very careful while doing the long division process about changing the signs of the terms in the quotient. The divisor should be multiplied by the quotient and the resulting whole polynomial should be subtracted from the dividend. If even a single sign is changed in the wrong way, the whole answer could be different and wrong.