Question

How do you use polynomial long division to divide \[(4{{x}^{2}}-x-23)\div ({{x}^{2}}-1)\] and write the polynomial in the form \[p(x)=d(x)q(x)+r(x)\]?

Answer 1

Hint: The relationship between dividend, divisor, remainder, and quotient will help us in verifying the question, whether it is right or wrong. If the dividend becomes equal to the relationship between dividend, divisor, remainder, and quotient, then the solution is right otherwise it is wrong.

Complete step-by-step solution:
we know that,
\[d(x)\overset{q(x)}{\overline{\left){\begin{align}
  &p(x) \\
 & \overline{r(x)} \\
\end{align}}\right.}}\]
Where,
\[d(x)=divisor\]
\[q(x)=quotient\]
\[p(x)=dividend\]
\[r(x)=remainder\]
There is a relation between dividend, divisor, remainder, and quotient and the relationship is given as shown below. The number which is being divided by divisor is called a dividend. The result that is obtained after dividing any two numbers is known as the quotient. The number by which the other number is divided is called the divisor, and the leftover is known as. remainder.
\[p(x)=d(x)q(x)+r(x)\]
So according to the question
Here,
\[\begin{align}
  & p(x)=4{{x}^{2}}-x-23 \\
 & d(x)={{x}^{2}}-1 \\
\end{align}\]
Which is given in the question.
We will divide the dividend with the divisor step by step because when we do the long division method with polynomials then it seems quite complicated. But here we will do step by step so that it will be easier for you to obtain the answer
step 1: write \[4{{x}^{2}}-x-23\] inside the bracket which is the dividend and write \[{{x}^{2}}-1\] outside the bracket which is the divisor.
Step 2:As our divisor is a polynomial so we will divide each of its terms one by one with the dividend so that it will be easy.
First, we will multiply \[{{x}^{2}}\] with \[4\] which will give us \[4{{x}^{2}}\] . As there is no term containing \[x\], so now only \[1\] will be multiplied by \[4\]which will give us the following results.
\[{{x}^{2}}-1\overset{4}{\overline{\left){\begin{align}
  & 4{{x}^{2}}-x-23 \\
 & 4{{x}^{2}}\text{ }-4 \\
 & \overline{\text{ }-x-19} \\
\end{align}}\right.}}\]
Now according to the above solution, the following values are obtained.
\[\begin{align}
  & p(x)=4{{x}^{2}}-x-23 \\
 & q(x)=4 \\
 & r(x)=-x-19 \\
 & d(x)={{x}^{2}}-1 \\
\end{align}\]
Now we will put these values in the below-given equation which is as follows
\[p(x)=q(x)d(x)+r(x)\]
This gives us the following results
\[\begin{align}
  & 4{{x}^{2}}-x-23=4({{x}^{2}}-1)+(-x-19) \\
 & \Rightarrow 4{{x}^{2}}-x-23=4{{x}^{2}}-4-x-19 \\
 & \Rightarrow 4{{x}^{2}}-x-23=4{{x}^{2}}-x-23 \\
\end{align}\]
So LHS becomes equal to RHS. This method is also used to determine whether we have done our solution right or wrong by proving them equal on both sides.
\[\begin{align}
  & 4{{x}^{2}}-x-23=4({{x}^{2}}-1)+(-x-19) \\
 & \Rightarrow 4{{x}^{2}}-x-23=4{{x}^{2}}-4-x-19 \\
 & \Rightarrow 4{{x}^{2}}-x-23=4{{x}^{2}}-x-23 \\
\end{align}\]

Note: If one number divides another number completely then the remainder is zero but if one number does not divide another number completely then we are left over with some remainder. Remainder always comes less than the divisor but if remainder comes greater than the divisor then it means that the division is incomplete.