Question

How do you divide $\dfrac{{{x}^{3}}-7x-6}{x+1}$ using polynomial long division?

Answer 1

Hint: This type of question is based on the concept of polynomial division. We can solve this question by the long division method or algebraic long method. We simply keep dividing the first term of the dividend (the polynomial to be divided) by the first term of the divisor. This gives us the first term of the quotient. We repeat this until no further division is possible.

Complete step by step solution:
In the given question, we have ${{x}^{3}}-7x-6$ as the dividend and $x+1$ as the divisor. We first arrange the indices in descending order and simplify the given equation.
$\Rightarrow \dfrac{{{x}^{3}}-7x-6}{x+1}$
Now we divide the ${{x}^{3}}$ term in the numerator by the $x$ term in the denominator to get the first term of the quotient. By doing that, we get the first term of the quotient as,
$\Rightarrow \dfrac{{{x}^{3}}}{x}={{x}^{2}}$
Let us start the long division method now.
$\Rightarrow x+1\overset{{{x}^{2}}}{\overline{\left){{{x}^{3}}-7x-6}\right.}}$
Now we multiply the divisor $(x+1)$ by ${{x}^{2}}$ and subtract the terms and we get,
$\Rightarrow x+1\overset{{{x}^{2}}}{\overline{\left){\begin{align}
  & {{x}^{3}}-7x-6 \\
 & {{x}^{3}}+{{x}^{2}} \\
 & \overline{\text{ }-{{x}^{2}}-7x} \\
\end{align}}\right.}}$
Now we divide the $-{{x}^{2}}$ term by the $x$ term to get the second term of the quotient which is
$\Rightarrow \dfrac{-{{x}^{2}}}{x}=-x$
Multiplying this with the divisor and subtracting the terms,
$\Rightarrow x+1\overset{{{x}^{2}}-x}{\overline{\left){\begin{align}
  & {{x}^{3}}-7x-6 \\
 & {{x}^{3}}+{{x}^{2}} \\
 & \overline{\text{ }-{{x}^{2}}-7x} \\
 & \text{ }-{{x}^{2}}-x \\
 & \overline{\text{ }-6x-6} \\
\end{align}}\right.}}$
Now we again divide $-6x$ term by $x$ to get the third term of the quotient,
$\Rightarrow \dfrac{-6x}{x}=-6$
Multiplying this with the divisor and subtracting again,
$\Rightarrow x+1\overset{{{x}^{2}}-x-6}{\overline{\left){\begin{align}
  & {{x}^{3}}-7x-6 \\
 & {{x}^{3}}+{{x}^{2}} \\
 & \overline{\text{ }-{{x}^{2}}-7x} \\
 & \text{ }-{{x}^{2}}-x \\
 & \overline{\text{ }-6x-6} \\
 & \text{ }-6x-6 \\
 & \overline{\text{ 0}} \\
\end{align}}\right.}}$
This remainder comes down to zero and we are left with no remainder. Therefore, the quotient is ${{x}^{2}}-x-6$ and the remainder is 0.
Hence the final answer obtained by dividing ${{x}^{3}}-7x-6$ by $x+1$, we get the quotient as ${{x}^{2}}-x-6$ with a remainder of 0.

Note: While solving this question, the students need to be careful while subtracting the terms after multiplying with the divisor since any confusion in the sign during subtraction can lead to a wrong answer. We can also solve this type of question by factorizing it but that is a more complicated way to get the solution.