Question

How do you divide $\dfrac{2{{x}^{2}}+x-16}{x-3}$ ?

Answer 1

Hint: In this question, we have to divide the given fractional number. Thus, we will apply the long-division method to get the solution. First, we will multiply the first term of the divisor by 2x to make it equal to the first term of the dividend. Then, we will multiply the second term of the divisor by 2x and place it below the second term of the dividend and change the sign of both the new terms. After, that we will make the necessary calculations and again apply the same steps as above, to get the required quotient and the remainder of the given problem.

Complete step by step solution:
According to the question, we have to divide the given fractional term.
Thus, we will use the long division method and the basic mathematical rules to get the solution.
The fractional term given to us is $\dfrac{2{{x}^{2}}+x-16}{x-3}$ -------------- (1)
Now, we know that both the numerator and the denominator of the given problem is in the form of polynomial, thus we will apply the long-division method in the algebraic fractional term, that is we get
$\Rightarrow x-3\overline{\left){2{{x}^{2}}+x-16}\right.}$
Now, we know that the divisor have the first term as x and the first term of the dividend is $2{{x}^{2}}$ , thus if we multiply the first term of the divisor with the 2x, we will get the first term of the dividend. The second term of the divisor is multiplied by the quotient which is to be placed below the second term of the dividend, and we will change the sign, thus we get
$\begin{align}
  & x-3\overset{2x}{\overline{\left){\begin{align}
  & +2{{x}^{2}}+x-16 \\
 & \underline{-2{{x}^{2}}\underset{+}{\mathop{-}}\,\text{6}x} \\
\end{align}}\right.}} \\
\end{align}$
On solving the above division, we get
$\begin{align}
  & x-3\overset{2x}{\overline{\left){\begin{align}
  & +2{{x}^{2}}+x-16 \\
 & \underline{-2{{x}^{2}}\underset{+}{\mathop{-}}\,\text{6}x} \\
 & \text{ 7}x-\text{16} \\
\end{align}}\right.}} \\
 & \text{ } \\
\end{align}$
Now, we will again apply the same method as done in the above step, we get
$\begin{align}
  & x-3\overset{2x+7}{\overline{\left){\begin{align}
  & +2{{x}^{2}}+x-16 \\
 & \underline{-2{{x}^{2}}\underset{+}{\mathop{-}}\,\text{6}x} \\
 & \text{ 7}x-\text{16} \\
 & \text{ }\underline{\underset{-}{\mathop{\text{+}}}\,7x\underset{+}{\mathop{-}}\,21} \\
\end{align}}\right.}} \\
 & \text{ } \\
\end{align}$
Thus, on further simplifying the above equation, we get
$\begin{align}
  & x-3\overset{2x+7}{\overline{\left){\begin{align}
  & +2{{x}^{2}}+x-16 \\
 & \underline{-2{{x}^{2}}\underset{+}{\mathop{-}}\,\text{6}x} \\
 & \text{ 7}x-\text{16} \\
 & \text{ }\underline{\underset{-}{\mathop{\text{+}}}\,7x\underset{+}{\mathop{-}}\,21} \\
 & \text{ }\underline{\underline{\text{ 5}}} \\
\end{align}}\right.}} \\
 & \text{ } \\
\end{align}$
Thus, we get remainder is equal to 5 and the quotient is equal to 2x+7.
Therefore, on dividing the fractional term $\dfrac{2{{x}^{2}}+x-16}{x-3}$ , we get the remainder equal to 5 and the quotient equals to 2x+7.

Note: While solving this problem, do the step-by-step calculations properly to avoid an error. You can also check your answer, using the formula $dividend=\left( divisor\times quotient \right)+remainder$ , that is simply we put the value of divisor, quotient, and the remainder in the above equation, and if it is equal to the dividend, then the solution is correct, otherwise you have made some mistake.