Question

How do you divide $\dfrac{-{{x}^{3}}+13{{x}^{2}}-x-5}{x-4}$?

Answer 1

Hint: If we can see that if any fraction form for any polynomials is not fully divisible, then we will get quotient polynomial and the remainder for the long division. We express the details for the long division and complete the division to get the solution.

Complete step by step answer:
We use the regular long division process to find the division of $\dfrac{-{{x}^{3}}+13{{x}^{2}}-x-5}{x-4}$.
We first complete the division and then find the quotient and the remainder.
We have been given a fraction form of polynomial as $\dfrac{-{{x}^{3}}+13{{x}^{2}}-x-5}{x-4}$ where $-{{x}^{3}}+13{{x}^{2}}-x-5$ is being divided with $x-4$.
We first explain the process for the given division then compete it.
The highest power of the dividend $-{{x}^{3}}+13{{x}^{2}}-x-5$ is 3 and for divisor $x-4$ is 1.
Therefore, at first, we multiply divisor with $-{{x}^{2}}$ to get $-{{x}^{3}}+4{{x}^{2}}$. On subtraction from $-{{x}^{3}}+13{{x}^{2}}-x-5$ we get $9{{x}^{2}}-x-5$. Now the coefficient of the remainder is 9. So, we multiply the divisor with $9x$ to get $9{{x}^{2}}-36x$. On subtraction from $9{{x}^{2}}-x-5$ we get $35x-5$. We end the division with a multiplying divisor with 35 to get $35x-140$. On subtraction from $35x-5$ we get 135.
$x-4\overset{-{{x}^{2}}+9x+35}{\overline{\left){\begin{align}
  & -{{x}^{3}}+13{{x}^{2}}-x-5 \\
 & \underline{-{{x}^{3}}+4{{x}^{2}}} \\
 & 9{{x}^{2}}-x-5 \\
 & \underline{9{{x}^{2}}-36x} \\
 & 35x-5 \\
 & \underline{35x-140} \\
 & 135 \\
\end{align}}\right.}}$

Therefore, the quotient for the division $\dfrac{-{{x}^{3}}+13{{x}^{2}}-x-5}{x-4}$ is $-{{x}^{2}}+9x+35$ and the remainder is 135.

Note: We have to remember that the highest power of the divisor is always less than or equal to the highest power of dividend. The same rule is applied for the quotient too. The sum of the quotients will be equal to the dividends.