Question

How do you divide $\dfrac{{{x}^{3}}+7{{x}^{2}}-3x+4}{x+2}$ and identify any restrictions on the variable?

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 have been given a fraction form of polynomial as $\dfrac{{{x}^{3}}+7{{x}^{2}}-3x+4}{x+2}$ where ${{x}^{3}}+7{{x}^{2}}-3x+4$ is being divided with $x+2$.
We use the process of long division to find the quotient polynomial and the remainder for the division.
we first explain the process for the given division then complete it.
The highest power of the dividend ${{x}^{3}}+7{{x}^{2}}-3x+4$ is 3 and for divisor $x+2$ is 1.
Therefore, at first, we multiply divisor with ${{x}^{2}}$ to get ${{x}^{3}}+2{{x}^{2}}$. On subtraction from ${{x}^{3}}+7{{x}^{2}}-3x+4$ we get $5{{x}^{2}}-3x+4$. Now the coefficient of the remainder is 5. So, we multiply the divisor with $5x$ to get $5{{x}^{2}}+10x$. On subtraction from $5{{x}^{2}}-3x+4$ we get $-13x+4$. We end the division with a multiplying divisor with $-13$ to get $-13x-26$. On subtraction from $-13x+4$ we get 30.
$x+2\overset{{{x}^{2}}+5x-13}{\overline{\left){\begin{align}
  & {{x}^{3}}+7{{x}^{2}}-3x+4 \\
 & \underline{{{x}^{3}}+2{{x}^{2}}} \\
 & 5{{x}^{2}}-3x+4 \\
 & \underline{5{{x}^{2}}+10x} \\
 & -13x+4 \\
 & \underline{-13x-26} \\
 & 30 \\
\end{align}}\right.}}$
Therefore, the quotient for the division $\dfrac{{{x}^{3}}+7{{x}^{2}}-3x+4}{x+2}$ is ${{x}^{2}}+5x-13$ and the remainder is 30.
The only restriction for the division is that the divisor can never be equal to 0.
So, $x+2\ne 0$ which gives $x\ne -2$.

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.