Question

How do you find the quotient of ${{x}^{3}}+6{{x}^{2}}-x-30$ divided by ${{x}^{2}}+8x+15$ ?

Answer 1

We have to divide a 3-degree polynomial equation with a 2-degree polynomial equation. From the basic knowledge of polynomial division, It is clear that the quotient will be a one-degree polynomial equation, that is, a linear equation in x-variable. Thus, we shall solve the problem by algebraic long division method.

Complete step-by-step solution:
${{x}^{3}}+6{{x}^{2}}-x-30$ divided by ${{x}^{2}}+8x+15$ can also be written as $\dfrac{{{x}^{3}}+6{{x}^{2}}-x-30}{{{x}^{2}}+8x+15}$.
The easiest method to perform division of higher order polynomials, that is, polynomials of degree three or more is the algebraic long division method.
This method has a special form of representation. The dividend, divisor and quotient are placed around the division box as follows:
$divisor\overset{quotient}{\mathop{\left| \\!{\overline {\, dividend \,}} \right. }}\,$
We start developing the quotient term-by-term. We find a term whose multiplication with the first of the divisor is equal to the first term of the dividend. Further we multiply all the terms of the divisor with the chosen term and place them below their respective degree term present in the dividend. This is one step. We keep simplifying and repeating this step until we obtain the remainder equal to zero or equal to a constant number. Another point to remember is that we have to subtract the terms coming in each step.

\begin{aligned} & {{x}^{3}}+6{{x}^{2}}-x-30 \\\ & {{x}^{3}}+8{{x}^{2}}+15x \\\ \end{aligned} \,}} \right. }}\,$$ Now we shall subtract these terms and get: $$\begin{aligned} & {{x}^{2}}+8x+15\overset{x-2}{\mathop{\left| \\!{\overline {\, \begin{aligned} & {{x}^{3}}+6{{x}^{2}}-x-30 \\\ & {{x}^{3}}+8{{x}^{2}}+15x+0 \\\ \end{aligned} \,}} \right. }}\, \\\ & \text{ }\overline{\text{ }-2{{x}^{2}}-16x-30} \\\ & \text{ }-2{{x}^{2}}-16x-30 \\\ & \text{ }\overline{0+0+0} \\\ \end{aligned}$$ We can stop now since we have obtained zero here. **Therefore, the quotient of ${{x}^{3}}+6{{x}^{2}}-x-30$ divided by ${{x}^{2}}+8x+15$ is $x-2$.** **Note:** We shall also note that the remainder of the division performed above is zero. Another method of dividing ${{x}^{3}}+6{{x}^{2}}-x-30$ by ${{x}^{2}}+8x+15$ is by factoring method. We can break up the 2-degree term into two such terms as ${{x}^{3}}+8{{x}^{2}}+15x-2{{x}^{2}}-16x-30$. We can now group the terms further into $\left( {{x}^{2}}+8x+15 \right)\left( x-2 \right)$. From here, we can see the two factors and obtain our quotient.