# Decompose the following fraction: \frac{x^2+2x+34}{x^3 \left( x^3 - 8\right)\left( x^4 -...

## Question:

Decompose the following fraction:

{eq}\frac{x^2+2x+34}{x^3 \left( x^3 - 8\right)\left( x^4 - 9\right)^2\left( x^2 + 6\right)} {/eq}.

## Partial Fractions:

We'll apply the method of Partial Fraction decomposition. We start by factoring the denominator as a product of real irreducible factors, with degrees one or two. For this problem, we'll set up an identity with a large number of unknowns (coefficients) and we'll use a CAS software to find the correct values of those unknowns.

## Answer and Explanation:

We'll factorize the numerator and denominator into linear or irreducible quadratic factors over the real numbers:

{eq}\begin{align*} \frac{x^2+2x+34}{x^3 \left( x^3 - 8\right)\left( x^4 - 9\right)^2\left( x^2 + 6\right)} &= \frac{x^2+2x+34}{x^3 \left( x - 2\right)\left( x^2+2x +4\right)\left( x^2 - 3\right)^2\left( x^2 +3\right)^2\left( x^2 + 6\right)} \\ &= \frac{x^2+2x+34}{x^3 \left( x - 2\right)\left( x^2+2x +4\right)\left( x - \sqrt{3}\right)^2\left( x + \sqrt{3}\right)^2\left( x^2 +3\right)^2\left( x^2 + 6\right)} \\ \end{align*} {/eq}

Now for the decomposition into partial fractions we write:

{eq}\begin{align*} \frac{x^2+2x+34}{x^3 \left( x^3 - 8\right)\left( x^4 - 9\right)^2\left( x^2 + 6\right)} &= \frac{x^2+2x+34}{x^3 \left( x - 2\right)\left( x^2+2x +4\right)\left( x - \sqrt{3}\right)^2\left( x + \sqrt{3}\right)^2\left( x^2 +3\right)^2\left( x^2 + 6\right)} \\ &=\frac{A}{x}+\frac{B}{x^2} +\frac{C}{x^3} +\frac{D}{\left( x - 2\right)}+\frac{Ex+F}{\left( x^2+2x +4\right)}+\frac{G}{\left( x - \sqrt{3}\right)}\\ &\qquad+\frac{H}{\left( x - \sqrt{3}\right)^2}+\frac{I}{\left( x + \sqrt{3}\right)}+\frac{J}{\left( x + \sqrt{3}\right)^2}+\frac{Kx+L}{\left( x^2 +3\right)}\\ &\qquad+\frac{Mx+N}{\left( x^2 +3\right)^2}+\frac{Px+R}{\left( x^2 + 6\right)}, \end{align*} {/eq}

where the coefficients {eq}A,\,B,\ldots,R {/eq} are real numbers to be determined.

Solving for these coefficients we get:

{eq}\begin{align*} A&=\frac{7}{5832},\quad B=-\frac{1}{1944},\quad C=-\frac{17}{1944},\quad D=\frac{1}{1120},\quad E=-\frac{6505}{51824864},\quad F=\frac{1555}{6478108},\\ G&=-\frac{8479 \sqrt{3}-66436}{143712144},\quad H=-\frac{314 \sqrt{3}+381}{1294704},\quad I=\frac{8479 \sqrt{3}+66436}{143712144},\quad J=\frac{314 \sqrt{3}-381}{1294704},\\ K&=-\frac{4919}{1724814},\quad L=-\frac{200}{287469},\quad M=-\frac{19}{6318},\quad N=-\frac{11}{4212},\quad P=-\frac{37}{918540},\quad R=-\frac{19}{153090}. \end{align*} {/eq}

Thus we have:

{eq}\begin{align*} \frac{x^2+2x+34}{x^3 \left( x^3 - 8\right)\left( x^4 - 9\right)^2\left( x^2 + 6\right)}&= \frac{\frac{7}{5832 }}{ x}-\frac{\frac{1}{1944}}{ {{x}^{2}}}-\frac{\frac{17}{1944}}{ {{x}^{3}}}+\frac{\frac{1}{1120}}{ x-2 }\\ &\qquad -\frac{\frac{6505}{51824864} x-\frac{12440}{51824864}}{ {{x}^{2}}+2 x+4 }+\frac{\frac{8479 \sqrt{3}+66436}{143712144}}{ \left( x+\sqrt{3}\right) }\\ &\qquad +\frac{\frac{314 \sqrt{3}-381}{1294704}}{ {{\left( x+\sqrt{3}\right) }^{2}}}+\frac{\frac{66436-8479 \sqrt{3}}{143712144}}{ \left( x-\sqrt{3}\right) }+\frac{\frac{-314 \sqrt{3}-381}{1294704}}{ {{\left( x-\sqrt{3}\right) }^{2}}} \\ &\qquad-\frac{\frac{4919}{1724814} x+\frac{1200}{1724814}}{ \left( {{x}^{2}}+3\right) }-\frac{\frac{38}{12636} x+\frac{33}{12636}}{ {{\left( {{x}^{2}}+3\right) }^{2}}}-\frac{\frac{37}{918540} x+\frac{114}{918540}}{ \left( {{x}^{2}}+6\right) }. \end{align*} {/eq}

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from High School Algebra I: Help and Review

Chapter 3 / Lesson 25