# Briefly explain why we cannot immediately apply the partial fraction decomposition technique for...

## Question:

Briefly explain why we cannot immediately apply the partial fraction decomposition technique for the following fraction. Then put the fraction into a form where we can apply the partial fraction decomposition technique. Do not find the decomposition.

{eq}\frac{6x^2 - 10x + 7}{2x^2 + 5x} {/eq}

## Partial Fractions

The integral of a proper rational function is obtained by decomposing the function into partial fractions.

{eq}\displaystyle \frac {p(x)}{(x-a)^n(x^2+b^2)^m}=\frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+...+\frac{A_n}{(x-a)^n}+\frac{B_1x+C_1}{x^2+b^2}+\frac{B_2x+C_2}{(x^2+b^2)^2}+...+\frac{B_mx+C_m}{(x^2+b^2)^m} {/eq},

where {eq}\displaystyle \text{degree}(p(x))<2m+n, \text{ and } A_i, (i=1,..,n), B_j, C_j, (j=1,...,m) \text{ are constants to be determined}. {/eq}

A rational function is proper if the degree of the numerator is strictly less than the degree of the denominator.

If the fraction is not proper, we will do the long division polynomials to obtain the proper form.

## Answer and Explanation:

We cannot apply the partial fraction decomposition to the integral {eq}\displaystyle \frac{6x^2 - 10x + 7}{2x^2 + 5x} {/eq} in the current form,

{eq}\displaystyle \boxed{\text{ because the rational function is not a proper function}} {/eq}.

The integrand function {eq}\displaystyle \frac{6x^2 - 10x + 7}{2x^2 + 5x} {/eq} is improper, since the degree of the numerator is greater not less than the degree of the denominator.

To be able to apply the partial fraction decomposition, we will do the division of the two polynomials

{eq}\displaystyle \begin{array}{c|ccc} &3 & &\\ \hline 2x^2 +5x & 6x^2 & - 10x & +7\\ \hline & 6x^2 & +15x & \\ \hline & 0 & -25x & +7 \end{array} {/eq}

Therefore, the quotient is {eq}\displaystyle 3, {/eq} and the remainder is {eq}\displaystyle -25x+7 {/eq} and the rational function is written now in the proper form as

{eq}\displaystyle \frac{6x^2 - 10x + 7}{2x^2 + 5x}=3+\frac{-25x+7}{2x^2 + 5x}. {/eq}

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from High School Algebra I: Help and Review

Chapter 3 / Lesson 26