Repeated factors

​​In a nutshell

A single fraction with two distinct linear factors in its denominator can be split into two separate fractions with linear denominators. This is what is known as splitting fraction into partial fractions. When the denominator contains repeated linear factors, you can split this into two or more separate functions.

Repeated factors

​​Procedure

Example 1

Split the following fraction into partial fractions $$\cfrac{10x^2 - 10x +17}{(2x+1)(x^2-6x + 9)}$$

First, factorise the denominator.

$$\cfrac{10x^2 - 10x +17}{(2x+1)(x-3)^2}$$​​

Now set $$A,\ B$$ and $$C$$ as numerators of your partial fractions.

$$\cfrac{10x^2 - 10x + 17}{(2x+1)(x-3)^2}= \cfrac{A}{2x+1} + \cfrac{B}{x-3}+\cfrac{C}{(x-3)^2}$$​​

Operate as usual, e.g. add the fractions (in order to do so, you need to find the common denominator)

$$\begin{aligned}\cfrac{10x^2 - 10x + 17}{(2x+1)(x-3)^2}&= \cfrac{A}{2x+1} + \cfrac{B}{x-3}+\cfrac{C}{(x-3)^2}\\\\&=\cfrac{A(x-3)^2}{(2x+1)(x-3)^2} + \cfrac{B(2x+1)(x-3)}{(2x+1)(x-3)^2} + \cfrac{C(2x+1)}{(2x+1)(x-3)^2}\\\\&=\cfrac{A(x-3)^2 + B(2x+1)(x-3)+ C(2x+1)}{(2x+1)(x-3)^2}\end{aligned}$$​​

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The numerators are equal, thus it is obtained

$$10 x^2 - 10 x + 17 = A(x-3)^2 + B(2x+1)(x-3) + C(2x +1)$$​​

Substitute $$x=3:$$​

​$$\begin{aligned}10 \times 3^2 - 10\times 3 + 17 &= A(3-3)^2 + B(2\times 3+1)(3-3) + C(2 \times 3 +1)\\90 - 30 + 17 &= 7C \end{aligned}$$

​​​

​$$\underline{C = 11}$$​​

​

Substitute $$x=-\cfrac{1}{2}:$$​

​$$\begin{aligned}10 \left(-\cfrac{1}{2}\right)^2 - 10 \left(-\cfrac{1}{2}\right) + 17 &= A\left( \left(-\cfrac{1}{2}\right)-3\right) + B\left(2 \left(-\cfrac{1}{2}\right)+1\right)\left( \left(-\cfrac{1}{2}\right)-3\right) + C\left(2 \left(-\cfrac{1}{2}\right) +1\right)\end{aligned}$$​​

​$$\begin{aligned}\cfrac{10}{4} + 5 + 17 &= -A\cfrac{49}{4}\\\\\cfrac{49}{2}& = -\cfrac{49A}{4}\end{aligned}$$​​

​​​

​$$\underline{A = -2}$$​​

Once the constants $$A$$ and $$C$$ are known, calculate B as it follows

$$\begin{aligned} 10 x^2 - 10 x + 17 &= 2(x-3)^2 + B(2x+1)(x-3) + 11(2x +1) \\10 x^2 - 10 x + 17 &= 2x^2 - 12x + 18 + 2Bx^2 -5Bx -3B + 22x +11 \end{aligned}$$​​

​​​

Equate $$x^2$$ terms, obtaining

​$$10x^2= 2x^2 + 2 Bx^2$$

​$$\underline{B=4}$$​​

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In conclusion:

$$\underline{\cfrac{10x^2 - 10x + 17}{(2x+1)(x-3)^2}= \cfrac{2}{2x+1} + \cfrac{4}{x-3}+\cfrac{11}{(x-3)^2}}$$​​

Example 2

Split the following fraction into partial fractions $$\cfrac{2x-39}{x^2(x+3)}$$​

​​

As the denominator is already factorised, set $$A, \ B, \ C\dots$$ as numerators of your partial fractions.

​$$\cfrac{2x-39}{x^2(x+3)} = \cfrac{A}{x}+ \cfrac{B}{x^2 } + \cfrac{C}{x+3}$$​​

Now find the common denominator and add the fractions. 

​$$\begin{aligned}\cfrac{2x-39}{x^2(x+3)}& = \cfrac{A}{x}+ \cfrac{B}{x^2 } + \cfrac{C}{x+3}\\\hspace{4mm}\\& = \cfrac{Ax(x+3)}{x^2(x+3)}+ \cfrac{B (x+3) }{x^2(x+3)} + \cfrac{Cx^2}{x^2(x+3)}\\\hspace{4mm}\\&=\cfrac{Ax(x+3) + B(x+3) + Cx^2}{x^2 (x+3)}\end{aligned}$$​​

The numerators are equal, thus it is obtained: 

$$\begin{aligned}2x - 39 &= Ax(x+3) + B(x+3) + Cx^2=\\ &= Ax^2 +3Ax +Bx + 3B + Cx^2\end{aligned}$$​​

Using the equating coefficients methods, you have: 

$$\begin{cases}\begin{aligned}0x^2 & = x^2(A+C) \\2x &= x(3A+B)\\-39 &= 3 B\end{aligned}\end{cases}$$​​

From the last equation you find that $$\underline{B=-13}$$. Therefore, in the second equation you have:

​$$2 = (3A+B) \implies \underline{A=5}$$​​

And in the first one:

$$0 = A+C \implies \underline{C=-5}$$​​

In conclusion:

​$$\underline{\cfrac{2x-39}{x^2(x+3)} = \cfrac{5}{x}- \cfrac{13}{x^2 } - \cfrac{5}{x+3}}$$​​