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Explainer Video

Tutor: Labib

Summary

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

1.
Factorise the denominator, if required.
2.
Set A, B, C…A, \ B, \ C \dotsA, B, C… as numerators of your partial fractions.
3.
Set the denominator of one of the fractions as the repeated single factor on its own. Set the denominator of another fraction as the factor repeated as it was in the original fraction.
4.
Add the fractions.
5.
Using substitution or by comparing coefficients, find the values of A,B,C...A, B, C...A,B,C...​​


Example 1

Split the following fraction into partial fractions 10x2−10x+17(2x+1)(x2−6x+9)\cfrac{10x^2 - 10x +17}{(2x+1)(x^2-6x + 9)}(2x+1)(x2−6x+9)10x2−10x+17​


First, factorise the denominator.

10x2−10x+17(2x+1)(x−3)2\cfrac{10x^2 - 10x +17}{(2x+1)(x-3)^2}(2x+1)(x−3)210x2−10x+17​​​


Now set A, BA,\ BA, B and CCC as numerators of your partial fractions.

10x2−10x+17(2x+1)(x−3)2=A2x+1+Bx−3+C(x−3)2\cfrac{10x^2 - 10x + 17}{(2x+1)(x-3)^2}= \cfrac{A}{2x+1} + \cfrac{B}{x-3}+\cfrac{C}{(x-3)^2}(2x+1)(x−3)210x2−10x+17​=2x+1A​+x−3B​+(x−3)2C​​​


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

10x2−10x+17(2x+1)(x−3)2=A2x+1+Bx−3+C(x−3)2=A(x−3)2(2x+1)(x−3)2+B(2x+1)(x−3)(2x+1)(x−3)2+C(2x+1)(2x+1)(x−3)2=A(x−3)2+B(2x+1)(x−3)+C(2x+1)(2x+1)(x−3)2\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}(2x+1)(x−3)210x2−10x+17​​=2x+1A​+x−3B​+(x−3)2C​=(2x+1)(x−3)2A(x−3)2​+(2x+1)(x−3)2B(2x+1)(x−3)​+(2x+1)(x−3)2C(2x+1)​=(2x+1)(x−3)2A(x−3)2+B(2x+1)(x−3)+C(2x+1)​​​​

​

The numerators are equal, thus it is obtained

10x2−10x+17=A(x−3)2+B(2x+1)(x−3)+C(2x+1)10 x^2 - 10 x + 17 = A(x-3)^2 + B(2x+1)(x-3) + C(2x +1)10x2−10x+17=A(x−3)2+B(2x+1)(x−3)+C(2x+1)​​


Substitute x=3:x=3:x=3:​

​10×32−10×3+17=A(3−3)2+B(2×3+1)(3−3)+C(2×3+1)90−30+17=7C\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}10×32−10×3+1790−30+17​=A(3−3)2+B(2×3+1)(3−3)+C(2×3+1)=7C​

​​​

​C=11‾\underline{C = 11}C=11​​​

​

Substitute x=−12:x=-\cfrac{1}{2}:x=−21​:​

​10(−12)2−10(−12)+17=A((−12)−3)+B(2(−12)+1)((−12)−3)+C(2(−12)+1)\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}10(−21​)2−10(−21​)+17​=A((−21​)−3)+B(2(−21​)+1)((−21​)−3)+C(2(−21​)+1)​​​


​104+5+17=−A494492=−49A4\begin{aligned}\cfrac{10}{4} + 5 + 17 &= -A\cfrac{49}{4}\\\\\cfrac{49}{2}& = -\cfrac{49A}{4}\end{aligned}410​+5+17249​​=−A449​=−449A​​​​

​​​

​A=−2‾\underline{A = -2}A=−2​​​


Once the constants AAA and CCC are known, calculate B as it follows

10x2−10x+17=2(x−3)2+B(2x+1)(x−3)+11(2x+1)10x2−10x+17=2x2−12x+18+2Bx2−5Bx−3B+22x+11\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}10x2−10x+1710x2−10x+17​=2(x−3)2+B(2x+1)(x−3)+11(2x+1)=2x2−12x+18+2Bx2−5Bx−3B+22x+11​​​

​​​

Equate x2x^2x2 terms, obtaining

​10x2=2x2+2Bx210x^2= 2x^2 + 2 Bx^210x2=2x2+2Bx2

​B=4‾\underline{B=4}B=4​​​

​

In conclusion:

10x2−10x+17(2x+1)(x−3)2=22x+1+4x−3+11(x−3)2‾\underline{\cfrac{10x^2 - 10x + 17}{(2x+1)(x-3)^2}= \cfrac{2}{2x+1} + \cfrac{4}{x-3}+\cfrac{11}{(x-3)^2}}(2x+1)(x−3)210x2−10x+17​=2x+12​+x−34​+(x−3)211​​​​



Example 2

Split the following fraction into partial fractions 2x−39x2(x+3)\cfrac{2x-39}{x^2(x+3)}x2(x+3)2x−39​​

​​

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

​2x−39x2(x+3)=Ax+Bx2+Cx+3\cfrac{2x-39}{x^2(x+3)} = \cfrac{A}{x}+ \cfrac{B}{x^2 } + \cfrac{C}{x+3}x2(x+3)2x−39​=xA​+x2B​+x+3C​​​


Now find the common denominator and add the fractions. 

​2x−39x2(x+3)=Ax+Bx2+Cx+3=Ax(x+3)x2(x+3)+B(x+3)x2(x+3)+Cx2x2(x+3)=Ax(x+3)+B(x+3)+Cx2x2(x+3)\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}x2(x+3)2x−39​​=xA​+x2B​+x+3C​=x2(x+3)Ax(x+3)​+x2(x+3)B(x+3)​+x2(x+3)Cx2​=x2(x+3)Ax(x+3)+B(x+3)+Cx2​​​​


The numerators are equal, thus it is obtained: 

2x−39=Ax(x+3)+B(x+3)+Cx2==Ax2+3Ax+Bx+3B+Cx2\begin{aligned}2x - 39 &= Ax(x+3) + B(x+3) + Cx^2=\\ &= Ax^2 +3Ax +Bx + 3B + Cx^2\end{aligned}2x−39​=Ax(x+3)+B(x+3)+Cx2==Ax2+3Ax+Bx+3B+Cx2​​​


Using the equating coefficients methods, you have: 

{0x2=x2(A+C)2x=x(3A+B)−39=3B\begin{cases}\begin{aligned}0x^2 & = x^2(A+C) \\2x &= x(3A+B)\\-39 &= 3 B\end{aligned}\end{cases}⎩⎨⎧​0x22x−39​=x2(A+C)=x(3A+B)=3B​​​​


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

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


And in the first one:

0=A+C ⟹ C=−5‾0 = A+C \implies \underline{C=-5}0=A+C⟹C=−5​​​


In conclusion:

​2x−39x2(x+3)=5x−13x2−5x+3‾\underline{\cfrac{2x-39}{x^2(x+3)} = \cfrac{5}{x}- \cfrac{13}{x^2 } - \cfrac{5}{x+3}}x2(x+3)2x−39​=x5​−x213​−x+35​​​​



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FAQs - Frequently Asked Questions

What is a repeated factor in a polynomial?

A repeated factor occurs when the polynomial can be expressed as linear factors such that a factor raised to an integer power is also a factor.

What is the first step when working with partial fractions with repeated denominators?

Factorising the denominator.

How do you calculate partial fractions with repeated factors?

Firstly, factorise the denominator. Next, set constants as the numerators of your partial fractions, such that one fraction's denominator is the repeated factor and another's is a single instance of the repeated factor. Then, complete your working as usual. Finally, in order to find the values of the constants, substitute x for the value of the root of the factor.

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