# PARTIAL FRACTIONS WITH REPEATED LINEAR FACTORS

As the name suggests, a partial fraction with repeated linear factors is one with repeated factors.

For example,
$\frac{5}{(x+1)^2}$ is a repeated linear fraction as it contains a repeated linear fraction.

To express this partial fraction, It's
$\frac{5}{(x+1)^2}=\frac{A}{x+1}+\frac{B}{(x+1)^2}$

NOT
$\require{cancel}\bcancel{\frac{5}{(x+1)^2}=\frac{A}{x+1}+\frac{B}{x+1}}$

👉This post extends our post on partial fraction👈

Example 1
Express $\frac{3y^2-17y+4}{y^2(y+4)}$ in partial fraction

Solution:
This has a repeated linear factors, hence,
$\frac{3y^2-17y+4}{y^2(y+4)}=\frac{A}{y}+\frac{B}{y^2}+\frac{C}{y+4}$

Adding the right hand side of the equation
$\frac{3y^2+17y+4}{y^2(y+4)}=\frac{A(y(y+4))+B(y+4)+C(y^2)}{y^2(y+4)}$

Equating the numerators
$3y^2+17y+4=A(y(y+4)+B(y+4)+C(y^2))$

Always take note of the Eqn that you derive after eliminating the fraction.

Next, let's expand the bracket and collect like term.
$3y^2+17y+4=Ay^2+4Ay+By+4B+Cy^2$
$3y^2+17y+4=Ay^2+Cy^2+4Ay+By+4B$
$(3)y^2+(17)y+4=(A+C)y^2+(4A+B)y+4B$

Comparing the coefficients
$3=A+C$
$17=4A+B$
$4=4B$

Let's find $B$
$4=4B$
$\frac{4}{4}=\frac{4B}{4}$
$1=B$
$B=1$

Substituting $1$ for $B$ in eqn 2
$17=4A+1$
$17-1=4A$
$16=4A$
$4A=16$
$\frac{4A}{4}=\frac{16}{4}$
$A=4$

Substituting $4$ for $A$ in eqn 1
$3=4+C$
$3-4=C$
$-1=C$
$C=-1$

Thus, the partial fraction decomposition is
$\frac{3y^2-17y+4}{y^2(y+4)}=\frac{4}{y}+\frac{1}{y^2}+\frac{-1}{y+4}$

Example 2
Write $\frac{9x^2}{(2x+1)(x-1)^2}$ in partial fraction

Solution:
$\frac{9x^2}{(2x+1)(x-1)^2}=\frac{A}{2x+1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}$

$\frac{9x^2}{(2x+1)(x-1)^2}=\frac{A(x-1)^2+B(x-1)(2x+1)+C(2x+1)}{(2x+1)(x+2)^2}$

Equating the numerators
$9x^2=A(x-1)^2+B(x-1)(2x+1)+C(2x+1)$

To obtain the value of $C$, we would eliminate $A$ and $B$. Hence, let $x$ be 1

$9(1)^2=A(1-1)^2+B(1-1)(2(1)+1)+C(2(1)+1)$
$9=C(3)$
$3C=9$
$\frac{3C}{3}=\frac{9}{3}$
$C=3$

To obtain $A$, we would need to eliminate $B$ and $C$. Therefore, let $x$ be $\frac{-1}{2}$
$9(\frac{-1}{2})^2=A(\frac{-1}{2}-1)^2+B(\frac{-1}{2}-1)(2(\frac{-1}{2}+1)+C(2(\frac{-1}{2}+1)$
$9(\frac{1}{4})=A(\frac{-1-2}{2})^2$
$\frac{9}{4}=A(\frac{9}{4})$
$\frac{9}{4}=\frac{9A}{4}$

Cross multiply
$9\times4=4\times9A$
$36=36A$
$36A=36$
$\frac{36A}{36}=\frac{36}{36}$
$A=1$

To obtain $B$, we need to eliminate $A$ and $C$. However, No number can successfully eliminate $A$ and $C$. Hence, let $x$ be zero

$0=A(1)+B(-1)(1)+C(1)$
$0=A-B+C$

We know that $A=1$ and $C=3$. Hence
$0=1-B+3$
$B=1+3$
$B=4$

Therefore the partial fraction is:
$\frac{1}{2x+1}+\frac{4}{x-1}+\frac{3}{(x-1)^2}$

Example 3
Write $\frac{y+4}{y^3-3y^2+4}$ in partial fraction.

Solution:
First, let's factorize the denominator
$\frac{y+4}{(y+1)(y-2)(y-2)}$

$\frac{y+4}{(y+1)(y-2)^2}=\frac{A}{y+1}+\frac{B}{y-2}+\frac{C}{(y-2)^2}$
$\frac{y+4}{(y+1)(y-2)^2}=\frac{A(y-2)^2+B(y-2)(y-1)+C(y+1)}{(y+1)(y-2)^2}$

Equating the numerator.
$y+4=A(y-2)^2+B(y-2)(y-1)+C(y+1)$

To obtain the value of $C$, let's eliminate $A$ and $B$. So, let $y=2$
$2+4=A(2-2)^2+B(2-2)(2-1)+C(2+1)$
$6=3C$
$\frac{3C}{3}=\frac{6}{3}$
$C=2$

To obtain $A$, Let $y$ be equal to $-1$
$-1+4=A(-1-2)^2+B(-1-2)(-1+1)+C(-1+1)$
$3=A(-3)^2$
$9A=3$
$\frac{9A}{9}=\frac{3}{9}$
$A=\frac{1}{3}$

Nothing can successfully eliminate $A$ and $C$. Hence, we say let $y$ be equal to $0$
$0+4=A(0-2)^2+B(0-2)(0-1)+C(0+1)$
$4=A(-2)^2+B(-2)(-1)+C(1)$
$4=4A-2B+C$

We knows that $A=\frac{1}{3}$ and $C=2$, hence
$4=A(\frac{1}{3})-2B+2$
$4=\frac{A}{3}-2B+2$

Let multiply through by $3$ so that the fraction is cleared
$4\times3=\frac{4}{3}\times3-2B\times3+2\times3$
$12=\frac{4}{\require{cancel}\bcancel{3}}\times\require{cancel}\bcancel{3}-6B+6$
$12=4-6B+6$
$12-4-6=-6B$
$2=-6B$
$\frac{-6B}{-6}=\frac{2}{-6}$
$B=\frac{-1}{3}$

Therefore the partial decomposition is:
$\frac{y+4}{(y+1)(y-2)^2}=\frac{1}{3(y+1)}+\frac{-1}{3(y-2)}+\frac{2}{(y-2)^2}$

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Example 4
Express $\frac{6y-11}{y^2-2y+1}$ in partial fraction.

Solution:
First, let's factorize the denominator
$\frac{6y-11}{(y-1)^2}$
$\frac{6y-11}{y-1}{y-1}=\frac{A}{y-1}+\frac{B}{(y-1)^2}$
$\frac{6y-11}{(y-1)^2}=\frac{A(y-1)+B(1)}{(y-1)^2}$

Equating the numerator.
$6y-11=A(y-1)+B(1)$

Let $y=1$
$6(1)-11=A(1-1)+B$
$6-11=B$
$-5=B$
$B=-5$

Let $x=0$
$6(0)-11=A(0-1)+B$
$-11=-A+B$

If $B=-5$, then
$-11=-A-5$
$A=-5+11$
$A=6$

Therefore, the partial fraction decomposition is:
$\frac{6y-11}{y-1}{y-1}=\frac{6}{y-1}+\frac{-5}{(y-1)^2}$
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