# 4 KILLER QUESTIONS ON PARTIAL FRACTION

Partial fraction is the reverse process of algebraic addition. We have already look at three cases of the partial fraction which are:

But today, We will be turning our attention to solving some rather hard question on partial fraction

Question 1
Express $\frac{1}{x^3-x}$ in partial fraction.

First, you will factorize the denominator
$\frac{1}{x(x^2-1)}$

$x^2-1$ is difference of squares, hence
$\frac{1}{x(x+1)(x-1)}$

This is a case of partial fraction with non-repeated linear factor. Hence,
$\frac{1}{x(x+1)(x-1)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1}$

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

The denominator are the same, hence, we equate the numerator.
$1=A(x^2-1)+B(x(x-1))+C(x(x+1))$

To get $C$ we need to eliminate $A$ and $B$, hence, let $x=1$
$1=A((1)^2-1)+B(1(1-1))+C(1(1+1))$
$1=C(2)$
$1=2C$
$\frac{2C}{2}=\frac{1}{2}$
$C=\frac{1}{2}$

To derive the value of $B$ we need to eliminate $A$ and $C$, hence, let $x=-1$
$1=A((-1)^2-1)+B(-1(-1-1))+C(-1(-1+1))$
$1=B(-1(-2))$
$1=2B$
$\frac{2B}{2}=\frac{1}{2}$
$B=\frac{1}{2}$

Now, let $x=0$
$1=A((0)^2-1)+B(0(0-1)+C(0(0+1))$
$1=A(-1)$
$1=-A$
$\frac{-A}{-1}=\frac{1}{-1}$
$A=-1$

Thus the partial fraction is:
$\frac{1}{x(x+1)(x-1)}=\frac{-1}{x}+\frac{1}{2(x+1)}+\frac{1}{2(x-1)}$

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

This can be factorize to:
$\frac{y+4}{(y+1)(y-1)(y^2+4)}$

This contains two non-repeated linear fraction and a irreducible quadratic. Accordingly,
$\frac{y+4}{(y+1)(y-1)(y^2+4)}=\frac{A}{y+1}+\frac{B}{y-1}+\frac{Cy+D}{y^2+4}$

Adding the right-hand side of the expression
$\frac{y+4}{(y+1)(y-1)(y^2+4)}=\frac{A(y-1)(y^2+4)+B(y+1)(y^2+4)+(Cy+D)(y^2-1)}{(y+1)(y-1)(y^2+4)}$

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

To get $B$, we need to eliminate $A$, $C$. and $D$. Hence, let $y=1$
$1+4=A(1-1)((1)^2+4)+B(1+1)((1)^2+4)+(C(1)+D)((1)^2-1)$
$5=B(2)(5)$
$5=10B$
$10B=5$
$\frac{10B}{10}=\frac{5}{10}$
$B=\frac{1}{2}$

To get $A$, we need to eliminate $B$, $C$. and $D$. Hence, let $y=-1$
-1+4=A(-1-1)((-1)²+4)+B(-1+1)((-1)²+4)+(C(-1)+D)((-1)²-1)
$3=A(-2)(5)$
$3=-10A$
$-10A=3$
$\frac{-10A}{-10}=\frac{3}{-10}$
$A=\frac{-3}{10}$

Now, let's insert the value of $A$ and $B$
$y+4=A(y-1)(y^2+4)+B(y+1)(y^2+4)+(Cy+D)(y^2-1)$
$y+4=\frac{-3}{10}(y-1)(y^2+4)+\frac{1}{2}(y+1)(y^2+4)+(Cy+D)(y^2-1)$
$y+4=\frac{-3}{10}(y^3-y^2+4y-4)+\frac{1}{2}(y^3+y^2+4y+4)+(Cy+D)(y^2-1)$
$y+4=\frac{-3y^3+3y^2-12y+12}{10}+\frac{y^3+y^2+4y+4}{2}+Cy(y^2-1)+D(y^2-1)$
$y+4=\frac{-3y^3+3y^2-12y+12}{10}+\frac{y^3+y^2+4y+4}{2}+Cy^3-Cy+Dy^2-D$

By simply multiplying through by 10, we will clear the fraction. Accordingly
10y+40=-3y³+3y²-12y+12+5y³+5y²+20y+20+10Cy³-10Cy+10Dy²-D

Collecting like terms
10y+40=-3y³+5y³+10Cy³-3y²+5y²+10Dy²-12y+20y+10Cy+12+20-D
(10)y+40=(-3+5+10C)y³+(3+5+10D)y²+(-12+20+10C)y+12+20-D

Comparing the coefficients
$0=-3+5+10C$
$0=3+5+10D$
$10=-12+20+10C$
$40=12+20-D$

From the first eqn.
$0=-3+5+10C$
$3-5=10C$
$-2=10C$
$\frac{10C}{10}=\frac{-2}{10}$
$C=\frac{-1}{5}$

From the second eqn...
$0=3+5+10D$
$-8=10D$
$\frac{-8}{10}=\frac{10D}{10}$
$D=\frac{-4}{5}$.

Hence, the partial fraction decomposition
$\frac{y+4}{(y+1)(y-1)(y^2+4)}=\frac{-3}{10(y+1)}+\frac{1}{2(y-1)}+\frac{-y-8}{5(y^2+4)}$

Question 3
Express $\frac{4y^2+5y-1}{4y^3-4y^2-7y-2}$ in partial fraction.

Let's factorize the denominator
$\frac{4y^2+5y-1}{(2y+1)^2(y-2)}$

This is a case of repeated linear factor, thus,
$\frac{4y^2+5y-1}{(2y+1)^2(y-2)}=\frac{A}{2y+1}+\frac{B}{(2y+1)^2}+\frac{C}{y-2}$

$\frac{4y^2+5y-1}{(2y+1)^2(y-2)}=\frac{A(2y+1)(y-2)+B(y-2)+C(2y+1)}{(2y+1)^2(y-2)}$

Equating the numerator
$4y^2+5y-1=A(2y+1)(y-2)+B(y-2)+C(2y+1)^2$

Let $y=2$
$4(2)^2+5(2)-1=A(2(2)-1)(2-2)+B(2-2)+C(2(2)+1)^2$
$16+10-1=C(5)^2$
$25=25C$
$\frac{25C}{25}=\frac{25}{25}$
$C=1$

Let $y=-0.5$
4(-0.5)²+5(-0.5)-1=A(2(-0.5)-1))(-0.5-2)+B(-0.5-2)+C(2(-0.5+1)²
$4(0.25)-2.5-1=-2.5B$
$1-3.5=-2.5B$
$-2.5=-2.5B$

$\frac{-2.5}{-2.5}=\frac{-2.5B}{-2.5}$
$B=1$

Now, let $y=0$
$4(0)^2+5(0)-1=A(2(0)-1)(0-2)+B(0-2)+C(2(0)+1)^2$
$-1=A(-1)+B(-2)+C(1)$
$-1=-A-2B+C$

Now, let insert the value of $B$ and $C$
$-1=-A-2(1)+1$
$-1=A-2+1$
$-1+2-1=A$
$0=A$
$A=0$

Thus, the partial fraction decomposition is:
$\frac{4y^2+5y-1}{(2y+1)^2(y-2)}=\frac{1}{(2y+1)^2}+\frac{1}{y-2}$

Related posts
Question 4
Relate $\frac{4y}{1-y^4}$ to partial fraction

The denominator, $1-y^4$ can be factorized to $(1+y)(1-y)(1+y^2)$. Thus
$\frac{4y}{(1-y)(1+y)(1+y^2)}$

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

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

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

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

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

Inserting $A$ and $B$
$4y=A(y^3+y^2+y+1)+B(-y^3+y^2-y+1)+(Cy+D)(1-y^2)$
$4y=1(y^3+y^2+y+1)-1(-y^3+y^2-y+1)+(Cy+D)(1-y^2)$
$4y=y^3+y^2+y+1+y^3-y^2+y-1+(Cy+D)(1-y^2)$
$4y=y^3+y^2+y+1+y^3-y^2+y-1+Cy-Cy^3+D-Dy^2$

Collecting like terms
$4y=y^3+y^3+Cy^3+y^2-y^2-Dy^2+y+y+Cy+1-1+D$
$(4)y=1+1+C)y^3+(1-1-D)y^2+(1+1+C)y+0+D$

Comparing the coefficients
$0=1+1+C$
$0=1-1-D$

$0=1+1+C$
$-2=C$
$C=-2$

$0=1-1-D$
$0=-D$
$D=0$.

Thus the partial decomposution is:
$\frac{y+4}{(y+1)(y-1)(y^2+4)}=\frac{1}{y+1}+\frac{-1}{y-1}+\frac{-2y}{y^2+4}$

We have now come to the end of our series on partial fraction. Meanwhile, if you have got questions, do well to ask our telegram community. Alternatively, you can ask our Facebook community.

Help us grow our readership by sharing this post