# PARTIAL FRACTION WITH IRREDUCIBLE QUADRATIC

So far, our multinational on partial fractions has been limited to only linear factors

However, not all partial fraction has a linear factor. Some may have an irreducible quadratic.

As the name suggests, a partial fraction with irreducible quadratic roots is a partial fraction that has a denominator made up of irreducible quadratic roots that are not factorizable over real numbers.

In plain English, an partial fraction with irreducible quadratic cannot be a factor or reduce.

In this case, the symbolic representation of the numerator will be like this: $Ax+B$, $Cx+D$, $Ex+F$.

๐This post extends our series on partial fraction๐

Example 1
Express $\frac{8y^2+12y-20}{(y-3)(y^2+y+2)}$ in partial fractions.

Solution:
This expression contains one linear factor and one irreducible quadratic. Hence, we represent it like this:
$\frac{8y^2+12y-20}{(y-3)(y^2+y+2)}=\frac{A}{y+3}+\frac{By+C}{y^2+y+2}$

$\frac{8y^2+12y-20}{(y-3)(y^2+y+2)}=\frac{A(y^2+y+2)+(By+C)(y+3)}{(y+3)(y^2+y+2)}$

Equating the numerator
$8y^2+12y-20=A(y^2+y+2)+(By+C)(y+3)$

To get $A$, We need to eliminate $B$ and $C$. Let $y=-3$
$8(-3)^2+12(-3)-20=A((-3)^2+(-3)+2)+(B(-3)+C)(-3+3)$
$72-36-20=A(9-3+2)$
$16=A(8)$
$8A=16$
$\frac{8A}{8}=\frac{16}{8}$
$A=2$

Now, let's $y=0$
$8(0)^2+12(0)-20=A((0)^2+(0)+2)+(B(0)+C)((0)+3)$
$-20=A(2)+(C)(3)$
$-20=2A+3C$

$A=2$, therefore
$-20=2(2)+3C$
$-20=4+3C$
$-20-4=3C$
$3C=-24$
$\frac{3C}{3}=\frac{-24}{3}$
$C=-8$

Let $y=1$
$8(1)^2+12(1)-20=A((1)^2+(1)+2)+(B(1)+C)(1+3)$
$8+12-20=A(1+1+2)+(B+C)(4)$
$0=A(4)+(B+C)4$
$0=4A+4B+4C$

We know that $A=2$ and $C=-8$
$0=4(2)+4B+4(-8)$
$0=8+4B-32$
$-8+32=4B$
$24=4B$
$\frac{4B}{4}=\frac{24}{4}$
$B=6$

Hence, the partial decomposition is:
$\frac{8y^2+12y-20}{(y-3)(y^2+y+2)}=\frac{2}{y+3}+\frac{6y-8}{y^2+y+2}$.

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

Solution:
Like the previous example, this one also has one linear factor and one quadratic factor. The only difference here is that this has quadratic roots before linear factor, hence.
$\frac{1}{(y^2+1)(y+1)}=\frac{Ay+B}{y^2+1}+\frac{C}{y-1}$

Adding the right-hand side of the equation
$\frac{1}{(y^2+1)(y+1)}=\frac{(Ay+B)(y-1)+C(y^2+1)}{(y^2+1)(y+1)}$

Since we have the same denominator, we would equate the numerator.
$1=(Ay+B)(y-1)+C(y^2+1)$

To eliminate $A$ and $B$, Let's $y=1$
$1=(A(1)+B)(1-1)+C((1)^2+1)$
$1=C(1+1)$
$1=2C$
$\frac{2C}{2}=\frac{1}{2}$
$C=\frac{1}{2}$

Substituting $\frac{1}{2}$ for $C$
$1=(Ay+B)(y-1)+\frac{1}{2}(y^2+1)$

Expanding the brackets
$1=Ay(y-1)+B(y-1)+\frac{y^2+1}{2}$
$1=Ay^2-Ay+By-B+\frac{y^2+1}{2}$

Let's clear the fraction my multiplying through by $2$
$(1\times2)=(Ay^2\times2)-(Ay\times2)+(By\times2)-(B\times2)+(\frac{y^2+1}{\require{cancel}\bcancel{2}}\times\require{cancel}\bcancel{2})$
$2=2Ay^2-2Ay+2By-2B+y^2+1$
$2=2Ay^2+y^2-2Ay+2By-2B+1$
$2=(2A+1)y^2+(-2A+2B)y-2B+1$

Now, we will turn to equate the coefficient method. The idea is that we equate this with the same coefficients
$0=2A+1$
$0=-2A+2B$
$2=-2B+1$

$0=2A+1$
$-1=2A$
$\frac{-1}{2}=\frac{2A}{2}$
$A=\frac{-1}{2}$

Now let's solve for $B$
$2-1=-2B$
$1=-2B$
$\frac{1}{-2}=\frac{-2B}{-2}$
$B=\frac{1}{-2}$

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

Example 3
Express $\frac{x^2-x-5}{x^3-x^2+4x-4}$ in partial fraction.

Solution:
First, we would factorize the denominator
$\frac{x^2-x-5}{(x-1)(x^2+4)}$

As usual, we will use symbolic character to represent the numerator
$\frac{x^2-x-5}{(x-1)(x^2+4)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+4}$

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

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

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

Now, let insert the value of $A$
$x^2-x-5=-1(x^2+4)+(Bx+C)(x-1)$
$x^2-x-5=-x^2-4+(Bx+C)(x-1)$

Expand the brackets
$x^2-x-5=-x^2-4+Bx(x-1)+C(x-1)$
$x^2-x-5=-x^2-4+Bx^2-Bx+Cx-C$
$x^2-x-5=-x^2+Bx^2-Bx+Cx-4-C$
$1(x^2)-1(x)-5=(-1+B)x^2+(-B+C)x-4-C$

By equating the coefficient
$1=-1+B$
$-1=-B+C$
$-5=-4-C$

$1=-1+B$
$1+1=B$
$B=2$

Let solve for $C$
$-5=-4-C$
$-5+4=-C$
$-1=-C$
$\frac{-1}{-1}=\frac{-C}{-1}$
$C=1$

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

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Example 4
Express $\frac{2y-3}{y^3+10y}$ in partial fraction.

Solution:
First, let's factorize the denominator
$\frac{2y-3}{y(y^2+10)}$

$\frac{2y-3}{y(y^2+10)}=\frac{A}{y}+\frac{By+C}{y^2+10}$

$\frac{2y-3}{y(y^2+10}=\frac{A(y^2+10)+(Bx+C)(y)}{y(y^2+10)}$

Equating the numerator
$2y-3=A(y^2+10)+(By+C)(y)$

Let $y=0$
$2(0)-3=A(0)^2+10)+(B(0)+C)(0)$
$-3=A(10)$
$10A=-3$
$\frac{10A}{10}=\frac{-3}{10}$
$A=\frac{-3}{10}$

Substituting the $\frac{-3}{10}$ for $A$
$2y-3=\frac{-3}{10}(y^2+10)+(By+C)(y)$
$2y-3=\frac{-3y^2+30}{10}+(By+C)(y)$

Expanding the brackets
$2y-3=\frac{-3y^2+30}{10}+By^2+Cy$

To clear the fraction, let's multiply through by $10$
$(2y\times10)-(3\times10)=(\frac{-3y^2+30}{\require{cancel}\bcancel{10}}\times\require{cancel}\bcancel{10})+(By^2\times10)+(Cy\times10)$
$20y-30=-3y^2+30+10By^2+10Cy$
$20y-30=-3y^2+10By^2+30+10Cy$
$(20)y-30=(-3+10B)y^2+(10C)y+30$

By equating the coefficients
$0=-3+10B$
$20=10C$

If $0=-3+10B$, then
$3=10B$
$\frac{10B}{10}=\frac{3}{10}$
$B=\frac{3}{10}$

Next, we solve for $C$
$20=10C$
$\frac{10C}{10}=\frac{20}{10}$
$C=2$

Accordingly
$\frac{2y-3}{y(y^2+10)}=\frac{-3}{10}+\frac{(\frac{3}{10})y+2}{y^2+10}$

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