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Partial fraction problem (x^2)/[(x-2)(x+3)(x-1)]

mathlearn

Active member
Jul 24, 2016
341
Hello Everyone , I need some help in solving this partial fraction $\frac{x^2}{(x-2)(x+3)(x-1)}$

I am using this method in which the partial fraction is broken in to 3 parts namely A,B &C

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

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

Common denominators cancel out resulting in,

${x^2}={A(x+3)(x-1)+B(x-2)(x-1)+C(x-2)(x+3)}$

Now what values should be substituted to $x$ in order to get the relevant values for A,B & C
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,736
Yes, we want:

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

Now, let's try the Heaviside cover-up method...cover up the factor $(x-2)$ on the LHS, and evaluate what's left for $x=2$ (the root of the factor being covered up...this value will be $A$:

\(\displaystyle A=\frac{2^2}{(2+3)(2-1)}=\frac{4}{5}\)

Likewise, we find:

\(\displaystyle B=\frac{(-3)^2}{((-3)-2)((-3)-1)}=\frac{9}{20}\)

\(\displaystyle C=\frac{1^2}{(1-2)(1+3)}=-\frac{1}{4}\)

Hence:

\(\displaystyle \frac{x^2}{(x-2)(x+3)(x-1)}=\frac{4}{5(x-2)}+\frac{9}{20(x+3)}-\frac{1}{4(x-1)}\)

This is essentially the same as going back to where you left off:

\(\displaystyle x^2=A(x+3)(x-1)+B(x-2)(x-1)+C(x-2)(x+3)\)

Let $x=2$:

\(\displaystyle 2^2=A(2+3)(2-1)+B(2-2)(2-1)+C(2-2)(2+3)=5A\implies A=\frac{4}{5}\)

Let $x=-3$:

\(\displaystyle (-3)^2=A((-3)+3)((-3)-1)+B((-3)-2)((-3)-1)+C((-3)-2)((-3)+3)=20B\implies B=\frac{9}{20}\)

Let $x=1$:

\(\displaystyle 1^2=A(1+3)(1-1)+B(1-2)(1-1)+C(1-2)(1+3)=-4C\implies C=-\frac{1}{4}\)
 

mathlearn

Active member
Jul 24, 2016
341
Thanks Mark, nicely explained
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Hello Everyone , I need some help in solving this partial fraction $\frac{x^2}{(x-2)(x+3)(x-1)}$

I am using this method in which the partial fraction is broken in to 3 parts namely A,B &C

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

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

Common denominators cancel out resulting in,

${x^2}={A(x+3)(x-1)+B(x-2)(x-1)+C(x-2)(x+3)}$

Now what values should be substituted to $x$ in order to get the relevant values for A,B & C
Are you under the impression that you have to do this is a specific way?

The simplest thing to do is to set x= 1, 2, and -3 so that the factors, x- 1, x- 2, and x+ 3, are 0: That gives three completely separated equations, one for A, another for B, and the last for C.

However, in fact, setting x equal to any three values gives three equations to solve for A, B, and C. For example, setting x= 0 gives $0= -3A+ 2B- 6C$. Setting x= -1 gives $1= -4A+ 6B- 6C$. Setting x= 3, $9= 12A+ 2B+ 6C$. You can solve those three equations for A, B, and C.

Or just multiply the formula on the right:
$x^2= Ax^2+ 2Ax- 3A+ Bx^2- 3Bx+ 2B+ Cx^2+ Cx- 6C= (A+ B+ C)x^2+ (2A- 3B+ C)x+ (3A- 2B+ 6C)$

Then, since these are to be equal for all x, The "corresponding" coefficients" must be equal: $A+ B+ C= 1$, $2A- 3B+ C= 0$, $3A- 2B+ 6C= 0$. Again, three equations to solve for A, B, and C.
 

BestMethod

New member
Dec 11, 2019
9
To find A,multiplying the both sides by (x-2);

x^2/(x+3)(x-1)=A + B(x-2)/(x+3) + C(x-2)/(x-1)

Now,by setting x=2,the expressions containing (x-2) vanish,so we get rid of B & C,

2^2/(2+3)(2-1)=A+0+0, then,

4/5=A

By the similar method we could find B & C.