Partial fraction (doubt)

Question

I have this partial fraction $$\displaystyle\frac{1}{(2+x)^2(4+x)^2}$$

I tried to resolve using this method: $$\displaystyle\frac{A}{2+x}+\displaystyle\frac{B}{(2+x)^2}+\displaystyle\frac{C}{4+x}+\displaystyle\frac{D}{(4+x)^2}$$ $$1=A(2+x)(4+x)^2+B(4+x)^2+C(4+x)(2+x)^2+D(2+x)^2$$

When x=-2 $$1=B(4-2)^2$$ $$B=\displaystyle\frac{1}{4}$$

When x=-4 $$1=D(2-4)^2$$ $$D=\displaystyle\frac{1}{4}$$

When x=0 $$1=A(2)(16)+B(16)+C(4)(4)+D(4)$$ $$1=A(32)+B(16)+C(16)+D(4)$$

Replacing the values for B y D $$1=A(32)+4+C(16)+1$$ $$1-4-1=A(32)+c(16)$$ $$-4=A(32)+C(16)$$

How I can get the values ​​of $A$ and $D$?

Answer 1

Try another value of $x$ as well, say $x=2$, giving $1 = 144A + 9 + 64C + 4$, or $144A + 96C = -12$. Combined with the equation you already have, $32A+16C=-4$, solving gives $A=-\frac{1}{4}$, $C=\frac{1}{4}$.

Answer 2

take LCM of second expression you get and sum it

{A(2+x)(4+x)^2 +B(4+x)^2+C(2+x)^2(4+x)+D(2+x)^2}/{(x+2)^2(x+4)^2}= your 1st expression

so you have {A(2+x)(4+x)^2 +B(4+x)^2+C(2+x)^2(4+x)+D(2+x)^2}=1

compare coefficients of x^3 ,x^2,x^1 and x^0 which on right side of aboveequation are equal to 0,0,0 and 1 respectively. you have

A+C=0

8A+B+4C+D=0

32A+8B + 20C+ 4D=0

32A + 16B + 16C +4D=1

soLVING YOU GET B=1/12, D=7/12 ,A =-1/6, C=1/6

Answer 3

If you clear fractions you get $$A(2+x)(4+x)^2+B(4+x)^2+C(4+x)(2+x)^2+D(2+x)^2=1$$

The easy ones are $x=-2$ which gives $B=\frac 14$ and $x=-4$ which gives $D=\frac 14$

Now equate the coefficients of $x^3$ on each side to give $$A+C=0$$ and set $x=0$ (or equivalently equate constant coefficients), to give $$32A+16B+16C+4D=1$$ which becomes $$16A+4+0+1=1$$

Where the $0$ comes from $16(A+C)=0$

I think it is the equating coefficients of $x^3$ - or using a value like $x=1$ which you've missed - you need four data points to identify the four unknown values.

Answer 4

Assign different values to $x$, like $-1, 0, +1, +2$ and solve the linear system of 4 equations in 4 unknowns.

$$\left[\begin{array}{cccc} 1/1& 1/1& 1/3& 1/9\\ 1/2& 1/4& 1/4& 1/16\\ 1/3& 1/9& 1/5& 1/25\\ 1/4& 1/16& 1/6& 1/36 \end{array}\right] \left[\begin{array}{c} A\\ B\\ C\\ D\end{array}\right]= \left[\begin{array}{c} 1/9\\ 1/64\\ 1/225\\ 1/576\end{array}\right]$$ $$\left[\begin{array}{c} A\\ B\\ C\\ D\end{array}\right]= \left[\begin{array}{c} -1/4\\ 1/4\\ 1/4\\ 1/4\end{array}\right]$$

Answer 5

Heaviside cover-up is an alternative technique worth exploring.

It directly gives $\begin{align}B = \frac{1}{(4+x)^2}\Bigg|_{x=-2} = \frac{1}{4} \text{ and } D = \frac{1}{(2+x)^2}\Bigg|_{x=-4} = \frac{1}{4}\end{align}\tag{1}$

Next, it requires that the $B$ and $D$ terms be moved to the other side.

The partial fraction identity then becomes

$\begin{eqnarray}\frac{A}{2+x} + \frac{C}{4+x} &= &\frac{1}{(2+x)^2(4+x)^2} - \frac{1/4}{(2+x)^2} - \frac{1/4}{(4+x)^2}\\&=&\frac{4-(4+x)^2-(2+x)^2}{4(2+x)^2(4+x)^2}\\&=&\frac{-2(8 + 6x + x^2)}{4(2+x)^2(4+x)^2}\\&=&\frac{-1/2}{(2+x)(4+x)}\end{eqnarray}$

which gives $\begin{align}A = \frac{-1/2}{(4+x)}\Bigg|_{x=-2} = -\frac{1}{4} \text{ and } C = \frac{-1/2}{(2+x)}\Bigg|_{x=-4} = \frac{1}{4}\end{align}\tag{2}$

Remark: Note how this technique eliminates the need to solve a system of linear equations for finding out the partial fraction decomposition coefficients (which can often be tedious).

See here for some more examples.