Easy way to find the partial fraction

Question

I always have trouble trying to find the partial fraction, especially for complicated ones.

For example, this is what I will do to find the partial fraction of $\displaystyle \frac{8x^3+35x^2+42x+27}{x(2x+3)^3}$

• $\displaystyle \frac{A}{x}+\frac{B}{2x+3}+\frac{C}{(2x+3)^2}+\frac{D}{(2x+3)^3}$
• $\displaystyle A(2x+3)^3+Bx(2x+3)^2+Cx(2x+3)+Dx = 8x^3+35x^2+42x+27 $
• When x = $0$ -> $\displaystyle 27A = 27, A=1$
• $\displaystyle (2x+3)^3+Bx(2x+3)^2+Cx(2x+3)+Dx = 8x^3+35x^2+42x+27$

After what I can do is to expand all the elements and group them based on their exponent, and solve the system of equation.

However, I remember seeing there exists an easier solution. Also given that there will be no calculator available on the exam, doing this way will take a long time and results in possible errors.

Does anyone have an easier way to solve this question or similar ones?

Thanks!

Answer 1

A suggestion may be in form of $$\displaystyle \frac{8x^3+35x^2+42x+27}{x(2x+3)^3}$$ look for $(2x+3)^3=8x^3+36x^2+54x+27$ so ,we can rewrite $$\displaystyle \frac{8x^3+35x^2+42x+27}{x(2x+3)^3}=\\\frac{(2x+3)^3-x^2-12x}{x(2x+3)^3}=\\ \frac{(2x+3)^3}{x(2x+3)^3}-\frac{x(x+12)}{x(2x+3)^3}=\\ \frac{1}{x}-\frac{(x+12)}{(2x+3)^3}=\\$$ can you take over ?

Answer 2

$A(2x+3)^3+Bx(2x+3)^2+Cx(2x+3)+Dx = 8x^3+35x^2+42x+27 $

With $ x = 0, 27 A = 27, A = 1 $.

This gives us $Bx(2x+3)^2+Cx(2x+3)+Dx = -x^2 - 12x $.

Dividing by $x$, this gives us $B(2x+3)^2+C(2x+3)+D = -x - 12 $.

With $x = - \frac{3}{2}$, $D = - 10 \frac{1}{2}$.

This gives us $ B(2x+3)^2 + C (2x+3) = -x - \frac{3}{2}$

Dividing by $(2x+3)$, this gives us $ B ( 2x+3) + C = - \frac{1}{2}$.

With $ x = - \frac{3}{2}$, this gives us $ C = - \frac{1}{2}$.

This gives us $ B ( 2x+3) = 0 $.

Dividing by $(2x+3) $, this gives us $ B = 0 $.

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With this process, you basically get rid of terms sequentially. Use the cover up rule to determine one variable, move that over, divide by the common terms, rinse and repeat.