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How do you decompose the following fraction?

$$\frac{2}{2x^2 (x+1)} = \frac{a}{2x^2} + \frac{b}{x+1}$$

i just want to know if I am on the right track or I am missing something(based on the above beginning).

Peter
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Lucas
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  • Not quite, choose $$\frac{Ax+B}{x^2}+\frac{C}{x}+\frac{D}{x+1}$$ First of all, cancel out the $2$ on the left side – Peter Aug 18 '17 at 13:38
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    @Peter A constant numerator is sufficient, i.e. you don't need $Ax+B$ but simply $A$ will do. Your suggestion will work too of course, but you don't need the extra coefficient. – StackTD Aug 18 '17 at 13:44
  • @StackTD You are right, we have $\frac{A}{x}$ and $\frac{C}{x}$, which we can sum up to $\frac{A'}{x}$ – Peter Aug 18 '17 at 13:59
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    @Peter Indeed. You only need a linear numerator for (natural powers of) irreducible quadratic factors in the denominator (i.e. with no real roots). – StackTD Aug 18 '17 at 14:01

1 Answers1

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Notice that you can cancel the common factor $2$ first.

Since $x=0$ is a double root, you need to take: $$\frac{1}{x^2 (x+1)} = \frac{a}{x^2} + \frac{b}{x} + \frac{c}{x+1}$$ Can you take it from there?

Sometimes, you can avoid having to solve a system (in your case of three linear equations in three unknowns) to find the coefficients/numerators with a few handy manipulations.

$$\begin{align} \frac{1}{x^2 (x+1)} & = \frac{1\color{blue}{+x-x}}{x^2 (x+1)} \\[5pt] & = \frac{1+x}{x^2 (x+1)} - \frac{x}{x^2 (x+1)} \\[5pt] & = \frac{1}{x^2} - \frac{1}{x (x+1)}\\[5pt] & = \frac{1}{x^2} - \frac{1\color{blue}{+x-x}}{x (x+1)}\\[5pt] & = \frac{1}{x^2} - \frac{1+x}{x (x+1)}+\frac{x}{x (x+1)}\\[5pt] & = \frac{1}{x^2} - \frac{1}{x}+\frac{1}{x+1} \end{align}$$

StackTD
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