How do you write the partial fraction decomposition of the rational expression (x-2) /( x^2+4x+3)?

Jan 14, 2016

$\frac{x - 2}{\left(x + 1\right) \left(x + 3\right)} = \frac{4}{x + 2} - \frac{3}{x + 1}$

Explanation:

$\frac{x - 2}{{x}^{2} + 4 x + 3}$

$\textcolor{red}{\text{Factorize the Denominator}}$
${x}^{2} + 4 x + 3 = \left(x + 1\right) \left(x + 3\right)$

Our expression becomes

$\frac{x - 2}{\left(x + 1\right) \left(x + 3\right)}$

This expression can be written as

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

Equating the Numerators
$x - 2 = A \left(x + 2\right) + B \left(x + 1\right)$

Now we have to solve for $A$ and $B$

Let us take $x = - 1$ for making $x + 1 = 0$ and thus removing $B$.

$- 1 - 2 = A \left(- 1 + 2\right) + B \left(- 1 + 1\right)$
$- 3 = A$
$A = - 3$

Now let us use $x = - 2$

$- 2 - 2 = A \left(- 2 + 2\right) + B \left(- 2 + 1\right)$
$- 4 = - B$

$B = 4$

Substituting the value of $A$ and $B$ in

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

$\frac{x - 2}{\left(x + 1\right) \left(x + 3\right)} = - \frac{3}{x + 1} + \frac{4}{x + 2}$
$\frac{x - 2}{\left(x + 1\right) \left(x + 3\right)} = \frac{4}{x + 2} - \frac{3}{x + 1}$