# How do you express (x²+2) / (x+3)  in partial fractions?

Feb 22, 2016

$\frac{x}{1} + \frac{- 3 x + 2}{x + 3}$

#### Explanation:

because the top quadratic and the bottom is linear you're looking for something or the form

$\frac{A}{1} + \frac{B}{x + 3}$, were $A$ and $B$ will both be linear functions of $x$ (like 2x+4 or similar).

We know one bottom must be one because x+3 is linear.

We're starting with
$\frac{A}{1} + \frac{B}{x + 3}$.
We then apply standard fraction addition rules. We need to get then to a common base.

This is just like numerical fractions $\frac{1}{3} + \frac{1}{4} = \frac{3}{12} + \frac{4}{12} = \frac{7}{12.}$

$\frac{A}{1} + \frac{B}{x + 3} \implies \frac{A \cdot \left(x + 3\right)}{1 \cdot \left(x + 3\right)} + \frac{B}{x + 3} = \frac{A \cdot \left(x + 3\right) + B}{x + 3}$.
So we get the bottom automatically.

Now we set $A \cdot \left(x + 3\right) + B = {x}^{2} + 2$
$A x + 3 A + B = {x}^{2} + 2$
$A$ and $B$ are linear terms so the ${x}^{2}$ must come from $A x$.
let $A x = {x}^{2}$ $\implies$ $A = x$
Then
$3 A + B = 2$
substituting $A = x$, gives
$3 x + B = 2$
or
$B = 2 - 3 x$
in standard from this is $B = - 3 x + 2$.
Putting it all together we have
$\frac{x}{1} + \frac{- 3 x + 2}{x + 3}$