# How do you write the partial fraction decomposition of the rational expression (5x - 1) / ((x - 2)(x + 1))?

Jan 6, 2016

$\frac{5 x - 1}{\left(x - 2\right) \left(x + 1\right)} = \frac{3}{x - 2} + \frac{2}{x + 1}$

#### Explanation:

The partial fraction decomposition suggests that the function can be broken down into the sum of two other functions, or;

$\frac{5 x - 1}{\left(x - 2\right) \left(x + 1\right)} = \frac{A}{x - 2} + \frac{B}{x + 1}$

Where we need to solve for $A$ and $B$. We can cross multiply to combine the terms on the right hand side over a common denominator. We get;

$\frac{5 x - 1}{\left(x - 2\right) \left(x + 1\right)} = \frac{A \left(x + 1\right) + B \left(x - 2\right)}{\left(x - 2\right) \left(x + 1\right)}$

We can now cancel the denominator on each side, leaving;

$5 x - 1 = A \left(x + 1\right) + B \left(x - 2\right)$

Now we can solve for $A$ and $B$. We can make one of the terms cancel out by choosing the right value for $x$. Lets try x=~1.

5(~1)-1 = A(~1+1) + B(~1-2)

The $A$ term goes away since it is multiplied by zero, leaving;

~6 = ~3B

Solving for $B$;

$B = 2$

We can substitute $B$ and solve for $A$, but it would be easier to do the same trick that we used to solve for $B$. Let $x = 2$.

$5 \left(2\right) - 1 = A \left(2 + 1\right) + B \left(2 - 2\right)$

This time, the $B$ term goes away;

$9 = 3 A$

$A = 3$

Now that we have our values for $A$ and $B$ we can plug into our first function and get;

$\frac{5 x - 1}{\left(x - 2\right) \left(x + 1\right)} = \frac{3}{x - 2} + \frac{2}{x + 1}$