# How do you express (3x+2) / (x^(2)+3x-4) in partial fractions?

Apr 16, 2018

The answer is $= \frac{2}{x + 4} + \frac{1}{x - 1}$

#### Explanation:

Perform the decomposition into partial fractions

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

$= \frac{A}{x + 4} + \frac{B}{x - 1}$

$= \frac{A \left(x - 1\right) + B \left(x + 4\right)}{\left(x + 4\right) \left(x - 1\right)}$

The denominators are the same, compare the numerators

$3 x + 2 = A \left(x - 1\right) + B \left(x + 4\right)$

Let $x = - 4$, $\implies$, $- 10 = - 5 A$, $\implies$, $A = 2$

Let $x = 1$, $\implies$, $5 = 5 B$, $\implies$, $B = 1$

Therefore,

$\frac{3 x + 2}{{x}^{2} + 3 x - 4} = \frac{2}{x + 4} + \frac{1}{x - 1}$

Apr 16, 2018

$\textcolor{b l u e}{\frac{2}{x + 4} + \frac{1}{x - 1}}$

#### Explanation:

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

Factor denominator:

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

For linear factors we expect the partial fraction form to be:

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

Adding $R H S$

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

Since this is an identity numerators are identical, so:

$3 x + 2 \equiv A \left(x - 1\right) + B \left(x + 4\right)$

We now proceed to find the values of A and B:

Let $x = - 4$

$3 \left(- 4\right) + 2 \equiv A \left(\left(- 4\right) - 1\right) + B \left(\left(- 4\right) + 4\right)$

$- 10 = - 5 A$

$A = 2$

Let $x = 1$

$3 \left(1\right) + 2 \equiv A \left(\left(1\right) - 1\right) + B \left(\left(1\right) + 4\right)$

$5 = 5 B$

$B = 1$

The partial fractions are therefore:

$\textcolor{b l u e}{\frac{2}{x + 4} + \frac{1}{x - 1}}$

Apr 16, 2018

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

#### Explanation:

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

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

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

or $\left(3 x + 2\right) = A \left(x - 1\right) + B \left(x + 4\right)$

Let $x = 1 \therefore 3 \cdot 1 + 2 = A \left(1 - 1\right) + B \left(1 + 4\right)$ or

$5 B = 5 \therefore B = 1$

Let $x = - 4 \therefore 3 \cdot \left(- 4\right) + 2 = A \left(- 4 - 1\right) + B \left(- 4 + 4\right)$ or

$- 5 A = - 10 \therefore A = 2 \therefore$

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