How do you express (4x^2 - 7x - 12) / ((x) (x+2) (x-3)) in partial fractions?

Dec 16, 2016

The answer is $= \frac{2}{x} + \frac{\frac{9}{5}}{x + 2} + \frac{\frac{1}{5}}{x - 3}$

Explanation:

Let's do the decomposition into partial fractions

$\frac{4 {x}^{2} - 7 x - 12}{x \left(x + 2\right) \left(x - 3\right)} = \frac{A}{x} + \frac{B}{x + 2} + \frac{C}{x - 3}$

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

Therefore,

$4 {x}^{2} - 7 x - 12 = A \left(x + 2\right) \left(x - 3\right) + B \left(x\right) \left(x - 3\right) + C \left(x\right) \left(x + 2\right)$

Let $x = 0$, $\implies$, $- 12 = - 6 A$, $\implies$, $A = 2$

Let $x = - 2$,$\implies$, $18 = 10 B$, $\implies$, $B = \frac{9}{5}$

Let $x = 3$, $\implies$, $3 = 15 C$, $\implies$, $C = \frac{1}{5}$

So,

$\frac{4 {x}^{2} - 7 x - 12}{x \left(x + 2\right) \left(x - 3\right)} = \frac{2}{x} + \frac{\frac{9}{5}}{x + 2} + \frac{\frac{1}{5}}{x - 3}$

Dec 16, 2016

$\frac{4 {x}^{2} - 7 x - 12}{x \left(x + 2\right) \left(x - 3\right)} = \frac{2}{x} + \frac{9}{5 \left(x + 2\right)} + \frac{1}{5 \left(x - 3\right)}$

Explanation:

Let $\frac{4 {x}^{2} - 7 x - 12}{x \left(x + 2\right) \left(x - 3\right)} \Leftrightarrow \frac{A}{x} + \frac{B}{x + 2} + \frac{C}{x - 3}$

Hence $\frac{4 {x}^{2} - 7 x - 12}{x \left(x + 2\right) \left(x - 3\right)} \Leftrightarrow \frac{A \left(x + 2\right) \left(x - 3\right) + B x \left(x - 3\right) + C x \left(x + 2\right)}{x \left(x + 2\right) \left(x - 3\right)}$

i.e. $4 {x}^{2} - 7 x - 12 \Leftrightarrow A \left(x + 2\right) \left(x - 3\right) + B x \left(x - 3\right) + C x \left(x + 2\right)$

Now putting $x = 0$, we get $- 6 A = - 12$ or $A = 2$

putting $x = 3$, we get $15 C = 3$ or $C = \frac{1}{5}$

and putting $x = - 2$, we get $10 B = 16 + 14 - 12 = 18$ or $B = \frac{9}{5}$

Hence $\frac{4 {x}^{2} - 7 x - 12}{x \left(x + 2\right) \left(x - 3\right)} = \frac{2}{x} + \frac{9}{5 \left(x + 2\right)} + \frac{1}{5 \left(x - 3\right)}$