# How do you express 3/ (x^2-6x+8) in partial fractions?

Jul 26, 2016

$\frac{3}{{x}^{2} - 6 x + 8} = \frac{3}{2 \left(x - 4\right)} - \frac{3}{2 \left(x - 2\right)}$

#### Explanation:

$\frac{3}{{x}^{2} - 6 x + 8} = \frac{3}{\left(x - 4\right) \left(x - 2\right)} = \frac{A}{x - 4} + \frac{B}{x - 2}$

Then, using Heaviside's cover up method, we find:

$A = \frac{3}{4 - 2} = \frac{3}{2}$

$B = \frac{3}{2 - 4} = - \frac{3}{2}$

So:

$\frac{3}{{x}^{2} - 6 x + 8} = \frac{3}{2 \left(x - 4\right)} - \frac{3}{2 \left(x - 2\right)}$

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Spelled out a little slower:

Given:

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

Multiply both sides by $\left(x - 4\right)$ to get:

$\frac{3}{x - 2} = A + B \frac{x - 4}{x - 2}$

Then let $x = 4$ to find:

$\frac{3}{4 - 2} = A + 0$

hence $A = \frac{3}{2}$

Similarly find $B = - \frac{3}{2}$

Note that this is actually division by $0$ in disguise, but is justifiable in terms of limits as $x \to 4$ or $x \to 2$.