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

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Answer 1 · 2016-07-26T12:56:26

$3/(x^2-6x+8) = 3/(2(x-4))-3/(2(x-2))$

$3/(x^2-6x+8) = 3/((x-4)(x-2)) = A/(x-4)+B/(x-2)$

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

$A=3/(4-2) = 3/2$

$B=3/(2-4) = -3/2$

So:

$3/(x^2-6x+8) = 3/(2(x-4))-3/(2(x-2))$

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Spelled out a little slower:

Given:

$3/((x-4)(x-2)) = A/(x-4)+B/(x-2)$

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

$3/(x-2) = A+B(x-4)/(x-2)$

Then let $x=4$ to find:

$3/(4-2) = A + 0$

hence $A=3/2$

Similarly find $B=-3/2$

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

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