How do you express $1/ [ (x-1)(x+2)(x-3)]$ in partial fractions?

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Answer 1 · 2018-06-25T20:30:48

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$1/( (x-1)(x+2)(x-3) )$

$= A/(x-1) + B/(x+2) + C/(x-3)$

$=> 1 = A(x+2)(x-3) +B(x-1)(x-3) +C(x-1)(x+2)$

Let $r = -2$ Solving to give $B= 1/15$

Let $r = 3$ Solving to give $C =1/10$

Let $r = 1$ Solving to give $A = -1/6$

Hence our new partial fraction form is...

$= - 1/(6(x-1)) + 1/(15(x+2)) + 1/(10(x-3))$

How do you express 1/ [ (x-1)(x+2)(x-3)] 1/ [ (x-1)(x+2)(x-3)] in partial fractions?