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

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Answer 1 · 2018-06-25T20:30:48

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$\frac{1}{(x - 1) (x + 2) (x - 3)}$ $1/( (x-1)(x+2)(x-3) )$

$= \frac{A}{x - 1} + \frac{B}{x + 2} + \frac{C}{x - 3}$ $= A/(x-1) + B/(x+2) + C/(x-3)$

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

Let $r = - 2$ $r = -2$ Solving to give $B = \frac{1}{15}$ $B= 1/15$

Let $r = 3$ $r = 3$ Solving to give $C = \frac{1}{10}$ $C =1/10$

Let $r = 1$ $r = 1$ Solving to give $A = - \frac{1}{6}$ $A = -1/6$

Hence our new partial fraction form is...

$= - \frac{1}{6 (x - 1)} + \frac{1}{15 (x + 2)} + \frac{1}{10 (x - 3)}$ $= - 1/(6(x-1)) + 1/(15(x+2)) + 1/(10(x-3))$

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