How do you express $\frac{2 x - 2}{(x - 5) (x - 3)}$ $(2x-2)/((x-5)(x-3))$ in partial fractions?

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Answer 1 · 2016-02-24T11:35:24

$\frac{2 x - 2}{(x - 5) (x - 3)} = \frac{4}{x - 5} - \frac{2}{x - 3}$ $(2x-2)/((x-5)(x-3))" "=" " 4/(x-5)- 2/(x-3)$

Write as: $\frac{2 x - 2}{(x - 5) (x - 3)} = \frac{A}{x - 5} + \frac{B}{x - 3}$ $" " (2x-2)/((x-5)(x-3))" "=" " A/(x-5)+B/(x-3)$

Thus: $\frac{2 x - 2}{(x - 5) (x - 3)} = \frac{A (x - 3) + B (x - 5)}{(x - 5) (x - 3)}$ $" " (2x-2)/((x-5)(x-3)) =(A(x-3)+B(x-5))/((x-5)(x-3))$

So: $2 x - 2 = A (x - 3) + B (x - 5)$ $" "2x-2" "=" "A(x-3)+B(x-5)$

$2 x - 2 = A x - 3 A + B x - 5 B$ $2x-2" "=" "Ax-3A+Bx-5B$

Collecting like terms

$2 x - 2 = (A + B) x - 3 A - 5 B$ $2x-2" "=" "(A+B)x -3A-5B$

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Comparing LHS to RHS

$2 x = (A + B) x so A + B = 2$ $2x=(A+B)x" so "A+B =2" "$ ............................(1)

$- 2 = - 3 A - 5 B so B = \frac{2 - 3 A}{5}$ $-2=-3A-5B" so "B=(2-3A)/5" "$ ...................(2)

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Substitute (2) into (1) giving:

$A + \frac{2 - 3 A}{5} = 2$ $A+(2-3A)/5=2$

$\frac{5 A + 2 - 3 A}{5} = 2$ $(5A+2-3A)/5=2$

$2 A = (2 \times 5) - 2 = 8$ $2A=(2xx5)-2 =8$

$A = 4$ $A=4" "$ .......................................(3)

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Substitute (3) into (1) giving:

$4 + B = 2$ $4+B=2$

$B = - 2$ $B=-2$
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$\frac{2 x - 2}{(x - 5) (x - 3)} = \frac{4}{x - 5} - \frac{2}{x - 3}$ $color(blue)((2x-2)/((x-5)(x-3))" "=" " 4/(x-5)- 2/(x-3))$

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Check: $4 (x - 3) - 2 (x - 5) = 4 x - 12 - 2 x + 10 = 2 x - 2$ $4(x-3)-2(x-5) = 4x-12-2x+10 = 2x-2$

Matching original numerator so ok!

How do you express (2x-2)/((x-5)(x-3))2x−2(x−5)(x−3)(2x-2)/((x-5)(x-3)) in partial fractions?