Question

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

Answer 1

`(2x-2)/((x-5)(x-3))" "=" " 4/(x-5)- 2/(x-3)`

Explanation:

Write as:`" " (2x-2)/((x-5)(x-3))" "=" " A/(x-5)+B/(x-3)`

Thus: `" " (2x-2)/((x-5)(x-3)) =(A(x-3)+B(x-5))/((x-5)(x-3))`

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

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

Collecting like terms

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

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Comparing LHS to RHS

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

`-2=-3A-5B" so "B=(2-3A)/5" "`...................(2)

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Substitute (2) into (1) giving:

`A+(2-3A)/5=2`

`(5A+2-3A)/5=2`

`2A=(2xx5)-2 =8`

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

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Substitute (3) into (1) giving:

`4+B=2`

`B=-2`
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`color(blue)((2x-2)/((x-5)(x-3))" "=" " 4/(x-5)- 2/(x-3))`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check: `4(x-3)-2(x-5) = 4x-12-2x+10 = 2x-2`

Matching original numerator so ok!