How do you express $(x-9)/[(x+5)(x-2)]$ in partial fractions?

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Answer 1 · 2016-03-02T14:27:19

$(x-9)/((x+5)(x-2))=2/(x+5)-1/(x-2)$

solution: set up the working equations first by using variables A and B.

$(x-9)/((x+5)(x-2))=A/(x+5)+B/(x-2)$

using LCD-least common denominator $=(x+5)(x-2)$ simplify

$(x-9)/((x+5)(x-2))=(A(x-2)+B(x+5))/((x+5)(x-2))$

$(x-9)/((x+5)(x-2))=(Ax-2A+Bx+5B)/((x+5)(x-2))$

rearrange the numerators

$(1*x-9*x^0)/((x+5)(x-2))=((A+B)*x+(-2A+5B)*x^0)/((x+5)(x-2))$

We can have the equations now to solve for A and B

$A+B=1$
$-2A+5B=-9$

Solve for A and B simultaneosly

$7B=-7$

$B=-1$ and $A=2$

God bless....I hope the explanation is useful..

How do you express (x-9)/[(x+5)(x-2)] (x-9)/[(x+5)(x-2)] in partial fractions?