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..