9x-20/(x+6)^2 partial fraction decomposition

Question

asked Nov 5, 2020 93.3k views

2 Answers

Kusalananda · Answer 1 · 2020-11-12T06:53:43+0000

We're looking for
a,b such that

$(9x-20)/((x+6)^2)=\frac a{x+6}+\frac b{(x+6)^2}$

Multiplying both sides by
(x+6)^2 gives us

9x-20=a(x+6)+b

Notice that when
x=-6 , the term involving
vanishes and we're left with

$9(-6)-20=b\implies b=-74$

Then

$9x-20=a(x+6)-74=ax+6a-74\implies a=9$

so that

$(9x-20)/((x+6)^2)=\boxed{\frac9{x+6}-(74)/((x+6)^2)}$