Question

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

Answer 1

We're looking for $a,b$ such that

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

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

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

Notice that when $x=-6$, the term involving $a$ 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

$\dfrac{9x-20}{(x+6)^2}=\boxed{\dfrac9{x+6}-\dfrac{74}{(x+6)^2}}$

Answer 2

Answer:

Find the partial fraction decomposition of 6x+8/x2-9x+8

Step-by-step explanation: