Why does my teacher say this isn't a valid partial fraction decomp?

[wise tree]

I think you would want to put

$\frac{x^3}{(x+1)^2(x-5)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}+\frac{C}{(x-5)}$

[pliant oyster]

Helixan

[wise tree]

Not

$\frac{x^3}{(x+1)^2(x-5)}-\frac{A}{(x+1)}+\frac{B}{(x+1)^2}+\frac{C}{(x-5)}$

[pliant oyster]

Helixan

[wise tree]

Then just solve for A, B, and C

[jovial sage]

This isnt valid for a reason

You forgot the integer part

the numerator is of degree 3

The denominator is of degree 3

Therefore there is an integer part

The decomposition is of the form

A + B/(x+1) + C/(x+1)^2 + D/(x-5)

A is not equal to zero

If you want to check this yourself, take the limit as x goes to infinity in your decomposition

The right side you wrote goes to zero

But the left side goes to 1

So clearly this is not the right decomposition

(this limit method allows you to find that A is actually equal to 1)