Using inverse Laplace transform to solve differential equation The differential equation is as follows- 
$$\frac{d^2 x}{dt^2} + 5 \frac{dx}{dt} + 6x = e^t $$
I use laplace transform to make it to become - $$X(s) = \frac{1}{(s-1)(s+3)(s+2)}$$
where $X(s)$ is the Laplace transform of $X(t)$
So now I am trying to find $X(t)$ using inverse transform.
From partial fractions- 
$X(s) = \frac{1}{(s-1)(s+3)(s+2)} = \frac{A}{s-1} + \frac{B}{s+3} + \frac{C}{s+2} $
Numerator - $ 1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3) $ 
I am stuck from here on how to carry on this partial fraction 
Can I sub all s values to be 0 ? 
For example 
$1 = A(0+3)(0+2)$
$1= B(0-1)(0+2) $
$1 = C (0-1)(0+3) $ 

So, if we have that
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
We can expand everything to get
$$1=A(s^2+5s+6)+B(s^2+s-2)+C(s^2+2s-3)$$
$$1=s^2(A+B+C)+s(5A+B+2C)+1(6A-2B-3C)$$
So we must have that $A+B+C=0$, $5A+B+2C=0$ and $6A-2B-3C=1$.
Or alternatively:
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
If we substitute $s=1$ we get that
$$1=A(1+3)(1+2)$$
With $s=-2$ we get that
$$1=C(-2-1)(-2+3)$$
And with $s=-3$ we get that
$$1=B(-3-1)(-3+2)$$
And here is a video about the partial fraction decomposition, you might find it helpful.
