express $\frac{42}{ s^2 + 7s}$ as a partial fraction express this rational function as a partial fraction
$$ \frac{42}{s^{2} + 7s}$$
So i factories 
$$ \frac{42}{s(s+7)}   =  \frac{A}{s}   +  \frac{B}{s +7} $$
$$42 =  A(s+7) + Bs$$
let s equal $-7$ 
$$42 = A(-7+7)   - 7b$$
$$b = -6 $$
To find $A$ let $s=0$
$$42 = A(0+7) +  B\cdot 0$$
$$42 = 7A $$
$$A = 6$$
$$\frac{42}{s^{2}+7s}   =  \frac{6}{s}   - \frac{6}{s +7}$$
Have I done this right? The example I have in my text always have nice factored base units. please help
 A: If you wonder whether your answer is right, you just go the other way and check:
$$
\frac{6}{s} - \frac{6}{s+7} \\= \frac{6(s + 7)}{s(s+7)} - \frac{6s}{s(s+7)} \\= \frac{6s + 42 - 6s}{s(s+7)} \\= \frac{42}{s^2 + 7s}
$$
It is very often a whole lot easier to check your answer than it is to find it in the first place. Making a habit of doing so is probably the most important single lesson if a good grade is what you're after. That does not make it a substitute for understanding what's going on, but it's a very good start.
As to whether what you've done is correct, firstly, it is very difficult to get the right answer using incorrect methods. If you follow the same procedure with two or three other, similar problems and still get the correct answer, you can be almost completely cetain that you're doing things the right way.
What you've done seems very reasonable. In this specific setting I haven't seen people letting $s$ equal $-7$ and $0$ like you've done, but it is correct and maybe even simpler to do it that way. At least as long as there are only two terms.
