Hi Rylee,

Factoring can be really hard to get a handle on. I know for me it took me a few years before I really got the hang of it!

This one, though, is easy. You don't have a coefficient at all on that b^{2} -- that means that when you factor this expression, you'll get something that looks like (b+m)(b+n), where the
*m* and *n* are constants of some kind. How do you find *m* and
*n*, though?

Well, expand that expression out with FOIL and compare it to your question. (b+m)(b+n) turns into b^{2} + bm + bn + mn, which can be rewritten with the Distributive Property as b^{2} + (m+n)b + mn.

Now, to compare it to b^{2} + 3b - 18: rewrite it as b^{2} + 3b** + -**18, so we can compare apples to apples, and match up the similar parts.

b^{2} & b^{2}: that's all good, we don't have to worry about that

3b & (m+n)b: okay, so we know that 3 = m+n.

-18 and mn: okay, so we know that -18 = m*n.

So we've discovered that m and n, the two constants in our factored expression, have to
**add up to 3** and **multiply to -18**. From experience, I can tell you that since they multiply to a negative number, exactly one of those two numbers has to be negative. Since they add to a positive, though, the bigger number will be positive, and the smaller will be negative. (If you want more explanation on this conclusion, just ask!)

So, make a list of factors of 18: 1&18, 2&9, 3&6. Okay, now apply that last conclusion we just made to these factors -- in each pair, make the smaller number negative. We have -1&18, -2&9, and -3&6.

All of these pairs multiply to -18 -- but which one adds to 3?

Find the pair that adds to -3, and plug those numbers in as m and n. Then, check: does your answer of the form (b+m)(b+n) multiply out to b^{2}+3b-18?

This can be used to factor any polynomial of degree 2 with no coefficient on its first term (all that just means, if you have an x^{3} or p^{7} in there, or there's a k^{2} but it's got a coefficient of 14 and is actually 14k^{2}, this won't work). If your polynomial is x^{2} + bx + c, find two numbers that add to
*c* and multiply to *b*, and then you can put them into your factors, (x+m) and (x+n). Easy!

If you have any questions, let me know :)