In the previous reading, we learned how to factor quadratics with no leading coefficient. Here we'll talk about how to factor quadratics in which the leading coefficient is something other than 1. Here's an example:

Example 1: Factor 2x² + 7x + 6.

Solution: We start out almost the same way as before, except now our two important numbers are +7 and +12. Can you see where I got those two numbers? +7 is the coefficient in the middle term, and +12 is the product of the first and last coefficient.

Just as before, we're going to look for two numbers that add to 7, and multiply to 12. But beware! When we find them, we're going to do something very different with them!

3 + 4 = 7
3 x 4 = 12

Here's what we're going to do: since 3 + 4 = 7, we can conclude that 3x + 4x = 7x. Agreed? Okay, so I'm going to take the 7x in our original expression and replace it with 3x + 4x.

2x² + 3x + 4x + 6.

Now don't combine like terms, because that'll get you right back where you started. Instead, factor whatever you can out of the first two terms, and factor whatever you can out of the last two terms:

x(2x + 3) + 2(2x + 3).

Conveniently, both parts of that expression have (2x + 3), which lets us do something quite beautiful:

(2x + 3)(x + 2).

And that's our answer!

Example 2: Factor 2x² + x - 10

Solution: We need two numbers that add to +1, and multiply to -20 (-20 = 2 x -10).

-4 + 5 = 1
-4 x 5 = 20

So now we write x as -4x + 5x in our original expression:

2x² -4x + 5x - 10

Grouping and factoring gives us:

2x(x -2) + 5(x - 2)

So the entire expression factors into

(x - 2)(2x + 5)

If you have doubts, you can always "FOIL" it out to see if you get the same quadratic that you started with.