Factoring Quadratic Expressions
Let's look closely at that last problem we just did. We said we could factor x2 + 8x + 15 as (x + 3)(x + 5). What do the numbers from the trinomial (the product) have to do with the numbers from the two binomials (the factors)? Two things, actually!
3 + 5 gives us 8, which was the coefficient from the middle term from the trinomial. And 3 × 5 gives us 15, the constant on the end of the trinomial. This means that if you can find two numbers that add to the middle term of your trinomial and multiply to the constant on the end, those are going to be the two numbers in your factored expression.
Let's see if we can apply this idea to a different problem; maybe factor x2 + 11x + 24. So, for our method to work, we need to find two numbers that add up to the middle coefficient (11) and that will also multiply to the constant on the end (24). Sometimes the answer is going to be obvious, but sometimes it will be hard to see. When it's hard to see right away, I like to write out all the different ways we can multiply to get the constant on the end and then see which one of those will add up to the number in the middle. Let's do that here to practice. We will always be able to do 1 times the number. In this case, (1 × 24). 24 is even, so 2 goes into it as well, so (2 x 12), I think 3, (3 x 8) and 4, (4 x 6), and that's it. I could do the other ones, (6 × 4), (0 × 3), but those are just the same, backwards, so those don't matter. So we have these four options - which ones add up to 11? Hey, 3 and 8!
That means that x2 + 11x + 24 factors into (x + 3)(x + 8). If you'd like to check your answer, you can quickly multiply it out and make sure it ends up back where we started! If we do that here, sure enough, we're all good!
One last example: factor 2x3 - 6x2 - 80x. This one looks quite a bit different. x3? We don't know how to do that. Sometimes, there are ways (maybe you'll learn those in a different class), but if you see that in this class, there's probably something tricky going on. Sure enough, looking at each term in our trinomial here, there's an x in it. Whether it's an x3, an x2 or just a plain old x, there's an x in each term, which means I can divide it out (or factor it out) of each term by doing the distributive property backwards.
When I pull that x out of each term, I can now write it in front and leave what I'm left with after I divide it out on the inside of the parentheses. So, I no longer have 2x3, but I have 2x2; instead of 6x2, I have 6x; and instead of 80x, I just have plain old 80. But now there's another different thing going on. There's a 2 in front of my x2. In a later lesson, we'll get to examples where you're going to have to deal with that 2, but it's also worth checking to see if I can do the same thing. Can I divide that 2 out of each term, just like I divided the xs out? Sure enough, again, we're going to be able to do that because 2, 6 and 80 are all even numbers and therefore divisible by 2.
Doing the distributive property backward and factoring a 2 out of each, bringing the 2 in front and then putting what I'm left with in parentheses, gives me 2x(x2 - 3x - 40). Now, even though I have a 2x out front, what's inside the parentheses is something we know how to do; it's a trinomial, it's got 1x2 and I can try to use my pattern.
That pattern says I need two numbers that add up to -3 and multiply to -40. So we actually have some negatives this time, but that won't actually change too much. Again, if the two numbers aren't obvious to me right away, I can write out all the factors of 40, like so. But since we need to multiply to -40, one of the numbers is going to have to be positive and one's going to have to be negative. So now we're looking for a pair of numbers that have a difference of 3, not a sum. Which set of two can I subtract to get -3? It looks like 5 - 8 would do the trick. Because I'm doing a +5 minus an 8, the two numbers that are going to work are +5 and -8.
That means I factor the x2 - 3x - 40 part as (x + 5)(x - 8), and the 2x is still hanging out in the front, so we get 2x(x + 5) (x - 8) as the answer.
The patterns we've talked about here are going to help you factor quadratic expressions that have a leading coefficient of 1. Like I mentioned before, sometimes we're going to have a 2 or 3 in front of the x2's. For those, you're going to have to use a different method. To find out about that method, you'll have to check out a future lesson.