-8x^{2}+25x-18
Hi Robert,
Let's think about what you need to do to solve this kind of problem. You may be familiar with what to do when you see something like this:
Let's think about what you need to do to solve this kind of problem. You may be familiar with what to do when you see something like this:
x^{2}+5x+6
We need to look for two numbers- we'll call them a and b. a and b need to multiply to 6, and add to 5. Why? It will make sense when we find them, trust me. We'll start by finding the factors of 6- we know they're 2 and 3. We're lucky, because 2 and 3 also add to 5. So, we can write the factored form of this polynomial as:
(x+2)(x+3)
How can we check? Well, if you do FOIL, you'll get x^{2}+ 2x +3x +6- which comes out to x^{2}+5x+6.
Now that we've done a simple example, let's move to the one you're asking about. We're going to do a similar process to start, but we're going to multiply the leading coefficient (-8) with the constant (-18). We do that and get 144. This means that our numbers, a and b, need to add to 25 and multiply to 144.
To find out which two numbers will add to 25 and multiply to 144, the way I would recommend is to try and factor 144 and see which numbers can add to 25.
1 + 144= 145. Nope. Too big
2 + 72= 74. Still too big
3 + 48= 51. No
4 + 36= 40 Nuh-uh.
6 + 24 = 30 We're getting warmer now...
8 + 18 =16 Warmer...!
9 + 16= 25 BINGO!
So, our numbers are 9 and 16. Now, we're going to rearrange the expression a little bit:
-8x^{2}+16x+9x-18
What have I done here? I've basically rewritten 25x as 16x+9x. No harm done right? It's still technically correct. Now, why have I done this? This is to help us with factoring the expression. Let's look at the left side for a minute
-8x^{2}+16x+9x-18
These terms have something in common, don't they? They have 8x in common. So, I can pull it out of those two terms. For a little extra cleanliness I'm also going to take out the negative from the first term. This rewrites the expression as follows:
-8x(x-2)+9x-18
Now, I'm going to do the same thing to the other side. Let's have a look at it:
-8x(x-2)+9x-18
These two terms have something else in common too, don't they? They have 9 in common! So, I'm going to factor it out of the right side:
-8x(x-2)+9(x-2)
Let's look at what we have:
-8x(x-2)+9(x-2)
You'll notice that both parts of this equation have (x-2) in common, right? This means that I can factor (x-2) out of both terms- it seems a little freaky to do, but it's allowed-we're basically treating it like any other number or single variable x. When we do that, we get:
(x-2)(-8x+9)->you can also write this part as (9-8x) if you want to.
To test it, we can FOIL it out:
-8x^{2}+9x+16x-18= -8x^{2}+25x-18
It works!
This method of factoring is known as Factoring by grouping, and we use it to help us with factoring quadratic expressions with a leading coefficient (the coefficient in front of the x^{2} term) that isn't 1, and with higher degree polynomials (x^{3} or x^{4}) when it can be helpful-but you probably won't see that for a while.
So, how do we do it for future problems? Let's review the steps:
1. Multiply the leading coefficient with the constant. Include negatives if they're there.
2. Find two numbers, a and b, that add to the middle term but multiply to the number you found in step 1. Do this by factoring the number you found in step 1 and finding two factors that will add to the middle term.
Keep in mind, sometimes one or both of your factors have to be negative to get the sum you're looking for.
3. Rewrite the expression, replacing the middle term with the two numbers you found, and write them in such a way that they're close to numbers that they have common factors with.
4. Factor out the common terms/factors from each half of the expression. You should have two of the same binomial (like x-2 in our case) left over.
5. Factor out that binomial and write what's left over as a new binomial.
6. Pat yourself on the back for a job well done :)
I hope this helps answer your question and helps you do similar problems in the future. Do let me know if you have any questions!