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>> Alright here is a quick tutorial on factoring quadratic trinomials when the leading coefficient
is a 1. Sounds much worse than it is. Really all I'm saying is how do you factor something
that kind of looks like this this X squared + 5x + 6 thing. And when I say looks like
this well I mean there's three terms, hence the word trinomial and the largest degree,
the largest exponent is a 2 and its quadratic and the leading coefficient is a 1. It juts
is 1X squared here, doesn't say 2X squared or 3X squared, or -5X squared, it's just 1X
squared. If that's not a 1 there we'll have to use a different method. But if you do have
a 1 there and you have a quadratic trinomial you can try to factor it. It won't always
factor but it'll be the easiest way to do what you need to do this stuff for later on.
So you should always check to see if it factors first and all of these will factor. So what
you do I just kind of named some terms so what I'm saying is the coefficient on the
X term, here's my X term right here the coefficient is the number in front of the X, let's call
that B. I'll even write that for this first one. B here would be 5 and the constant term,
the coefficient on the constant term, I think that's a little redundant is this 6 right
here, C is equal to 6. And so what you want to do in step 2 is you want to find 2 numbers
that multiply to 6 in this case and add up to 5. You might be able to just do that in
your head but sometimes the numbers will get bigger and it'll be a little bit more difficult
to do that. So the way I do it, something that'll help when these numbers get larger
is look at the C, don't look at the B, don't look at the 5 just look at the C and find
all possible numbers really all possible integers cause I only want to deal with whole numbers
that multiply to 6. In this case that's either 1 times 6 and 2 times 3. You might be like
wait what about 12 times a half, yeah that works but I only want to deal with integers
with whole numbers. Ok what about -2 times -3? You're right I only want to deal with
positive integers and I'll deal with the signs later. Some people will list -1, -6, -2, -3
also here. But for me it makes the list too long so I don't even bother listing them.
I just keep in mind that there could also be -1's here. So anyways I have this list
right here and what I want to do is find two numbers that add up to 5 and multiply to 6.
Well here's two that multiply to 6 but they add up to 7. Here's two that multiply to 6
and they add up to 5, it looks like those are the ones that I want. So then what I can
do is I can factor this, I can express my answer as X + the first number. So that first
number was 2 times X + that second number which is 3 and what I've done is I've factored
this. If you want you can check your answer by multiplying these back together. Let's
see maybe I'll do that down here. The way you multiply polynomials like this a lot of
people like to foil, which works although I prefer to use this box method here. Foiling
works out great for examples like this when you're doing X+2 times X+3. The problem is
when you have longer polynomials when they're not just binomials foiling won't work, the
acronym doesn't make sense so I like to use this box method cause it'll work for any polynomials
you're trying to multiply. All you do is you write one of the polynomials across this way,
it could be as long as you want, and one of them down this way and then fill in each of
the boxes. X times X is X squared. X times 2 is 2X, X times 3 is 3X and 3 times 2 is
6 and then all you got to do is add up everything that's inside the boxes. X squared and combine
like terms here I'll write it out in steps. Add up everything in the boxes and then note
that you can combine like terms you got 2X and you got a 3X so really what you have is
5X. And sure enough X squared + 5X+6 is what I was supposed to get, I just checked my answer.
You don't have to check your answer that's why I wrote optional here but I just want
to show you how it works. So I'll do a few more and kind of work through these a little
bit faster. Same idea here I want two numbers that multiply to -6 this time and add to 5.
What I would do is I would multiply all of the numbers or list all of the numbers that
multiply to 6, which you might be like wait but you don't want 6, your C is not equal
to 6, your C is equal to -6. It's like right but I know when I look at this I just know
that either the 1 has to be negative or the 6 has to be negative. So when I look at this
list I'm thinking -1 + 6, or +1 -6. -2 +3 or +2 -3. I want them to add to +5 so what
I'll need is +6 and -1. See that +6 and -1 add to +5 multiply -6. So when I'm writing
this I'll say +6 and -1. So the way I write it up is like this, of course you have some
freedom and you can do this however you want, just be a little bit careful with those signs
there. If you want to check your answer you can, just like we did down here. Moving onto
the third one, you might have a problem when you see this third one cause I did say that
the leading coefficient has to be a 1. Don't forget that you can always factor out the
greatest common factor before you even get started. So when you see this thing right
here what I was hoping you'd notice was that 3 is common to all three of our terms here.
Before we even start we can factor out a 3, if I take out a 3 from 3X squared I'm left
with X squared. If I take out a 3 from -6X I'm left with -2X and if I take out a 3 from
-45 I'm left with -15 and so what we've now done is we've taken care of the 3 and what
I'm hoping you'll see is that in these parenthesis here we have a quadratic trinomial whose leading
coefficient is 1. So we can factor that quadratic trinomial, we need two numbers that multiply
to -15 and add to -2 and I think those will end up being -5 and +3. If it helps you to
list them all go for it but if you can just see it you don't even have to list them. Alright
and in this last example it's probably a little bit more difficult than something I should
be doing on a tutorial video but maybe it's worth showing. This doesn't quite fit the
type of problem we're talking about. I guess example 3 didn't really fit this type of problem
until we factored out a 3 and made things work. Kind of the same idea with example 4
here. This isn't a quadratic trinomial because it's a forth degree polynomial here, not a
second degree polynomial. So at first glance this doesn't look like a problem that'll fit
this mold but it turns out that it does if you do a little substitution. If you change
all of the X squareds into a different letter maybe Y then what you have here is really
X squared squared. X to the fourth is the same as X squared squared so I can write it
as Y squared + 7 and then instead of writing X squared here I'll write Y because Y is X
squared + 6. So really what I can do is I can do this little substitution and now what
I have here is a quadratic trinomial whose leading coefficient is 1. Yeah it doesn't
Xs anymore it has Ys but that's ok I can deal with that. So what I'd want to do is find
two numbers that multiply to 6 and add to 7, +6, +7 in this case. Looking at my list
here let's see +1 and +6 looks like it'll do the trick. So this factors into Y+1 times
Y+6 but you might have a problem with that you're like wow how does X to the fourth +
7X squared + 6 factor into this? Well you're right it doesn't I mean it does and it doesn't.
We don't want it to have Ys in it; we want it to have Xs in it. But we can fix that.
We know that Y is equal to X squared, that was the substitution we did here. So we could
substitute back, we can replace all of these Ys with X squares. So really X to the fourth
+ 7X squared + 6 is equal to X squared + 1 times X squared + 6 if you do this little
substitution method. So I just want to show you one more little example, this last one
probably a little bit beyond what you should be doing if you're just learning this stuff
now but maybe something to come back to later when you get really good at the rest of these
factoring methods. I guess that's the end of this video.