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Polynomial Equations in Factored Form

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Polynomial Equations in Factored Form

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Everything we've done with polynomials so far has been
either adding or multiplying them.
But what if we wanted to go the other way?
What if we actually wanted to factor a polynomial?
Let me show you what I mean.
So let's have the polynomial 3x to the
third power minus 21x.
When I say factor it, I essentially want to break it
up into the product of, maybe polynomials, it could be a
mononomial, which is just one term.
And to do that over here, we could factor out a 3x here.
So if you factor out a 3x this is equal to 3x times-- what's
3x to the third divided by 3x?
Let me write this way actually.
This is the same thing as 3x times 1/3 x times 3x to the
third minus 21x.
Notice, 3x times 1/3x, I'm not changing the number.
This is 3x/3x, this is 1.
But the reason why I picked 3x is because I saw that both of
these terms are divisible by 3x.
And you're going to see that when I divide
both of them by 3x.
So this is going to be the same thing as 3x times 3x to
the third over 3x minus 21x/3x.
And now this is going to be equal to 3x.
What's this first term equal to?
The 3's cancel out.
x to the third divided by x.
We learned in our exponent rules, you could just subtract
the exponents there.
That's x times x times x divided by x, just going to be
x times x, or just x squared.
So it's going to be, this first term here is just going
to be an x squared.
And then the x's cancel out here.
21 divided by 3 is 7.
x squared minus 7.
And if you multiply them out, you're going to see you're
going to get 3x to the third minus 21x.
Now I went through all of these steps really, just to
show you exactly what I'm doing.
But as we progress, you're going to see that your brain
will just automatically say, hey, both of these are
divisible by 3x, let me factor out a 3x, and then divide both
of these terms by 3x to get x squared minus 7.
Let's do another one.
Let's say we had 4x to the third plus 10x
squared minus 2x.
So when I look at this, immediately what pops out is
that all of these terms are divisible by 2, and all of
them are divisible by x.
So let me factor out a 2x.
So this is equal to 2x times-- and when we factor out a 2x
let's divide each of these terms by 2x.
4x third divided by 2x.
4 divided by 2 is 2.
And then x third divided by x is x squared.
And then we have plus 10x squared divided by 2x.
10 divided by 2 is 5.
x squared divided by x is x.
And then finally, negative 2x divided by 2x is
just negative 1.
And you know, if you don't believe me, just
multiply this out.
Distribute the 2x and you will get exactly this.
So all we see, we try to look for something that's divisible
into every term.
We factor it out, factoring it out, we put it outside of a
parentheses, so to speak.
And then we divide each of those terms by that factor.
We did that explicitly up here.
Now you're probably saying, hey Sal, why are we even
worried about doing this?
And the answer is, is it really helps us solve a lot of
certain types of polynomial equations.
Let me show you what I mean.
Let's say we have the polynomial equation 9x squared
is equal to 27x.
And we want to make sure that we get all of the x's that
satisfy this equation.
So let me subtract 27x from both sides.
So I get something that equals 0.
And this will be a general trend whenever you're doing
polynomial equations, very oftentimes you will.
The left-hand side becomes 9x squared minus 27x.
The right-hand side becomes 0.
And now over here, we can see that both of these terms are
divisible by 9x, right?
27 is divisible by 9, so is 9.
And they're both divisible by x.
So let's factor out a 9x.
So 9x squared divided by 9x is just an x.
And then 27x divided by 9x.
That's negative 27/9 is negative 3.
And then the x divided by x is just a 1.
So this is 9x times x minus 3-- you can multiply it out to
verify that these are the same thing-- is going
to be equal to 0.
Now this part is interesting.
When I take the product of two numbers, and I say they're
equal 0-- let me do a little aside here.
If I were tell you that a times b is equal to 0, what
does that tell us?
That tells us that either a or b or both are 0.
At least one of these two have to be 0.
Maybe both of them are going to be equal to 0.
So we're doing the exact same thing here.
We're taking the product-- you can view that as a, you could
view that as b-- we're taking the product of two numbers and
saying that that product is equal to 0.
So that tells us that either 9x is going to be equal to 0,
or x minus 3 is going to be equal to 0.
Either of these, or both of these, are going to equal 0.
This we know how to solve.
Divide both sides by 9.
You get x is equal to 0.
This we know how to solve.
Add 3 to both sides.
And you're left with x-- because these cancel out-- is
equal to 3.
So this equation here actually had two solutions. x could be
0, and you can verify that.
0 is equal to 0.
Or x could be equal to 3.
9 times 9 is the same thing as 27 times 3.
And you might have been able to figure that out in your
head, but when you do it this way you make sure that you get
all of the solutions to the problem.
Let's do one more like this.
Let's say I had 4a squared plus a is equal to 0.
Once again, we can factor an a out.
Both of these terms are divisible by a, so this is the
same thing as a times 4a plus 1, right?
When I factor an a out, I took the a out of the parentheses,
and then I divide each of these terms by a.
And that's going to be equal to 0.
So that means either a is equal to 0, or 4a plus 1 is
equal to 0.
Well, that's one solution right from the get go.
And this one right here, we could
subtract 1 from both sides.
We get 4a is equal to negative 1.
And then you could divide both sides by 4.
And we get a is equal to negative 1/4.
Now I said that was the last one, let me do one more.
Because I want to show you, here, I'm factoring and then
solving, but I want to show you how useful a factored
polynomial is.
So I'm going to start with one that's already factored.
So say I had x minus 5 times 2x plus 7 times 3x minus 4, is
equal to 0.
And if you're bored, you can actually multiply these out
and you'll find that you get a third degree polynomial.
But if you want to solve that being equal to 0,
it'd be very difficult.
But this, we're going to see, is not too bad.
As we said, if we have three numbers, if we have a times b
times c equal to 0, that means that at least one of these are
0, maybe more than one.
Either a or b or c, or some combination of them,
are equal to 0.
So once again, I have three numbers.
This could be viewed as an a.
This could be viewed as a b.
This could be viewed as a c.
I'm multiplying them and they're equal to 0.
So that means that either x minus 5 is equal to 0.
Or 2x plus 7 is equal to 0.
Or 3x minus 4 is equal to 0.
And we can solve each of these independently.
Add 5 to both sides of this equation, and you get x is
equal to 5.
So that's one solution to this equation.
Over here, subtract 7 from both sides.
You get 2x is equal to negative 7.
Divide both sides by 2.
You get x is equal to negative 7 over 2.
That's another solution.
And then finally, add 4 to both sides of this equation.
You get 3x is equal to 4.
Divide both sides by 3.
You get x is equal to 4/3.
So this third degree polynomial, if you were to
multiply it out, actually had three solutions to that
polynomial equalling 0.
And we're going to see, these are actually called roots of
the polynomial, the x values that make that polynomial
equal to 0.
But hopefully, you've just gotten an appreciation for
one, how to factor a polynomial in this video, and
also, why it's useful for solving polynomial equations.

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Polynomial Equations in Factored Form

Polynomial Equations in Factored Form