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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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