WEBVTT
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What about you?
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Alright guys ready?
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What we're going to do is we're going to work on using the division of two polynomials.
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So for division of two polynomials, what we're going to do is we're going to set it up just like any other long division problems.
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If you guys can remember when we're doing long division, let's see if I have room over here.
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Seven goes into what?
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24.
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Seven.
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So you can say seven goes into 14, goes into one time, and then I multiply the one time seven.
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So you get two times seven.
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Two times seven is 14, and then I subtract these two rows to get me zero.
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Seven doesn't go into zero, so I bring down the next number which is seven.
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Seven goes in seven.
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One time, right?
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This is actually a problem actually.
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Just double check, I'll show it still.
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I was like, oh I need to divide this.
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Well let me just double check my answer.
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I did it in my head, but you can always use long division.
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The whole process still works.
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So the first thing we're going to set our division on dividend is the exact same way.
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Let me say x plus two divided by x to the fourth plus five x cubed plus six x squared minus x minus two.
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So we got all the stuff we needed to divide in, right?
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A lot of stuff.
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Well remember, what we get, if you look at this, we have a binomial, right?
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Well this number two, this doesn't divide into any of these x values, doesn't.
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Two can't go into x to the fourth, right?
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You can't divide out of x to the fourth, it doesn't go into it.
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It really only goes into the constant.
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So what we're going to work on though is, so we're going to see what we can divide our x value
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into each one of these terms and let's see what we're going to get.
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So I say x goes into x to the fourth, we say it's going to go into x cubed times.
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But what I mean is it goes into x cubed times.
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Well if I do x times x cubed, I'm going to get back x to the fourth.
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So I do x, x cubed times x, and I get x to the fourth.
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All right, so you want to see what you can divide back in there, it goes x cubed times.
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Now you've got to make sure that when you multiply, you multiply the x cubed times x
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and also the x cubed times two.
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So x cubed times two is going to give me two x cubed.
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Now here's where a lot of people forget.
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Remember we're subtracting our top row from our second row.
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You've got to make sure you subtract everything.
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Because I'm going to put a nice little negative sign out here.
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Let me just say negative x to the fourth, it also means subtract two x cubed.
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So I draw my line x to the fourth minus x to the fourth is obviously zero x to the fourth, right?
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And zero x to the fourth is just zero.
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5x cubed minus two x cubed is going to give you three x cubed.
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All right.
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So that we can say x goes into three x cubed, three x squared times.
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And I multiply three x squared times x.
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That's going to give me three x cubed.
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Three x squared times two gives me six x.
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Yes, you bring three x is going to give you down to six x.
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And I write down this six x squared minus x.
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All right.
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So now I'm going to subtract again these two rows.
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So I say three x cubed minus three x cubed is zero.
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And three six x squared minus six x squared is going to be zero.
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All right.
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So then I bring down a negative two.
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I'm sorry, negative x minus two.
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So x goes into negative x, negative one times.
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Negative one times x is a negative x.
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Negative one times two is a negative two.
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Again, we're going to subtract these two rows.
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Let me put this negative side of negative.
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So just being very careful guys when you're doing this,
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because this is the word that wears double negatives
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where the last two to make a mistake.
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A negative x minus a negative x.
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These two negatives cancel out to make it a positive, right?
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So this is end I'm going to give us a zero.
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Negative two minus a negative two is just like negative two plus two,
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which also gives us a zero.
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So therefore we do not have a remainder.
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And our final portion is going to be x cubed plus three x squared minus one is our final answer.
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Is there a problem with our new remainder?
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Yes, we'll go with one.
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Absolutely.
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But for this one, whenever you get zero at the end, we're going to have no remainder.
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So therefore this binomial evenly divides into our final.
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Any questions?
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Yes?
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I was wondering what number that was.
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That's question in the end.