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Robert McKeown: Here we have a multivariate

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expression. That means that we've got more than

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one unknown variable, we've got x, which is an

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unknown variable y, which is an unknown

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variable, and W, which is an unknown variable. I

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like this question, because it's going to bring

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us back to our exponent rules. Remember our

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exponents rules, said that if we have x A

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divided by x B, that's equal to x A minus B. And

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if we have x A times x B, that's going to be

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equal to x A plus B. And notice that the base

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are is the same. To make this look a little bit

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clear, I can divvy it up so I can write this as

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four y squared divided by four w, y cubed plus

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eight, x to the six y to the five divided by

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four w y to the three. And I can do that just to

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make it a little bit easier to look at. And if I

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do that, I get the fours those just divide into

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one, I'm going to get why six minus three

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overdub w plus two x to the six y five minus

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three over w. And simplifying a little bit, I'm

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going to have y three plus two x six, y two. And

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that whole thing over who just brought the

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denominator together. And I can go a little bit

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further, I could factor out a y two. So I've got

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y plus two x six over w. Yeah, the question is

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asking us to simplify. So we want to simplify it

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as much as possible. So here I am on ALEKS, I'm

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going to plug in our simplification. So we have

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y squared, and then a bracket y plus two x to

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the power of six. And I'll hit write and make

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sure I get my brackets looking as correct as

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possible. And then maybe I'll highlight this

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whole thing, then I'm going to hit this button

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here and create a fraction for our W. And if I

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do that, and I click on the check button. Let's

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see if we get the right answer. And we did so we

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simplify it as much as possible. And we use

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those exponent rules that I just showed you.

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We're asked to simplify the following

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expression. One thing that makes this expression

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easier to simplify is that in the numerator, the

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denominator is already in its lowest common

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denominator form, that's the three x right here.

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And similarly in the denominator, it's already

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in its lowest common denominator form. Since

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that's that is the case, I'm going to start by

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putting it from its division form into its

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multiplicative form. So we're going to have the

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numerator remaining the same. And it's gonna be

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multiplied by the flipped denominator. So we're

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gonna have four x cubed in the numerator and 15

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minus five x in the denominator. I see lots of

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things that we can do here to make this simpler,

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make this expression simpler. One thing, notice

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that in this numerator here, what can I do with

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that? Well, I can use the or write it in the

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difference of squares formula like that, we'll

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see what happens. We'll see what happens there.

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Maybe actually, what I'll do is I'll just take

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the difference of squares right away. So we've

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got three minus x three plus x like that.

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And notice that I've got an x here and three x

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is over there. So I could just rewrite this as

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three and four x squared up there. And then with

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the denominator down here, I've got a five and a

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15. So I can factor out five, and I'm left with

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three minus x. Now we can see that this three

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minus x here and this three minus x, they're

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they're going to cancel out. And we're going to

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be left with three plus x over three times four

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x squared over five.

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Well, the last thing I think we can do is we

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could write this as four x squared, three plus

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x. So I haven't actually changed the numerator,

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I've just moved one factor to the left, and then

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the denominator three times five, that's 15. And

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I don't think we can simplify it any further. So

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why don't we put this into ALEKS and see if we

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have the right answer. So here we are an ALEKS,

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I'm going to enter the answer that we came up

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with, I've got four x squared, and create a

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bracket three plus x. And then I'm going to

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highlight this whole thing, press this button

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over here, the square with a line below it, and

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then another square below that line, and type in

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15, click the check button. And we simplify this

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expression. Let's try some polynomial long

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division. Now, it's going to be a lot like the

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long division, you may remember from grade

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school, I'm going to write three x minus two,

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this term here, and I'm going to put one of

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these little tables like that. And I'm going to

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write the entire numerator inside it. And this

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is what's called an algorithm. So this is just

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an algorithm. It doesn't have any, there is no

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intuition why it works. It just works into an

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algorithm. It's like, you know, we take

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algorithms, we take a computer programs, you

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know, it sucks it, we do it because it works.

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And so if I want to know how many times the

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polynomial three x minus two goes into this

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larger, higher order polynomial, then I start by

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looking at the first term, that's three x minus

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two, and I asked myself, how many times is three

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x go into 12, x cubed, and so on basically

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saying, well, what's 12 x? three over three x?

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Well, it's equal to four x squared. So I'm going

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to write well, I'll write it up here, four x

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squared. Now I'm going to multiply this four x

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squared by the first term and the second term.

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And so I'm going to get and remember we negative

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negative 12 x cubed minus eight x squared, but

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that's a minus. And so it's gonna be a plus,

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because it's minus a minus is a plus. I draw a

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bar like that, as we always want 12 x cubed

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minus 12. x cubed is zero. Now I've got this

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negative 23. or excuse me, I didn't write it

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correctly, I should write it as x squared like

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that. plus eight, that's going to give us minus

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15 x squared.

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Now, how many times does three x go into

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negative 15 X squared? Well, I'll write up here

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and I'll say negative 15 x divided by three x.

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Well, that's going to be negative five x So I'll

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write negative five x up here. Negative five x.

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multiplied by this thing. No, I better not, I

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don't want to forget my negative sign. I'm going

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to get plus 15. x goes to negative and a

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negative is a plus. That gives me zero. What

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about the next term, I've got negative five x

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multiplied by negative two. So I've got a

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negative and a negative and a negative. So

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that's gonna give me negative 10 X. Oh, I wrote

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in the wrong space, excuse me. Negative 10 x.

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And now I want to know what four x minus 10 x is

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equal to, well, that's going to be equal to

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negative six. How many times does three x go

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into negative six? Well, it goes in there

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negative two times. So I'll write a negative two

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up here. Now I've got negative two times three

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x, so that's plus six x. What about the last

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term, I've got negative two times negative two

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minus so a minus a minus and a minus is a minus.

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So I'm going to have minus four and plus one

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minus four is equal to minus three. Now, how

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many times has three x? Go into three x? Well,

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we're going to say zero times. And this here is

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it gonna be our remainder.

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And I'm going to call this thing here, our

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quotient. So let's plug these answers into ALEKS

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and see if we got the right answer. Here we are

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on ALEKS for x squared minus five x minus two.

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And our remainder is minus three. So I'll click

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the check button. And let's see if I did it

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correctly. And I did would have been very easy

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for me to make a sign error. So I'm a little bit

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relieved to see the correct button. Correct

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icon. But we did it. So our next question is

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asked us to solve for y. Normally, I would like

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to start with a lowest common denominator. But

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I'm going to approach this slightly differently.

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Because I see, I've got this y plus one, and I'd

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like to move it. So I'm going to multiply both

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sides of this expression by y plus one. When I

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do that, we get negative six is equal to

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negative six times y plus one minus y plus one

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over y minus one. Why did I do that? Well, now

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it's very obvious to me what I need to do to

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find the lowest common denominator here. If I

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multiply these two terms by y minus one, all the

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terms will have the same denominator. So why

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don't I go ahead and do that, when I multiply

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the numerator and denominator, these two by y

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minus one, I get negative six y minus one over y

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minus one is equal to negative six y plus one.

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And it's y negative one over y negative one

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minus y plus one, y minus one. And now that they

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all have the same denominator, I can get rid of

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them by multiplying this whole expression, both

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sides by y minus one. So we've got negative six

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y minus one is equal to negative six y plus one,

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y minus one minus y plus one. Now might not be

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obvious how to proceed. If I look at each of the

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terms, there are no common factors. So we've got

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a y plus one over here. And a y plus one over

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here, but there's no y plus one over there. So

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what I'm going to do is I'm going to open up the

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brackets, I'm going to multiply through the

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brackets. And after I do that, I'm going to

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collect like terms. So let's get started, we've

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got negative six y, plus one. And then here,

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I've got negative six, multiplied by, well,

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that's a difference of squares, I know that's

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going to be y squared minus one minus y minus

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one. And there's a hint and the question, which

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is that we're doing quadratic factoring. And so

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I'm going to set this question up as a quadratic

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equation, which means it should be equal to

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zero. So I'm going to move all the terms to the

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right hand side. So I haven't haven't finished

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and we open this bracket. So I've got six y

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squared plus six minus y minus one. And then I'm

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going to move these over, so I get plus six y

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minus one.

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And I can see right away that I made a mistake,

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I made a mistake. What mistake did I make, I was

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going too quickly. And that should have been a

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plus six. So I want to have a negative six over

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here. Now let me collect like terms. So there's

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only one y squared term, so we have negative six

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y squared. The six minus six is equal to zero.

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And I'm going to be left with plus five y minus

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one. And this whole thing, of course, is equal

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to zero. Now the question is asking us to solve

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for y. And now that we've got an answer,

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quadratic form, if we can factor it, or if we

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use the quadratic equation, we can find the

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solution for y. Now, we probably suspect, how

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many solutions are there going to be for a

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quadratic equation? Typically, there's two,

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there's going to be two answers for why. So

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looking at this expression, I want to factor it,

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I always find it challenging to factor an

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expression if the coefficient in front of the

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square term is not one. And in this case, it's

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negative six. So I can look at I can play around

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with it, I can do some trial and error to get

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the right answer. But you also might want to use

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the quadratic formula. To remind you the

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quadratic formula says that y is going to be

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equal to negative v plus minus the square root

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of v squared, I got rid of that minus sign in

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front of the B. And I've added the equation

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right there. So you could use this quadratic

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formula, and I'll label it to find the answer to

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the problem, I'm going to factor it by trial and

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error. And I know that three times two is equal

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to six. And I also know that three plus two is

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equal to five. So I'm good at right three, why

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here like that, I'm going to write to why

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they're like that. And if I want to get negative

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one, the only way I can get negative one is to

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have a plus one and a negative one. And if I

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want three y plus two y and I've got a negative

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three y, I'm going to need to have my negative

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sign there, my plus one there. And that's how

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you could factor this problem without using the

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quadratic equation. If you multiply this out,

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you're going to get the original result. Keep

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going for the answer. So the answer is, if this

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thing is equal to zero, or this thing is equal

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to zero, then that y that gets us to that result

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is the answer. So negative three y plus one is

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equal to zero. Well, then y must be equal to

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one, negative one third. And if we have two y is

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equal, or excuse me, two y minus one is equal to

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zero, then y is equal to one half. And so my

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answers are y is equal to negative one third, or

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y is equal to one half both these solve, solve

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the the equation. Well, I can see right away,

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because I'm plugging it in that I made a mistake

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over here, it should be y is equal to one third.

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when y is equal to one third, then we get

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negative one plus one, which is equal to zero,

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and y was equal to one half as well. So I almost

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made a mistake, but I caught myself. Now let's

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go to ALEKS and make sure that I've done this

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correctly. So here we are said if there is no

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solution, click on no solution. But if there's

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more than one solution, separate them by commas.

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So I have one over three as one of my answers.

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I'll provide a comma in there, and one over two,

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and I'm using my keyboard to navigate along. Now

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ready for the moment of truth. Let's click the

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check button and see if I got the answer.

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Correct.

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And happily, I did.