Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Simplifying rational expressions. First we need to get them into monomials over monomials, and then we're going to reduce. So if we looked at 7A cubed, that is a monomial. There aren't any pluses or minuses. 21 A, it's also a monomial. So what we really need to do is we need to think about reducing the seven 21st portion, and that's really just a fraction and that reduces to 1/3. You could think of seven and 21 as 3 * 7. So the 7 / 7 would reduce a cubed is a * A * A / a single a. So in this case we would have the sevens reduce because anything over itself is 1 and one of the A's on top will reduce with one of the A's on bottom. So we'd end up with a ^2 / 3. If we look at this next example, it's top is a monomial, but it's bottom is not. We could factor a three out of the bottom and get a two X -, 3 left. Now that is a monomial, so we could think of this 21 on top as 3 * 7 / 3 times the quantity two X - 3. The three on top and the three on bottom are going to reduce or cancel because anything over itself is 1. So we end up with 7 / 2 X -3. That two X -, 3 you can leave in parentheses, but you don't have to. This next one, neither the numerator or the denominator are monomials. So we're going to start by factoring out a three on the top, and we're going to factor an X out on the bottom. Now when we look, the X plus sevens are exactly the same, so they're going to be what actually cancels in this example, leaving us 3 / X here. I would always recommend factoring on a negative so that your leading coefficient or the coefficient with the variable is positive if possible. So I'm going to take out a -1 here. And what that really does is it changes all the signs. The -5 a turns positive, the +6 turns negative in the denominator. Let's see, there's a two that could come out. No, there's not. There's nothing that can come out, so this one can't reduce any further. I think that I miswrote it. Let's try this one again with a different number. Let's try ten, a -, 12 instead of 21. If we took a -1 out of the top, that gives us five a -, 6. Now if we took a two out of the bottom, that gives us five a -, 6. The 5A minus sixes are exactly the same. Anything over itself is 1, so those can cancel, leaving a negative a half 25 -, P ^2. That difference of squares 5 -, P five plus P You could, if you wanted to, factor out a -1 and then make it P -, 5 and P + 5. They're equivalents. It won't matter. P ^2 + 10, P plus 25, that's a binomial squared, and it's P + 5 * P + 5. Now, addition doesn't matter which way we do it. So 5 + P and P + 5 are really the same thing. So this one's going to reduce to 5 -, P / P + 5. If we do multiplication, all we're going to do is we're going to get everything to a monomial first, and then we're going to cancel things top and bottom. So a ^2 -, 1 is really the difference of squares, A plus one, a -, 1 this 2 -, 5 A. Once again, I'm going to recommend taking out a negative so that we can have the leading coefficient, the 5A being a positive term, 515 A -6. We have a three that can come out, which would leave us five, A - 2 and a ^2 + 5 A -6 A+ 6 * A - 1. Now we're going to look for anything on the top that has the exact same thing on the bottom. This A -, 1 on top and an A -, 1 on bottom. Doesn't matter where on top or where on bottom, as long as one's on the top. Ones on the bottom, the 5A minus twos are going to also cancel, so our final answer is going to be a negative three a + 1 / a + 6. We don't leave a negative on the bottom ever. So if there's a negative in the bottom, we pull it up in front and put it in front of the coefficient. So -3 A+ 1 / a + 6. This next one little more complex, X ^3 -, 27. That's going to be X - 3, X squared plus three X + 9 because it's the difference of cubes X to the 4th -9 X squared. There's an X ^2 in common, leaving X ^2 - 9, but X ^2 - 9 is really, really just the difference of squares. So this is going to turn into X ^2 X + 3 X -3, and let's forget about this X ^2, X ^2 - 9 because we factored it further. In the next, we're going to be able to pull out an X ^3, and that leaves us X ^2 - 6 X +9. That's that binomial formula. So it's going to be X - 3 * X - 3 because we factored it. Let's ignore that one because it factored further. And then X ^2 + 3 X +9. Are there 2 numbers that multiply to give us 9 and add to give us 3? And the answer's no. So all we can do is pull a one out of that. And what that does is it takes it from a trinomial to a monomial. Now we want to look, is there anything top and bottom exactly the same? Well, here's an X -, 3 and there's an X -, 3. Here's an X ^2 + 3, X plus 9, and X ^2 + 3 X +9. And that looks like we have an X ^3 over an X ^2. So even though those won't go all the way right away, if I have three of them on top and two of them on bottom, one on top and one on bottom is going to reduce, leaving just a one of them up on top. So our final answer is going to be X * X -, 3 ^2 / X + 3. If we wanted to do division with these, we're going to flip. We sometimes think about skip the first term, flip the second term, and multiply instead of divide. So skip, flip, multiply 16 A to the 7th over 3B to the fifth times 6B over 8A cubed. Now we just need to reduce things top and bottom. Does this 16 have anything on the bottom it can reduce with? And the answer is yes, it can reduce with the 8. So we could think of 16 as 2 * 8. So the 8 on top and the 8 on bottom would cancel, leaving just A2. If we look at the eight of the 7th, well, there's seven on top and there's three on bottom. If there's seven on top and three on bottom, three of them top and bottom will cancel, leaving us an A to the fourth on top. If we look at the six and the three, we could think of six really as 2 * 3. So the three on top and the three on bottom would cancel, leaving A2 on top. And finally, we have AB on top and AB to the fifth on bottom. Well, one on top will cancel with one on bottom, leaving us four of them left on the bottom. So the final answer here is going to be two A to the fourth times 2 or 4A to the fourth over B to the fourth because everything reduced this next example, they're not monomials, so we're going to get them into monomials. 1st X ^3 + 8 Y cubed is the cube formula X + 2 Y X ^2 -, 2 XY plus four y ^2. The denominator 2 numbers that are going to multiply to give us this four on the outside, but add to give us 5. So it's going to be 4 and 12X plus yx plus 2Y. We're going to flip it. So we're going to have a 2 come out, which will leave us four X ^2 -, y ^2, and that's the difference of squares. So that can factor further. So we need to think about that turning into two X + y, two X -, y. If we do that, remember we're going to ignore this piece here because we've just factored it further. The top, we can pull out an X and we'd get X ^2 -, 2 XY plus four y ^2. Well, that one goes further too. That's going to be X -, 2 Y times X -, 2 Y. Oh no, it's not. That doesn't go further. I'm sorry. There weren't 2 numbers that multiplied to give us 4 and add to give us -2 So now that's as far as we can factor. We're going to start reducing. Here's an X + 2 Y on top and X + 2 Y on bottom, and X ^2 -, 2 XY plus four y ^2 on top and on bottom. Here's a two X + y and a two X + y. So when we get this all cancelled, we're going to get 2 times two X -, y on top with an X on bottom. Sometimes they get even more complex. This one, we're going to have to foil it out before we can factor. Now, the monomial was already in the bottom, so that's nice. But the top was not a monomial, so we multiply it out. We get X ^2 -, X -, 6 / X + 1, X plus two X + 3. Now we have to factor that numerator, and that numerator is going to factor into X -, 3 X +2 all over that common denominator. So now we can see that the X plus twos will cancel, leaving us X - 3 / X + 1 * X + 3. Is our final answer. This next one, the top is difference of squares, but that bottom's got 5 terms. So we need to regroup them so that the m ^2 + t ^2 + 2 Mt are going to be together. And the reason we're going to do that is that's going to give us that binomial square and then we have that m + t. So when we look at this one, that m ^2 + 2 Mt plus t ^2 is going to factor into m + t quantity squared. Those last two terms only have a one in common, so we're going to just be able to factor out an m + t Once again, the numerator is already factored, but that denominator is at 2 terms now, and each of those terms have an m + t in common. If we take an m + t in common out, we get one m + t left here, and we get a + 1 left here because the m + t is going to go away. So now top and bottom, there's an m + t So our answer is going to be m -, t / m + t + 1. The last example, we're going to group as the difference of two cubes and the difference of two squares in the numerator. The bottom is a binomial quantity squared, so it's just X -, y quantity squared. That top X -, y * X ^2 plus XY plus y ^2 plus X -, y * X + y from the difference of squares all over X -, y quantity squared. That numerator is now down to two terms, but we need it to be the one we need a monomial. So those two terms each had an X -, y in common. If I take an X -, y out, we get X ^2 plus XY plus y ^2 + X + Y all over X -, y ^2. So we'd get X ^2 plus XY plus y ^2 + X + Y all over X -, y as our final answer. Thank you and have a wonderful day.