Algebra (all content)
- Factoring with the distributive property
- Factoring polynomials by taking a common factor
- Taking common factor from binomial
- Taking common factor from trinomial
- Taking common factor: area model
- Factoring polynomials: common binomial factor
- Factor polynomials: common factor
- Factoring by common factor review
- Factoring polynomials: common factor (old)
- Factoring polynomials: common factor (old)
An old video where Sal factors 20u²v - 10uv² as 10uv(2u-v). Created by Sal Khan and Monterey Institute for Technology and Education.
Want to join the conversation?
- from2:11on I am completely confused. And yes I have watched it multiple times.(11 votes)
- Factor the terms. The gcf, in this situation, is 10uv. You put all the rest in parentheses. Does that help?(7 votes)
- At4:30Did he say that "U" divided by "U" is "U"? cuz isnt it 1?(4 votes)
You are correct. Sal misspoke. He wrote the answer correctly but he said it incorrectly.
Its another case where Sal is not perfect. He is just like you and me; We all make mistakes.(6 votes)
- why wouldnt you use 4 times 5 instead of 2 times 2 times 5?(5 votes)
- Because when Sal was doing it, remember, both 10 and 20 divide into 10, and 2*5=10,
so he broke the 4 apart into 2 twos. But he could've done it like you, but later, he'd have to
split the four.(3 votes)
- @4:05why didn't Sal write 20u^2v/10uv^2? why did we drop the exponent?(3 votes)
- That did confuse me a bit, but the reason why is because we have to put the GCF in the bottom for it to be the same as 10uv(2u-v). If we did 20u^2v/10uv^2 along with the other part, we would be left with our answer being 10uv(2u/v-1), which is not the same as 10uv(2u-v).(2 votes)
- What grade or school level would be doing this type of math?(1 vote)
- How do you factor a problem like: 81 − a²?
- Then what would be 2*2*5*u*u*v - 2*5*u*v*v?(1 vote)
- You could factor out all the terms that are in both
There is a 2, 5, u and v in both so you would factor out those
(2*5*u*v)/(2*5*u*u) * (2*2*5*u*u*v - 2*5*u*v*v)
which if you distribute the demoninator you get
(2*5*u*v)*[ (2*2*5*u*u*v/)/(2*5*u*u) - (2*5*u*v*v)/()/(2*5*u*u) ]
Then you canel out the common terms to get
(2*5*u*v)*[(2*u*v) - (v)]
= (2*5*u*v)*(2*u*v - v)(3 votes)
- instead of 2x2x5 cant you do 4x5 instead!(1 vote)
If you factored 20 to 4*5, you might get the wrong answer.
Assume you had 20x + 10y and needed to factor out the largest common factors.
You could factor the 20x to 4*5*x, you would factor the 10y to 2*5*y.
Then you would only see the common factor 5 so your answer would be
If instead you would reduce to prime factors factoring the 20x to 2*2*5*x, and factor the 10y to 2*5*y.
You would see the commom factors of 2*5. Then your answer would be
This would be the correct answer if you are to factor out everything possible.
If you find the prime factors, you don't miss something, so you want to find the prime factors.
I hope that helps.(3 votes)
- solve 4/(x+4)-2/(x+1)(1 vote)
- "2(x-2)/(x+4)(x+1), x =! -1,-4" This is an interesting graph and looks kind of like a band gap filter response. Please check this answer so that it satisfies any criteria you may be working under.(1 vote)
We're asked to factor 20u squared v minus 10uv squared. And when they say factor a binomial like this, an expression that has two terms like this, they really mean break it up into the product of one or more terms. So let's see if we can do that. And the easiest way we can do that, is say hey, is there any common factor? In particular, let's find the greatest common factor of each of these terms and then divide that out. Or you can almost imagine, un-distribute it out, and I'll show you what I'm talking about in a second. And eventually you'll be able to do this in your head, but we'll work through it step by step right now. So what is 20u squared v if we factor it out? 20u squared v, if we do into the prime factorization, it is 20 is 2 times 2 times 5. Right? That's 2 times 10, that's 20. u squared is times u times u and then v is just one v. So we just rewrote 20u squared v into kind of a product of its smallest components, its most fundamental components, prime numbers and just u's and v's. Now let's do the same exact thing for the 10uv squared. So we'll put that minus sign there so that we haven't fundamentally changed the expression. 10 is, if we break it down to its prime factors, 2 times 5. Then we have one u times one u times v times v. That's what v squared is, times v times v. Now what's the greatest common factor of these two terms right here? Well let's see. They both have one 2. They both have one 2 right there. Maybe I'll circle that one, I could have circled that one. They both have one 5-- that's one 5, one 5. They both have one u, one u there, one u there. This one has two, but only this one has one, and they both have at least one v. So the greatest common factor is 2 times 5 times u times v. So I could rewrite this expression, I can kind of un-distribute the 2 times 5 times u times v, and what'll we get? If we wrote 2 times 5 times u times v, and we say that's going to be-- this expression is equal to this times what? Well if you factor the 2 times 5 times u times v out, all you're going to be left with in this first term is the 2 times u, so 2u here. And in the second term, all you're going to be left with is a v. Right? All this other stuff gets factored out. All you're going to be left with is a v. Hopefully you see, if I multiply 2 times 5 times u times v times 2u, I'm going to get this first term here. So if I were to distribute it, I would get this first term. And if I multiply 2 times 5 times u times v times this v over here, I'm going to get this second term. So this expression, and that expression is the exact same thing. We have factored it out, now we can simplify it a little bit. 2 times 5 times u times v we rewrite as 10uv. And then inside the parentheses, we of course have a 2u and then a minus v. And we're done! We have factored the expression. Now you won't be doing it to this granular level, but this is the best way to think about it. Eventually you're going to say, hey, wait, look, the largest number that divides both of these is a 10. Because you could see 10 goes into the 20, 10 goes into 10. And, let's see, a u goes into both of these, and a v goes into both of these. So let me factor out a 10uv, and then if I divide this thing by 10uv, I'm going to be left with 2u. And if I divide this by 10uv, I'm going to just be left with a v. So that's another way to think about it. Let me do that right now, so we could say that this is the same thing. Another way of approaching it, you could have said that this is the same thing as-- Well, the largest number that divides both of these is 10uv, and that's going to be times 20u squared v over 10uv minus this thing. 10uv squared over 10uv. This expression and this are obviously the same thing. If I were to distribute the 10uv it would cancel out with each of these in the denominator right there. So they're the same thing, but we can do is we can simplify this. We could say that 20 divided by 10 is just 2, u squared divided by u is just a u, v divided by v is just 1, 10 divided by 10 is 1, u divide by u is u, v squared divided by v is just a v to the first power. So you're left with 10uv times the quantity 2u minus v. Either way you get the same answer.