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### Course: Algebra (all content)>Unit 10

Lesson 12: Factoring polynomials by taking common factors

# 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?

• from on I am completely confused. And yes I have watched it multiple times.
• Factor the terms. The gcf, in this situation, is 10uv. You put all the rest in parentheses. Does that help?
• At Did he say that "U" divided by "U" is "U"? cuz isnt it 1?
• Zack,
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.
• why wouldnt you use 4 times 5 instead of 2 times 2 times 5?
• 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.
• @ why didn't Sal write 20u^2v/10uv^2? why did we drop the exponent?
• 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).
• What grade or school level would be doing this type of math?
(1 vote)
• in Brasil 8th
• How do you factor a problem like: 81 − a²?
.
(1 vote)
• This is called difference of squares.
(81 - a^2) = (9^2 - a^2)= (9-a)(9+a)
• 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)
(1 vote)
• Mia,
If you factored 20 to 4*5, you might get the wrong answer.
For example:
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
5*(4x+2y).

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
10(2x+y).
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.
(2a-5)^2 is the same as  (2a-5) (2a-5)