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## Algebra 1 (Eureka Math/EngageNY)

### Unit 4: Lesson 2

Topic A: Lessons 1-2: Common factor- Greatest common factor of monomials
- Greatest common factor of monomials
- Greatest common factor of monomials
- 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

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# Taking common factor from binomial

Sal factors 8x²y+12xy² as (4xy)(2x+3y) by taking out the

**greatest common factor***.***.**## Want to join the conversation?

- what if it was a subtraction problem like 8x^2y-12xy^2 would the answer still be 4xy(2x+3y) or 4xy(2x-3y)?(10 votes)
- Your 2nd is correct. You can verify it by redistributing the 4xy. Your 1st version doesn't recreate the original polynomial.

Note: you could factor out -4xy. If you did, you would then have: -4xy(-2x+3y)

Hope this helps.(21 votes)

- how would you do 6x^2+11x-10(5 votes)
- What two numbers multiply to be 6*-10 = -60 (which tells us one is positive and one is negative) and subtract to be 11? We end up with 15 and - 4, so 6x^2 + 15x - 4x - 10, then get GCF of first two and of second two to get a common factor, try it and see.(12 votes)

- why is factoring important?(3 votes)
- You can use factoring to help solving quadratic or even higher degree equations a lot of times without using the proper formula like (-b+-sqrt(b^2-4ac))/2a saving a lot of time and unnecessary calculation.

For example: 4x^3-2x^2-6x = 0

you can solve:

2x(2x^2-x-3)=0 //factoring 2x

2x(2x^2+2x-3x-3)=0 //using grouping method

2x(2x(x+1)-3(x+1))=0 //factor 2x from the 1st 2 term; 3 from the 2nd 2 term

2x(2x(x+1)-3(x+1))=0 //factoring (x+1)

(2x)(x+1)(2x-3)=0

Applying rule: A product is zero when some of its factor is zero.

Either one of the 3 must be 0.

I. 2x=0 -> x=0

II. x+1=0 -> x=-1

III. 2x-3=0 -> x=3/2

So you just solved a cubic equation without using any higher college level math. Not working always but certainly an useful skill to learn in high school math.(11 votes)

- Can you use this method for polynomials with more than two or three terms?(3 votes)
- Yep. Think of it like distributing in reverse, which is kinda what factoring is. so 12x^6+8x^4-4x^2 = 4x^2(3x^4+2x^2-1). So you can distribute that 4x^2 and see it turns back into the original expression. But more importantly it shows that method works.(4 votes)

- 15a^2+12a^3

how do you work this on out(3 votes)- what goes into a^2 and a^3? a^2 because a^2 * 1 = a^2 and a^2 * a = a^3, so we can factor out an a^2 so this gets us a^2 (15 + 12a)

Now we deal with the 15 and 12? what number goes into both of them? 3. if you have trouble with that you need to work on your factors. so if we factor out a 3 we divide both by 3. This leaves us with:

3a^2 (5 + 4a)

You can try distributing again to check. Now, since 5 and 4a do not have any common factors, this is as simple as it gets.

You can find factors by writing a number in prime factorization. If that is not something you get let me know and I can explain it.(3 votes)

- How come when you factor two of the binomials like 30k^5+6k^2 the even exponent can just be taken out of the 30k^5 and be written now as 30k^3? Why wouldn't you divide instead of subtract?(2 votes)
- We can't and aren't using subtraction. k^5 - k^2 are unlike terms. So, subtraction is not possible.

It is division!

k^5/k^2 = (k*k*k*k*k) / (k*k)

Cancel / divide out 2 instances of "k". You end up with k^3.

This is a basic property of exponents. When you divide and the factors have a common base, you subtract the exponents.

Hope this helps.(5 votes)

- (4xy)(2x) + (4xy)(3y) = (4xy)(2x+3y), I don't really understand how they are equivalent, I understand how to do it but is there any other way of demonstrating it turning into that without drawing two arrows and saying you factor out the 4xy?(1 vote)
- First you need to understand the distributive property (you might already know this):

𝑎(𝑏 + 𝑐) = 𝑎𝑏 + 𝑎𝑐

For any three quantities 𝑎, 𝑏, and 𝑐. Going from the LHS to the RHS is known as*distributing*, while the reverse process (going from the RHS to the LHS) is known as*factoring*. For instance let 𝑎 = 3, 𝑏 = 5, and 𝑐 = 7. This gives us:

3(5 + 7) = 3 • 12 = 36

But we can also use the distributive property to get the correct answer as well:

3(5 + 7) = (3 • 5) + (3 • 7) = 15 + 21 = 36

We get the correct answer both ways thanks to the distributive property.

Now we make the substitutions:

𝑎 = 4𝑥𝑦

𝑏 = 2𝑥

𝑐 = 3𝑦

And we get:

4𝑥𝑦(2𝑥 + 3𝑦) = (4𝑥𝑦)(2𝑥) + (4𝑥𝑦)(3𝑦)

As desired. Comment if you have questions!(6 votes)

- What is the factored form of this problem x2+8x+15?(2 votes)
- Find 2 numbers that add to be 8 and multiply to be 15, then put them in ( x +
*__*)(x +*__*). 15 only has two choices.(4 votes)

- At4:30, why didn't Sal do (4xy)^2 (2x+3y)

instead of (4xy) (2x+3y)?

He started with*two*4xy's but ended up only adding one to the problem. Why is that?(1 vote)- It was (4xy)(2x)+(4xy)(3y)

4xy is a common factor, he simply factored it out to get (4xy)(2x+3y). If you distribute this, you'll get the expression above. (4xy)² is only when you multiply (4xy) by (4xy), but it is not the case here.(5 votes)

- how you work with 9m3-27m

because it have the negative sign i try to do it but i don´t understand....(2 votes)- You factor out a common factor of 9m to get: 9m(m^2-3)

Hope this helps.(2 votes)

## Video transcript

- [Voiceover] So we're told
to factor the polynomial below by its greatest common monomial factor. So what does that mean? So we have these two terms,
and I want to figure out their greatest common monomial factor, and then I want to express this with that greatest common
monomial factor factored out. So how can we tackle it? Well, one way to start is I can look at just the constant terms, or not the constants, the
coefficients, I should say. So I have the 8 and the 12. Then I could say, "Well, what is just the greatest
common factor of 8 and 12?" The gcf of 8 and 12. And there are a lot of
common factors of 8 and 12. They're both divisible by 1,
they're both divisible by 2, they're both divisible by 4. But the greatest of their
common factors is going to be 4. So that is equal to 4. So, let me just leave that there. And then we could think
about, "What is...?" Well, let me actually
write it right over here. I'll put a 4 here. And now we can move on to the powers of x. We have an x squared and we have an x. And we could say, "What is the largest power
of x that is divisible into both x squared and x?" Well, that's just going to be x. x squared is clearly divisible by x, and x is clearly divisible by x, but x isn't going to have a
larger power of x as a factor. So this is the greatest
common monomial factor of x squared and x. Now we do the same thing for the ys. So, we have a y and a y squared. If we think in the same terms the largest power of y that's
divisible into both of these is going to be just y to
the first power, or y. And so 4xy is the greatest
common monomial factor. And to see that we could
express each of these terms as a product of 4xy and something else. So this first term right over here, so let me pick a color, so this term right over
here we could write as 4xy. That one's actually,
that color's hard to see, let me pick a darker color. We could write this right
over here as 4xy times what? And I encourage you to pause
the video and think about that. Let's see, 4 times what
is going to get us to 8? Well, 4 times 2 is going to get us to 8. x times what is going
to get us to x squared? Well, x times x is going
to get us to x squared. And then y times what
is going to get us to y? Well, it's just going to be y. So 4xy times 2x is
actually going to give us this first term. So actually just let me rewrite
a little bit differently. So it's 4xy times 2x is this first term, and you can verify that. 4 times 2 is going to be equal to 8, x times x is equal to x squared, and then you just have the y. Now let's do the same
thing with the second term. And I just want to do this to show you that this is their largest
common monomial factor. So the second term, and I'll do this in a
slightly different color, do it in blue. I want to write this as the product of 4xy and another monomial. So 4 times what is 12? Well, 4 times 3 is 12. x times what is x? Well, it's just going to be 1, so we don't have to write a times 1 here. And then y times what is y squared? It's going to be y times y is y squared. And you can verify. If you multiply these two you're
going to get 12 xy-squared. 4 times 3 is 12, you get your x, and then y times y is y squared. So, so far I've written
this exact same expression but I've taken each of those
terms and I've factored them into their greatest common monomial factor and then whatever is left over. And now I can factor the 4xy out. I can actually factor it out. So this is going to be equal to, if I factor the 4xys out, you can kind of say I
undistribute the 4xy. I factor it out. This is going to be equal
to 4xy times 2x plus -- when I factor 4xy from here
I get the 3y left over. Plus 3y, and we're done. And you can verify it. If you were to go the other way, if you were to distribute this 4xy and multiply it times 2x,
you would get 8 x-squared y. And then when you distribute
the 4xy onto the 3y you get the 12xy-squared. And so we're done. This right over here is our answer. The answer is going to be 4xy, which is the greatest
common monomial factor, times 2x plus 3y.