Main content

## Algebra 2

# Taking common factor: area model

CCSS.Math:

Sal finds the measures of a rectangle whose area is 12x⁴+6x³+15x² by taking the greatest common factor.

## Want to join the conversation?

- How is x squared divisible into x cubed? I have been confused with this since the last video in this area. I have seen other explanations of this, but I still do not understand.(11 votes)
- x cubed is x*x*x. x squared is x*x. So you have x*x*x/x*x. You have 3 x's in the numerator and 2 in the denominator.

You can cancel out 2 x's in the top and 2 x's in the bottom. That leaves you with just 1 x remaining in the top.

Alternately, if the bases are the same, you can simply subtract the powers. x^3 / x^2. You can subtract 2 from 3 and that leaves you with x^1 or just x.(12 votes)

- I'm confused. How can x^2 be a part of the greatest common monomial facter? How does x^2 go into x^3 ? Shouldn't the greatest common monomial factor in this situation be 3x because x can go into all variables?(7 votes)
- I think you're confusing about division of exponent with same base. Rules of exponent state that a^m/a^n = a^(m-n).

So x^4 / x^2 = x^(4-2)= x^2

x^3 / x^2 =x^(3-2)=x^1 = x

x^2 / x^2= x^(2-2) = x^0 = 1(13 votes)

- At2:29Sal says, "What are the non-prime factors of each of these numbers?" Why non-prime factors specifically?(4 votes)
- Sal is showing two different ways to find the greatest common factor.

The first way, at1:30,*he does*prime factorization.

The second way, at2:26, he uses non-prime factorization.

So if he didn't make the specification in the second example of "non-prime factors" he would just be repeating what he had shown at1:30.(6 votes)

- When you have factored everything out, and get get (3x^2)(4x^2+2x+5), could you just add it up to get (3x^2)(11x^3), and then 33x^5?(3 votes)
- No. You can't add unlike terms. 4x^2 + 2x + 5 are all unlike terms. If you are unfamiliar with like vs unlike terms, here is a link to help you: https://www.khanacademy.org/math/algebra-basics/core-algebra-expressions/core-algebra-manipulating-expressions/v/combining-like-terms(7 votes)

- So I have a question...

How can area models and Diamond problems help you factor challenging expressions?(3 votes)- Area models cannot help you factor out challenging expressions, but it can help you
**visualize**your results.

If you look at the video carefully, Sal does not use the area model to help you factor out the equation, he just uses it to show that (3x^2)(4x^2 + 2x + 5) = 12x^4 + 6x^3 + 15x^2.(2 votes)

- How is the least common of x is x^2(1 vote)
- You are not finding LCM, you are finding GCF.(4 votes)

- please help factor this using the Quadratic AC method. 8x^6+10x^4-3(2 votes)
- I believe that that question is unfactorable(2 votes)

- what should I do if I don't

understand something even if someone helps me(1 vote)- You should ask again and they will most likely help you out. If you need help, feel free to ask!(3 votes)

- At4:00why is Sal dividing the area values by the length?(1 vote)
- The question asked for the width of the rectangle and to find width, you divide the area by the length(3 votes)

- Well, I yet have a question

Could anyone factorise:

x^4+x^2+1(2 votes)- since x^4 = (x^2)^2 let's pretend x^2 = a so now we have a^2 + a + 1. I'm saying right now you will only get a complex factorization.

To factor I am going to use completeing the square.

a^2 + a + 1 = 0

(a + .5)^2 + 3/4 = 0

(a + .5)^2 = -3/4

a + .5 = +/-sqrt(3)/2 i

a = +/- sqrt(3)/2 i - .5

so now we know it factors into (a - [sqrt(3)/2 i - .5])(a - [-sqrt(3)/2 i - .5]) but we made a = x^2 so it becomes (x^2 - [sqrt(3)/2 i - .5])(x^2 - [-sqrt(3)/2 i - .5]) Then you could factor those but it's already pretty complicated(1 vote)

## Video transcript

- [Voiceover] The
rectangle below has an area of 12x to the fourth plus 6x to the third plus 15x squared square meters. And we can see the area right over here and they broke it up; this
green area is 12x to the fourth, this purple area is 6x to the third, this blue area is 15x squared. You add 'em all together,
you get this entire rectangle which would be the combined
area is 12x to the fourth plus 6x to the third plus 15x squared. The length of the rectangle in meters, so this is the length right over here that we're talking about, so we're talking about this distance. The length of the rectangle
in meters is equal to the greatest common monomial
factor of 12x to the fourth, 6x to the third, and 15x squared. What is the length and
width of the rectangle? I encourage you to pause the video and try to work through it on your own. Well the key realization
here is that the length times the width, the
length times the width is going to be equal to this area. And if the length is the
greatest common monomial factor of these terms, of 12x to the fourth, 6x third, and 15x squared, well then we can factor that
out and then what we have left over is going to be the width. So let's figure out what
is the greatest common monomial factor of these three terms. And the first thing we
could look at is let's look at the coefficients. Let's figure out what's
the greatest common factor of 12, six, and 15. And there is a couple
of ways you could do it. You could do it by looking
at a prime factorization, you could say alright, well 12 is two times six which is two times three, that's the prime factorization of 12. Prime factorization of six
is just two times three. Prime factorization of 15 is three times five. And so the greatest common factor, the largest factor that's
divisible into all of them, so let's see, we can
throw a three in there. Three is divisible into all of them. And that's it 'cause we
can't say a three and a two. A three and a two would be
divisible into 12 and six, but there's no two
that's divisible into 15. We can't say a three and a five
'cause five isn't divisible into 12 or six; so the
greatest common factor is going to be three. Another way we could have done
this is we could have said where are the non-prime factors
of each of these numbers. 12 you could have said, OK I can get 12 by saying one times 12 or two times six or three times four. Six you could have said, let's see, that could be one
times six or two times three. So those are the factors of six. And then 15 you could have
said well one times 15 or three times five. And so you say the greatest common factor? Well three is the largest
number that I've listed here that is common to all
three of these factors. So once again, the greatest common factor of 12, six, and 15 is three. So when we're looking at the
greatest common monomial factor the coefficient is going to be three. And then we look at these powers of X. We have X to the fourth. I'm using a different color. We have X to the fourth, X to the third, and X squared. Well what's the largest power of X that's divisible into all of those? Well it's going to be X squared. X squared is divisible
into X to the fourth and X to the third and of
course X squared itself. So the greatest common
monomial factor is 3x squared. This length right over here, this is 3x squared. So if this is 3x squared,
we can then figure out what the width is. If we were to divide 12x to
the fourth by 3x squared, what do we get? Well 12 divided by three is four and X to the fourth divided by X squared is X squared. Notice 3x squared times 4x squared is 12x to the fourth. And then we move over
to this purple section. If we take 6x to the third
divided by 3x squared, six divided by three is two. And then X to the third
divided by X squared is just going to be X. And then last but not least, we have 15 divided by three is going to be five. X squared divided by
X squared is just one, so it's just gonna be five. So the width is going to be 4x squared, plus 2x, plus five. So once again, the length,
we figure that out, is the greatest common
monomial factor of these terms. It's 3x squared. And the width is 4x squared plus 2x, plus five. And one way to think about
it is we just factored this expression over here. We could write that. Actually I wanna see the original thing. We could write that 3x squared times 4x squared, plus 2x, plus five which is the entire width,
well that's going to be equal to the area. That's going to be equal
to our original expression: 12x to the fourth power plus 6x to the third plus 15x squared.