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## Algebra (all content)

### 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 4x⁴y+8x³y as 4x³y(x+2). Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• What is the difference between the LCM , LCD and GCF ?
• LCM is the lowest common multiple. This means that you find the smallest number that can be divided by both. For example, the LCM of 4 and 3 is 12, and the LCM of 3 and 6 is 6.

LCD is the lowest common denominator. This is useful when adding or subtracting fractions, as you need a LCD to add/subtract them. You would find it in the same method that you would find the LCM. for example, the LCD of 1/2 and 1/3 would be 6. You would change the denominator of both fractions to six and then alter the numerator by the same factor as the denominator. So, 1/2 would become 3/6 and 1/3 would become 2/6. They can now be added or subtracted.

GCF is the largest number that both numbers can be divided by. There really isn't a simple way to find it, but you just try to find the largest common factor to both numbers. For example, the GCF of 12 and 18 is 6, and the GCF of 24 and 25 is 1.

hope this helps!
• I'm trying to understand how to factor polynomials using GCF but I still don't get it. Can you help me understand it?
• Andria,
Welcome to Khan Academy.
Try watching the vidoe again. Sometimes it takes a couple times for things to click.
Sal explains it much better than I can.
You are using the Distributive Property. x*(a+b) = xa +xb, but in reverse that is:
xa + xb = a*(a+b).
You want to find the most common factors that make up the x. And this GCF is then placed outside the parenthesis and everything else is left inside the parenthesis.
If you had just numbers such as
2*3*5*5*7 + 2*3*9
you would factor out everything that is common to both. In this case 2*3 and place that outside the parenthesis so you would get
2*3(5*5*7 + 9).
If instead they were letters and numbers such as
x*x*y*5*3 + x*y*y*5*2
You would still find the all of the common factors, in this case x, y and 5 and place them outside the parenthesis
x*y*5(?+?)
and leave everytning else inside
x*y*5(x*3 + y*2)
I hope that helps
• I have to do this x4-x2 of course the 4 And 2 are exponents the book said that if possible factor the polynomial completely ?
• Just factor out x^2 from each term in the original polynomial, which becomes x^2(x^2-1). Factoring further it becomes x^2(x+1)(x-1). Hope this helps.

Another way to try this one is using substitution, which can be helpful for tricky factoring. An easy way to tackle this problem is to substitute the lowest exponent value of x (in this case x^2) as another variable, such as y. Then, at the very end of the problem, we can put all our y-variables back into x's. So, set x^2 = y. Now the polynomial becomes y^2 - y^1. Factor out a y^1. Now the polynomial becomes: y(y-1). This cannot be factored any further. Now put back the x's, noting that y=x^2. So, the polynomial becomes: x^2(x^2 - 1) This can be factored further: x^2(x+1)(x-1)
• Shouldn't the title for this video be "How to factor a quadratic binomial"? Or is quartic a term I don't know? Genuinely asking.
• IT is a term but it is rarely used. Another on of the synonyms for 4.
• At , why does he divide by 4x^3 y?
• Because 4 is the greatest common factor between the numbers in each term, the two terms share the variable x and you always have to take the smallest exponent so you get x^3, and the two terms also share the variable y (like before you take the smallest exponent) so you get just y. I hope this makes sense.
• Well, in the video, it never says that you should subtract the exponents when you find your GCF, because I was told that you should subtract the exponents. I understand everything else, but that part of the video when he uses the GCF. Does anybody know the answer to my question?
• What is the difference between a binomial and a monomial?
• A binomial has two expressions, such as 4x-18, while monomials are just one such as 6x
• So if I've been given y^2(2y - 1), 3(2y - 1) to find the GCF, is the GCF just (y^2 + 3), or do I have to do something more? Can I also get a link to the new version of this if there is one?
• Could someone link me to the newest version of this? I can't find it.
• Why did you add the ( x+2)? Didn't the question say to factor the GCF? So Wouldn't the answer be 4x^3y?

## Video transcript

Factor out the greatest common factor. And the expression they give us is 4x to the fourth y plus 8x to the third y. When they say to factor out the greatest common factor they're essentially telling us, find the greatest common factor of 4x to the fourth and 8x to the third y and factor it out of this expression. Or kind of undistribute it. And to find that greatest common factor-- and I always put it in quotes when we speak in algebraic terms. Because we don't really know what x and y are, whether they're positive or negative or whether they're greater than or less than 1. So it's not always going to be the greatest absolute number. But it's the greatest in it contains the most terms of these two expressions, these two monomials. So if we were to essentially factor out 4x to the fourth y it would look like this. We would do the prime factorization of 4, which is just 2 times 2, times x to the fourth, which is x times x times x times x, times y. We just expanded it out as a product of its basic constituents. Now let's do the same thing for 8-- I'll color code it-- 8x to the third. Let me do it in similar colors. So in this situation we have 8x to the 1/3 y. So the prime factorization of 8 is 2 times 2 times 2. It's 2 times 2 times 2. Prime, or I should say the factorization of x to the third, or the expansion of it, is just times x times x times x. x multiplied by itself three times. And then we are multiplying everything by a y here, times y. So what factors are common to both of these? And we want to include as many of them as possible to find this greatest common factor. So we have two 2's here, three 2's here. So we only have two 2's in common in both of them. We have four x's here, only three x's here. So we only have three x's in common. Three x's and three x's. And we have a y here and a y here. So y is common to both expressions. So the greatest common factor here is going to be 2 times 2. So it's going to be 2 times 2 times x times x times x times y. Or 4x to the third y. So this is what we want to factor out. So that means we can write this thing as-- if we factor out a 4x to the third y, where essentially we have to divide each of these by 4x to the third. We're factoring it out. So let me rewrite this. So this is 4x to the fourth y plus 8x to the third y. And we're going to divide each of these by 4x to the third y. And hopefully this makes sense to you. If we were to multiply this out, we would distribute this 4x to the third y on each terms. And then it would cancel with the denominator. You would have the same thing in the numerator and denominator. And then you would get this expression over here. So hopefully this makes sense that these are the exact same expression. But when you write it this way, then it becomes pretty clear that this is 4x to the third y. And then you just simplify each of these expressions. 4 cancels with 4. x to the fourth divided by x to the third is x. y divided by y is just 1. So you have x plus 8 divided by 4 is 2. x to the third divided by x to the third is 1. Y divided by y is 1. So x plus 2. Another way to see what's left over when you factor it out is if you were to take out the common factor. So we took out this and this. What was left over in 4x to the fourth y when we took this stuff out? When we undistributed it? Well, the only thing that was left was this x right over here. Let me do that in another color. The only thing that was left was this x. So that's why we just have that x over there. When we factored everything out of the 8x to the third y we factored all this other stuff out. We factored out 4x to the third y. We factored it out, so all we had left was the 2. Now in general, you don't always have to go through this process. You could have done it this way, but this really hopefully makes it clear exactly what we're doing. You could have said, look, 4x to the fourth y plus 8x to the third y. You could have said, well let's see, the largest number that's divisible into both 4 an 8 is 4. So let's factor out a 4 out here. The largest multiple of x that's divisible into x to the fourth and x to the third, well that's going to be x to the third. And you put an x to the third out here. And you say, well, the largest thing that's divisible both into y and y is just y. So you could have done it a little bit faster in your head. So you factor out a 4x to the third and you say, OK, if I take out a 4 out of here, then this becomes a 1. If I take an x to the third out of x to the fourth I'll just have an x left over. And then if I take a y out of the y then I just have a 1 there. So this term becomes x. And then if I take a 4x to the third y out of here, if I take a 4 out of an 8 I just have a 2 left over. If I take an x to the third out of x to the third, that's just 1. If I take y out of y, that's just one. So I'm just left with x plus 2. Eventually you'll just do this in your head a little bit faster. But hopefully this makes everything clear.