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### Course: Algebra basics > Unit 7

Lesson 4: Factoring polynomials by taking common factors- 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

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# Taking common factor from trinomial

Break down the process of taking common factors from trinomials. Learn how to identify the greatest common factor of a trinomial expression and use it to simplify the expression. Follow along as Sal factors 4x⁴y-8x³y-2x² as 2x²(2x²y-4xy-1) by taking the greatest common factor. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- At1:15... I know this is right, I just need an explanation.

How does x^2 go into x^3 and x^4?(15 votes)- 5 goes into 15, 3 times (5*3=15)

8 goes into 16, 2 times (8*2=16)

9 goes into 54, 6 times(9*6=54)

x^2 goes into x^3, x times (x^2 * x=x^3)

x^2 goes into x^4, x^2 times (x^2 * x^2 = x^4)(25 votes)

- why we have taken only smallest no.(5 votes)
- I'm assuming you meant the smallest number--we take the smallest number out of 2 , 4, and 8 because 2 is the largest number that can divide EACH of the numbers evenly. If we wanted to use 2, 4, or 8--or any number, really--we could, but it would make it more complicated because we would have fractions.(13 votes)

- I tried to do this for a question, and Khan Academy said I was incorrect, any tips?

(3b^5)+(15b^4)-(18b^7) = 3(b^5 + 5^4 - 6b^7)(4 votes)- First, you lost the variable in the middle term of your answer.

Next, you need to factor out the**greatest**common factor. You found the numeric portion, however, you didn't look at the variables. The greatest common factor must include some number of b's because all the terms have b's.

Give it a try. If you still can't get the right answer, comment back.(12 votes)

- how would you factor 4a^2+12ab+9b^2?(7 votes)
- If you take a closer look at the expression, you might be able to notice something. The first term, 4a^2, can be represented as (2a)^2. On the other hand, the third term, 9b^2, can be represented as (3b)^2. Rewriting the expression with these new representations, you would get (2a)^2+12ab+(3b)^2. Now if you look at it, you may be able to notice that it is a perfect square in the form of a^2+2ab+b^2, where "a" is 2a and "b" is 3b. This matches up, because a^2+2ab+b^2 = (2a)^2+2(2a)(3b)+(3b)^2 = 4a^2+12ab+9b^2. Knowing this, we can factor the expression in the form (a+b)^2 where "a" is 2a and b is 3b, so your final answer would be (2a+3b)^2.(1 vote)

- Wait, x cubed is divisible by x squared? I'm confused.(3 votes)
- Yes, x^3 is divisible by x^2.

x^3 = x*x*x and x^2 = x*x

They share a common factor os x^2. If you did that out your are left with x. This is just like if you have 8/4. 8 is 2^3 and 4=2^2. When you divide 8 by 4 you get 2.

Hope this helps.(10 votes)

- I commonly see in posts and comments, below, this symbol --that I've inclosed within quotation marks, " ^ ". It usually comes after a letter and before a number. Is this symbol used to express an exponent?(6 votes)
- yes, it is hard to write an exponent on a computer so the use the "^" symbol and then the number.(3 votes)

- what if there is no GCF(3 votes)
- That's ok. Many polynomials will not have a common factor. You will learn several factoring techniques to use when factoring polynomials. Your first step should always be to factor out the GCF if one exists.(7 votes)

- How would we factor x^2+4x+3 then?(4 votes)
- How do you factor 8k^2 - 16k + 6(4 votes)
- Using x, start with seeing all even numbers, so factor out a 2 to get 2(4x^2-8x+3). One way is to multiply ac to get 12 (slide the 4 which will later be used for dividing) and factor the related equation of 2(x^2-8x+12)=2(x-6)(x-2). To finish slide the 4 back by dividing and reduce to get 2(x-6/4)(x-2/4)=2(x-3/2)(x-1/2) then the denominators go to the front to end up with 2(2x-3)(2x-1)

So another example would be 10x^2 - x - 2.

Multiply ac and move to get x^2 - x - 20

Factor (x-5)(x+4), divide by 10 (x-5/10)(x-4/10)

reduce (x-1/2)(x+2/5) and move denominator (2x-1)(5x+2)(3 votes)

- We can, also for our convenience, choose to take "y^0" as the common factor for all the powers of "y".(3 votes)
- You could, but since y^0=1, you would use 1 instead of y^0. And, factoring out a 1 is generally done only when factoring by grouping. Factoring out a GCF of 1 from a trinomial, doesn't help you to factor the trinomial to prime polynomials.(5 votes)

## Video transcript

We're told to factor 4x to the
fourth y, minus 8x to the third y, minus 2x squared. So to factor this, we need to
figure out what the greatest common factor of each
of these terms are. So let me rewrite it. So we have 4x to the fourth y,
and we have minus 8x to the third y, and then we have
minus 2x squared. So in the other videos, we
looked at it in terms of breaking it down to its simplest
parts, but I think we have enough practice now to be
able to do a little bit more of it in our heads. So what is the largest
number that divides into all of these? When I say number I'm talking
about the actual, I guess, coefficients. We have a 4, an 8, a 2. We don't have to worry about the
negative signs just yet. And we say, well, the largest,
of, the largest common factor of 2, 8 and 4 is 2. 2 goes into all of them, and
obviously that's the largest number that can go into 2. So that is the largest number
that's going to be part of the greatest common factor. So let's write that down. So it's going to be 2. And then what's the greatest,
I guess, factor, what's the greatest degree of x that's
divisible into all three of these? Well, x squared goes into all
three of these, and obviously that's the greatest degree of
x that can be divided into this last term. So x squared is going to be the
greatest common x degree in all of them. 2x squared. And then what's the largest
degree of y that's divisible into all of them? Well, these two guys are
divisible by y, but this guy isn't, so there is no
degree of y that's divisible into all of them. So the greatest common factor
of all three of these guys right here is 2x squared. So what we can do now is we can
think about each of these terms as the product of the 2x
squared and something else. And to figure that something
else we can literally undistribute the 2x squared, say
this is the same thing, or even before we undistribute the
2x squared, we could say look, 4x to the fourth y is the
same thing as 2x squared, times 4x to the fourth
y, over 2x squared. Right? If you just multiply this
out, you get 4x to the fourth y. Similarly, you could say that
8x to the third y-- I'll put the negative out front-- is the
same thing as 2x squared, our greatest common factor,
times 8x to the third y, over 2x squared. And then finally, 2x squared
is the same thing as if we factor out 2x squared-- so we
have that negative sign out front-- if we factor out 2x
squared, it's the same thing as 2x squared, times 2x squared,
over 2x squared. This is almost silly what I'm
doing here, but I'm just showing you that I'm just
multiplying and dividing both of these terms by 2x squared. Multiplying and dividing. Here it's trivially simple. This just simplifies to 2x
squared right there, or this 2x squared times 1. That simplifies to 1, maybe
I should write it below. But what do these simplify to? So this first term over here,
this simplifies to 2x squared times-- now you get 4 divided
by 2 is 2, x to the fourth divided by x squared
is x squared. And then y divided by 1 is
just going to be a y. So it's 2x squared times 2x
squared y, and then you have minus 2x squared times,
8 divided by 2 is 4. x to the third divided
by x squared is x. And y divided by 1, you can
imagine, is just y. And then finally, of course,
you have minus 2x squared time-- this right here
simplifies to 1-- times 1. Now, if you were to undistribute
2x squared out of the expression, you'd
essentially get 2x squared times this term, minus this
term, minus this term. Right, if you distribute this
out, if you take that out of each of the terms, you're going
to get 2x squared times this 2x squared y, minus 4xy,
and then you have minus 1, minus 1, and we're done. We've factored the problem. Now, it looks like we
did a lot of steps. And the reason why I kind of of
went through great pains to show you exactly what we're
doing is so you know exactly what we're doing. In the future, you might
be able to do this a little bit quicker. You might be able to do many
of the steps in your head. You might say OK, let me
look at each of these. Well, the biggest coefficient
that divides all of these is a 2, so let me put that 2,
let me factor 2 out. Well, all of these are divisible
by x squared. That's the largest
degree of x. Let me factor an
x squared out. And this guy doesn't have a y,
so I can't factor a y out. So let's see, it's going to be
2x squared times-- and what's this guy divided
by 2x squared? Well 4 divided by 2 is 2. x to the fourth divided by
x squared is x squared. y divided by 1, there's no
other y degree that we factored out, so it's just
going to be a y. And then you have minus
8 divided by 2 is 4. x to the third divided
by x squared is x. And then you have y divided
by say, 1, is just y. And then you have minus
2 divided by 2 is 1. x squared divided by x squared
1, so 2x squared divided by 2x squared is just 1. So in the future, you'll do it
more like this, where you kind of just factor it out in your
head, but I really want you to understand what we did here. There is no magic. And to realize that there's no
magic, you could just use the distributive property to
multiply this out again, to multiply it out again, and
you're going to see that you get exactly this.