Algebra 1 (Eureka Math/EngageNY)
- 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
Taking common factor from trinomial
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.
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- Could some one please drop some knowledge on me and help me factor 8x^3 - y^6 ???? do i need to cut it down to 8x^3 - 2y^3? this doesn't look right to me. Pre-reply thanks guys :)(35 votes)
- [I realize the post is quite old, but in case it helps:]
Note that you have a difference of cubes, which means you can use the well known, established pattern for factoring your binomial. See: <https://www.khanacademy.org/math/algebra/polynomials/dividing_polynomials/v/difference-of-cubes-factoring>
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
Restate each term as a perfect cube and substitute for "a" and "b" (ignore the signs):
8x^3 - y^6:
a = (2x)^3 = 8x^3
b = (y^2)^3 = y^6
(2x - y^2)[(2x)^2 + (2x)(y^2) + (y^2)^2]
Which, when further simplified, yields Jason's answer:
(2x - y^2)(4x^2 + 2xy^2 + y^4)
One common mistake that catches me is mixing up exponentiation properties. For example: mixing up (a^b)^c = a^(b*c) with a^b * a^c = a^(b+c).(28 votes)
- Why is x^3 divided by x^2 just x?(9 votes)
- x^3 is the same as x*x*x, whilst x^2 is the same as x*x. Obviously if you multiply X^2 by x then its the same as x*x*x (x^3). In which case we can assume dividing x*x*x/x*x = x. Im sorry i couldn't give a better explanation but i hope that helps a little. also if you want proof. Lets say x= 10. we get 1000/100=10 wich is true. Watch the videos on exponents if you want more info.(32 votes)
- At1:15... I know this is right, I just need an explanation.
How does x^2 go into x^3 and x^4?(12 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)(18 votes)
- why did sal multiply and divide 2x squared?(7 votes)
- It may not be the best way to describe/ think about it. Do you understand how 4x^4y can be rewritten as (2x^2)(2x^2y)? It's basically that thinking, and then using the method of (5x^2+4xy+6x) = x(5x+4y+6) If you don't get that relation try working n reverse and distribute the x. Factoring a lot of times can be thought of as reverse distributing. Does that help?(7 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.(10 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.(8 votes)
- how would you factor 4a^2+12ab+9b^2?(5 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?(5 votes)
- yes, it is hard to write an exponent on a computer so the use the "^" symbol and then the number.(3 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'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.