Multiplying and factoring expressions
Factoring simple expressions
None
Factor expressions using the GCF
Factor expressions using the GCF
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ok so i have to find the GCF of 18 and 25. Both of them r not prime but they have no common factors epxect for 1 which doen't count in our classroom! Help! Please, Thanks! I said 1 doesn't count cause my teacher said we werent allowed to use it But thanks for the opinions everyone.
When two numbers are relatively prime (i.e., they have no common factors between them besides one), then the GCF is one.
Yes, one is allowed :)
Yes, one is allowed :)
First get the prime factors of each number:
18 = 2 x 9
18 = 2 x 3 x 3
25 = 5 x 5
These have no common factors so the greatest common factor is 1:
GCF(18, 25) = 1
18 = 2 x 9
18 = 2 x 3 x 3
25 = 5 x 5
These have no common factors so the greatest common factor is 1:
GCF(18, 25) = 1
exactly that is the answer
I think they are also coprimes ( Numbers which have GCD as 1 ).
One should work, I don't see anything else it could be.
The answer is 1
the answer would be prime
yes one can be used
Why don't you add the ones from the xs and ys?
Because one times something is just that something. Writing 1x is the same as writing x.
because its not possible to add them because u don't know what they are, if u multiply them is become xys^2
Because you can't add xs and ys
you dont do this because they are not like terms the way i remember is Cheese Lettuce and Tomatoes Combine Like Terms
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!
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 dont get the difference of the Greatest Common Factor (GCF) and the Least Common Denaminator (LCD) ? is it the same formula to find it ?
well, LCM is the first multiple that is common in both numbers*
GCF is the largest factor that divides BOTH the numbers
the procedure for finding the GCF is different from LCM finding
if you prime factorize both the numbers, the PRODUCT OF ALL the COMMON factors is the GCF, while on the other hand LCM is the product of all the common factors and the remaining factors
eg: GCF and LCM of 12 and 9
12 = 2 * 2 * 3 and 9 = 3 * 3 here the common factor is 3 and is common only ONCE.
(HCF IS 3) now multiply 3 * 2 * 2 * 3 (i'm taking 3 only once because it is common only once) 36 is the lcm
GCF is the largest factor that divides BOTH the numbers
the procedure for finding the GCF is different from LCM finding
if you prime factorize both the numbers, the PRODUCT OF ALL the COMMON factors is the GCF, while on the other hand LCM is the product of all the common factors and the remaining factors
eg: GCF and LCM of 12 and 9
12 = 2 * 2 * 3 and 9 = 3 * 3 here the common factor is 3 and is common only ONCE.
(HCF IS 3) now multiply 3 * 2 * 2 * 3 (i'm taking 3 only once because it is common only once) 36 is the lcm
No, you are only using GCF to factor these types of polynomials. Look up GCF and LCM on Khan Academy videos and you will see the difference.
No, you're thinking of the LCM (Least Common Multiple) .
The LCD is the LCM of the denominators (assuming the fractions are already in their simplest form)
The LCD is the LCM of the denominators (assuming the fractions are already in their simplest form)
umm i dont think so
I have to do this x4x2 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^21). Factoring further it becomes x^2(x+1)(x1). 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 yvariables 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(y1). 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)(x1)
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 yvariables 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(y1). 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)(x1)
My friend who has all As told me to put can't be factored for an answer
Thanks i got what you're saying but didnt U mean i had to square it or am i just off base can u also help me with this one m^25m35
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
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
At 5:09 he says it as if he is subtracting, how does taking a 4 out of an 8 result in a 2? Is he saying we are dividing? Then at 4:57 why is there nothing resulted instead of a 1?
It's not subtracting. He is using division.
You divide the coefficient to divide (which is what the lesson is, basically), and you subtract exponents to divide, so Sal did want that two, he just misspoke when talking about his operations
Sal made a mistake there. 84=4
how do you factor x^336? Im having trouble.
your gonna have to figure it out
x times x times x 36, if thats what your looking for
I have to study for a math test tomorrow and I'm struggling with factoring things out. I also have to memorize sixteen different axioms and propertys. Any tips??
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?
you aren't supposed to subtract the exponents
These videos are great, but I'm just wondering one thing: why did Sal say stuff like X+X+X+X when he could have said X times 4? I mean, I think a good number of us viewers know that 4x=x times 4.
He didn't say x+x+x+x, he said x TIMES x times x times x. Very different.
X + X + X + X is equal to 4X.
X * X * X * X is equal to x^4.
X + X + X + X is equal to 4X.
X * X * X * X is equal to x^4.
I think that many viewers DO understand that 4x=x times 4. However, some may not. If a student has a "gap" in their understanding, this extra explanation might help to fill it. Also, sometimes it helps to understand the concept when it is explained further.
This confused me way more about gcf! D:
That is true.
find the GCF of 55, 22, 33
Aleta,
Break the numbers down to their prime numbers
55 = 5*11
22 = 2*11
33 = 3*11
So the only common factor in all three numbers is 11
Break the numbers down to their prime numbers
55 = 5*11
22 = 2*11
33 = 3*11
So the only common factor in all three numbers is 11
My problem is 14x^2 + 55x  21 Help! I'm clueless.
Wait, does factoring out GCF the same as just normal factoring?
Would the answer be the same if the question asked "factor 4x^4 y+8x^3 y"?
Would the answer be the same if the question asked "factor 4x^4 y+8x^3 y"?
Yes, it's the same thing. When people ask you to factor something, you can just assume that they mean factor out the GCF.
can you help me factor x2  5xs  24s2
x^2  5xs  24s^2
Think of this similar problem:
x^2  5x  24
= (x8)(x+3)
Your problem is almost the same, just add letter s:
x^2  5xs  24s^2
= (x8s)(x+3s)
Think of this similar problem:
x^2  5x  24
= (x8)(x+3)
Your problem is almost the same, just add letter s:
x^2  5xs  24s^2
= (x8s)(x+3s)
a^2+1516 factor the following
Simplify this question of 2/3(30+1) =
2/3 (30 + 1) =
Use the distributive property:
20 + (2/3) = 20 2/3 or 62/3
or
1. Complete the operation on the inside of the parenthesis first, and then multiply.
2/3 ( 30 + 1)
2/3 (31) = 20 2/3 or 62/3
Hope this helps! Terrell
Use the distributive property:
20 + (2/3) = 20 2/3 or 62/3
or
1. Complete the operation on the inside of the parenthesis first, and then multiply.
2/3 ( 30 + 1)
2/3 (31) = 20 2/3 or 62/3
Hope this helps! Terrell
Is there another way to say Greatest Common Factor?
Some people call it the H.C.F (Highest Common Factor).
maybe u could also say largest common factor (LCF) or biggest (BCF)
how do you this problem 5x50
Answer: 5(x10)
Solution:
Think about this way what is their GCF?
If we factor 5 we get 5 and 1. and when we factor 10 we get 2 and 5. So their GCF is 5. Now that we know the GCF, we factor it out to get 5(x10).
Solution:
Think about this way what is their GCF?
If we factor 5 we get 5 and 1. and when we factor 10 we get 2 and 5. So their GCF is 5. Now that we know the GCF, we factor it out to get 5(x10).
What is
x2 − 5x − 6?
x2 − 5x − 6?
I assume you mean x^2  5x  6 and "what is" is asking for the solution to the equation.
We can solve using the quadratic equation!
[b ± √(b^2  4ac)] / 2a
FYI √ is square root and ± is plus or minus
If you don't know what a, b, and c are, that's okay. a is coefficient of x^2, b is the coefficient of x, and c is the last number. This equation only works with equations like these.
a = 1
b = 5
c = 6
(5) = 5
{5 ± [√ (5^2)  4(1)(6)] } ÷ 2(1)
{5 ± [√ 25 + 24]} ÷ 2
{5 ± √49} ÷ 2
{5 ± 7} ÷ 2
5 + 7 = 12
5  7 = 2
12 ÷ 2 = 6
2 ÷ 2 = 1
So your xintercepts (solutions) are at 6 and 1.
In other words the final factored form is
(x6)(x+1) < This is the answer
You're probably thinking, "You said the answers were 6 and 1, but you put x6 and x+1. Yes I did! That's the correct format. When you set x to 6 in the equation, you get zero and same if you put in 1. Hope this helps :)
We can solve using the quadratic equation!
[b ± √(b^2  4ac)] / 2a
FYI √ is square root and ± is plus or minus
If you don't know what a, b, and c are, that's okay. a is coefficient of x^2, b is the coefficient of x, and c is the last number. This equation only works with equations like these.
a = 1
b = 5
c = 6
(5) = 5
{5 ± [√ (5^2)  4(1)(6)] } ÷ 2(1)
{5 ± [√ 25 + 24]} ÷ 2
{5 ± √49} ÷ 2
{5 ± 7} ÷ 2
5 + 7 = 12
5  7 = 2
12 ÷ 2 = 6
2 ÷ 2 = 1
So your xintercepts (solutions) are at 6 and 1.
In other words the final factored form is
(x6)(x+1) < This is the answer
You're probably thinking, "You said the answers were 6 and 1, but you put x6 and x+1. Yes I did! That's the correct format. When you set x to 6 in the equation, you get zero and same if you put in 1. Hope this helps :)
how can i solve a^3 +6a^211a
your right the a= A(A*2+6A11)
Well,A^3 = A * A * A
and 6A^2 = 2 * 3 * A * A
and 11A = 11 * A
so we have as a common factor, A
So, It turns into A(A^3/A + 6A^2/A  11A/A)
= A(A^2 + 6A  11)
(I may be wrong)
and 6A^2 = 2 * 3 * A * A
and 11A = 11 * A
so we have as a common factor, A
So, It turns into A(A^3/A + 6A^2/A  11A/A)
= A(A^2 + 6A  11)
(I may be wrong)
what if it is a 2 instead of a 4 or higher? would you say 1 x 1 x whatever...
If there were a two in the first polynomial instead of a four (2x^4y instead of 4x^4y), you would factor it out as: 2*x*x*x*x*y.
1*1*x*x*x*x*y would just be x^4y, the ones are redundant.
1*1*x*x*x*x*y would just be x^4y, the ones are redundant.
I don't get it why 4x3y?
are there any more example videos on this subject
trying to solve this equation: x^34x^2+4x16
Explain how to factor a monomial GCF out of terms in a polynomial
question what do you do when you have a problem with a variable squared a number and variable and then a number by itself?
At 2:29, how do you get 4x^3y
80+65
distributive property write each sum as a product of the GCF of the two numbers
distributive property write each sum as a product of the GCF of the two numbers
Austin
Break each number into its primes
80 = 2*2*2*2*5
65 = 5*13
The only common factor is 5 so factor out a five
5(16+13)
I hope that helps make it click for you.
Break each number into its primes
80 = 2*2*2*2*5
65 = 5*13
The only common factor is 5 so factor out a five
5(16+13)
I hope that helps make it click for you.
where do you find simplifying
look it up in the search box
how would you solve 80xy  245x^2
1. Identify the common factors of 80xy and 245x^2
Factors of 80xy are 5 * 2 * 2 * 2 * 2 * x * y
Factors of 245x^2 are 5 * 7 * 7 * x * x
2. Take the common elements out of this list of factors and determine the number to factor out. Here the number is 5x.
3. Determine how many times 5x goes into 80xy and 245x^2 by dividing these numbers by 5x, then write your result in parentheses: 5x(16y  49x)
5x(16y  49x) is your answer
Factors of 80xy are 5 * 2 * 2 * 2 * 2 * x * y
Factors of 245x^2 are 5 * 7 * 7 * x * x
2. Take the common elements out of this list of factors and determine the number to factor out. Here the number is 5x.
3. Determine how many times 5x goes into 80xy and 245x^2 by dividing these numbers by 5x, then write your result in parentheses: 5x(16y  49x)
5x(16y  49x) is your answer
Can you please explain how do i factorize 8(4x+5y)2 +12(4x+5y)
the 2 in above expression is square ..
the 2 in above expression is square ..
If I have a more complicated expression that becomes 11g(2^2+x^2) how do I simplify that? I was told only when you are subtracting two squares you can use (a+b)(ab)
not always
you can use (a+b)(ab)
you can use (a+b)(ab)
At 3;51 my teacher told me something entirely different. i don't understand either onecan you show me a simpler
how do you factor by grouping trinomials?
Do you have a solution for factors of 9 than GCF and factor of 12 than GCF if you do please tell me
at 2:40 why do you divide?
how can i factor with two terms?
Is there anyway I can find a video related to the FOIL method? Please anyone respond.
Ya just type in the foil method in the search box
Do you have any videos that tell how to do this when you have a polynomial with four terms, our math teacher shows what you're doing when you have two terms, but when there are four or more, she shows something COMPLETELY different. Thanks :)
There are several factoring techniques. Remove a GCF from a polynomial can be done with a polynomial of 2 or more terms. However, not all polynomials have GCF factors. So, other techniques are needed. There is a technique called grouping this is used for a polynomial of 4 terms. Check out video titled: Example: Basic grouping.
Is there a more simple explanation to this video? Im only in 6th grade and have a test
ples dont delet this coment
I have different answer for question find GCF for 4x^4y+8x^3y =4xy(x^3+2x^2)
ok, but what do you do when 4x^48x^3y? (subtracted)
On one of the mastery challenges, it says to use this same method to rewrite the expression 60+44 as the product of the greatest common factor of 60 and 44 and the sum of the remaining numbers. How do you do that exactly.
60 = 2 * 2 * 3 * 5 and 44 = 2 * 2 * 11 so the GCF = 2 * 2 = 4
So rewrite 60 + 44 as 4*(15 + 11)
So rewrite 60 + 44 as 4*(15 + 11)
when two or more numbers are fractions, im presuming only fractions and zero are considered factors, is that right?
wasn't this vid different a few days ago?
why do we hae to do Gfc
What is the steps i take is i have to factor a fraction out instead of a whole number? I.E. (1/2y^2 9/2y 11) I can get as far as 1/2(y^29y*****) I dont know if i factor a 1/2 from the 11 or not.
Of course you factor 1/2 from the 11. Why would you factor it out of the first two terms but not the last one? What is 11 divided by 1/2?
why cant you use 4 as the GCF?
Well you can if the two numbers were 4 and some multiple of 4 but in this case, the numbers involve variables so you have to account for the variables as well as the numbers you are trying to find the greatest common factor.
Question: Pam had two pieces of cloth the same length. She cut the first one into 8 equal parts. She cut the second one into 12 equal parts. How many parts from the second piece equal the same length as 4 parts from the first.
Suppose you have a question like y^3 +y^2 y, won't the GCF be y^3?
No, it might be more clear to you, if you would write power as multiplication:
y*y*y + y*y  y. If you look at each of those numbers, the smallest number has only one y, so you can factor only y: y(y^2+y+1)
GCF is y.
y*y*y + y*y  y. If you look at each of those numbers, the smallest number has only one y, so you can factor only y: y(y^2+y+1)
GCF is y.
What is the fastest way to factor any problem. EX: Polynomials, Trinomials, Binomials Ect.
Thanks, Herobrine
Thanks, Herobrine
Start by factoring out the Greatest Common Factor, then it depends on the makeup of the problem to decide whether it can be factored or which method to use. Sorry no better answer here.
OK, so I understand how to factor binomials, but what exactly is the point. When would this be needed?
i have a question...when a math prob is like this: 6y^3 (to the power of..) do they mean to put 2 'y's' or three extra?
can you use a gcf calculator to find gcf because i found a website with a gcf calculator. This is the website: http://www.mathsisfun.com/greatestcommonfactortool.html.
You could, but it wouldn't work for the variables like x and y.
For example, if you typed in 5x and 15x, they would have no idea what to do and they wouldn't give you an answer. But it would work for the numbers alone.
For example, if you typed in 5x and 15x, they would have no idea what to do and they wouldn't give you an answer. But it would work for the numbers alone.
The GCF of two numbers x and y is 3. Can you predict the GCF of 4x and 4y? Explain your answer.
12, since you multiplied both by 4, that would be part of the GCF.
For example if I where to have 3x squared 3x+60 and it asked me find the GCF of the expression. Then factor the expression. How would I do this because it is different then simply factoring a quadratic expression
Hi, it is probably just 3 as the GCF: 3( x^2 x +20 ). 3 goes into all the coefficients, and we can't take out any x's because 60 doesn't have one.
how would u solve this problem
75x^5y and 135x^2y^3
75x^5y and 135x^2y^3
Factor it out? 75x^5y+135x^2y^3
=> 15x^2y(5x^3+9y^2)
=> 15x^2y(5x^3+9y^2)
i still dont get this pls help : /
Watching earlier videos helps a lot too. Reading other people's questions and answers on the videos help a lot too.
Watch the video one more time and play close attention. That helps for me.
completely factor the polynomial.
4x^3+12x^2+4x+12
4x^3+12x^2+4x+12
how would you factor the equation 9x^3  3x^2  3x +1 = 0?
So first, Factor any constants out of the polynomial
4(x^3+3x^2+x+3)
Now, use a rarely used factoring method: grouping.
4(x^3+x+3x^2+3)
Now realize that if you factor an x out of the first two terms and 3 out of thast two, they become the same thing.
4(x(x^2+1)+3(x^2+1))
Now, you have x of something and 3 of the same something, so add them together to get 3+x of something.
4(x+3)(x^2+1)
And we're done!
Sal
4(x^3+3x^2+x+3)
Now, use a rarely used factoring method: grouping.
4(x^3+x+3x^2+3)
Now realize that if you factor an x out of the first two terms and 3 out of thast two, they become the same thing.
4(x(x^2+1)+3(x^2+1))
Now, you have x of something and 3 of the same something, so add them together to get 3+x of something.
4(x+3)(x^2+1)
And we're done!
Sal
How to use zeros to write a polynomial function
I have to find the lcm and gcf of xy^2 x xy^ 7. How do I do this and what impact does the negative seven have on the answer?
I have watched this video a few times. Every time I think I have it down I get thrown off. Can someone please help me understand how to factor polynomials using GCF?
I'm a bit confused; not about the actual solving, but the meaning of finding the GCF and factoring. When I was younger, I learned that the GCF is basically the biggest term that can go into a set of numbers. For example:
GCF of 8 and 10 is 2
But this video explains it further with more "complications". For example:
GCF of 8 and 10 is 2(4+5)
So could someone tell me which one is the "real" GCF?
GCF of 8 and 10 is 2
But this video explains it further with more "complications". For example:
GCF of 8 and 10 is 2(4+5)
So could someone tell me which one is the "real" GCF?
Hello, maybe I can help. In your first set, they are just 2 separate numbers, 8 & 10. In the second example you give it is an expression, 8+10, and not just 8 & 10. It may not make sense what the difference is, but remember that we are working with variables whose values we don't know. What is the GCF of 8 and 10? It is 2 as you said already. What is the GCF of 8x+10? It is 2, and the factored expression then becomes 2(4x+5). Another way to think of the difference is that your first example is to teach you how to find the GCF, while the second is the _application_ of the GCF to _factor_ an expression.
Where is the exercise?
How do you factor 10x^2+25x
you would take out the GCF between 10 and 25. The multiples of 10 are 1,2,5,10 and the multiples of 25 are 1,5,25. The greatest one that has the same multiple in each of the number is 5. So you would factor out the 5 which the equation would 5(2x^2+5). This is the furthest you can go unless you were to factor out by solving a square.
So, I get how to find the GCF of a polynomial, but how do you find the greatest common monomial factor of a polynomial? Is it sort of the same thing? Or is it different?
I am having math trouble with my math homework with GCF can anyone help me it`s 15y^2240 and I need factor by using GCF please help me.
The GCF is, in other words, is the greatest number that goes into both 15 and 240. In this case the GCF is 15. It goes into both 15 and 240. So
15y^2240. Factor out 15.
15(y^216). This equals to 15(y4)(y+4).
15y^2240. Factor out 15.
15(y^216). This equals to 15(y4)(y+4).
I am factoring polynomials, and trinomials using the GCF first, and then grouping. I can't find a video for this. Help please. Thanks!
I lost you at about 3:45, I don't understand the things after that.... __
What would the GCF of 12p^2 and 30q^5 be?
Grace,
12p^2 when factored to its primes is 2*2*3*p*p
30q^5 when factored to its primes is 2*3*5*q*q*q*q*q
The only common factors in both are 2*3.
So the Greatest Common Factor = 6
I hope that helps make it click for you.
12p^2 when factored to its primes is 2*2*3*p*p
30q^5 when factored to its primes is 2*3*5*q*q*q*q*q
The only common factors in both are 2*3.
So the Greatest Common Factor = 6
I hope that helps make it click for you.
when is the GCF of two or more numbers equal to one of the numbers?
How do I factor 25a^2 9b^2
How do you factor out 60x+20x^2+45?
OK, so you could factor out the 4x to the 3rd y using the first step, but then for the second step when Sal shows that you can take the remaining x and the 2 and put them in parenthesis, where did he get the plus sign from exactly? Is it because 2 is a positive factor and not a negative one?
How do I find the sum of the numbers 56+64 as the product of their GCF and another sum
What if one polynomial has a 5 in it and the other has a 10, lets say the question is "5ly + 10l*2 y*2. Can I write "5 times l times y and for the other one 5 time 2 times l times l times y times y? Like can I use different numbers?
How do I work exponents problems on this page
So why would we rewrite 2 x 2 x X x X x X x y?
how to factor x2  5xs  24s2
(x8s)(x+3s) Has to multiply to AC and Add up to B
Okay, I know this isn't for homework but I am really confused when it comes to solving a simple trinomial using the GCF method. For example if i have 2x^2 + 20x + 50, I am so used to the grouping method I just dont understand how to solve it. Like I understand that the GCF is 2, but then... I don't really know, and my homeschool videos are awfully confusing, can you please help? Thankyou.
how do you factor 8x4
You look for the greatest common factor of 8x and 4. 8x = 2*2*2*x 4 = 2*2 Now ask yourself what are the parts that show up in BOTH factorizations. Answer is 4. So you factor out the 4 by dividing each of the terms by 4.
4(2x  1)
4(2x  1)
So, if i have a problem: 45x^5  20x^2 Answer: 5x^(9x^34) Correct?
Yes, if the question is: how can you factor this expression.
At 8:19, What if I had 120 and 90? Would the GCF be 30? Im confused how you would know for sure.
Oops I didn't mean to say that I thought he was actually asking for the time...
how would you solve with parenthesis x(y+8) + 5(y+8). it throws me off
Factor out y+8
=> (y+8)(x+5)
=> (y+8)(x+5)
Graphing polynomials
how can i solve a^3 +6a^211a
Just factor it out. Watch the videoagain if you do not understand.
What's the difference between the LCM , LCD and GCF ?
?
?
LCM is least common multiple.LCD is least common divisor.
how do you convert a CGF that is negative into a positive GCF?
6k^315k^216k+40 GCF and figure out
watching this video, when I got to the point of 2:28 I tried doing this for my own problem of
16b(^4)+8b(^2)+20b
and it didn't work... I got 4*4*b. did I do something wrong?
16b(^4)+8b(^2)+20b
and it didn't work... I got 4*4*b. did I do something wrong?
Well, the GCF (Greatest Common Factor) of the numbers: 8, 16, and 20 is 4. The greatest exponent of b that can fit into all three is b^1. So knowing this, you will get
4b (4b^3 + 2b + 5)
When you got 4 * 4 * b, that didn't work because you had too many 4s. Try doing a prime factorization of 8, 16, and 20:
8: 2 *2 *2
16: 2* 2* 2* 2
20: 2*2*5
They all have two 2s in common which is 4!
4 *4 is 16, which would result be 2*2*2*2.
8 doesn't have four 2s. 20 doesn't have 4 2s either.
So that is why your original answer was incorrect.
Hope this helps!
4b (4b^3 + 2b + 5)
When you got 4 * 4 * b, that didn't work because you had too many 4s. Try doing a prime factorization of 8, 16, and 20:
8: 2 *2 *2
16: 2* 2* 2* 2
20: 2*2*5
They all have two 2s in common which is 4!
4 *4 is 16, which would result be 2*2*2*2.
8 doesn't have four 2s. 20 doesn't have 4 2s either.
So that is why your original answer was incorrect.
Hope this helps!
16b(^4)+8b(^2)+20b
=4b (4b(^3)+2b+5)
=4b (4b(^3)+2b+5)
can somebody help me out with this problem? 2v^220v
Try factoring out the common terms.
The first term is 2*v*v
The second term is 1*2*2*5*v
So the common factors appear to be 2 and v
Try factroing out a 2v
I hope that helps.
The first term is 2*v*v
The second term is 1*2*2*5*v
So the common factors appear to be 2 and v
Try factroing out a 2v
I hope that helps.
I have the problem 9x^2  30xy + 25y^2 they dont have a gcf, and if i multiply 2*2*2*2*2*2*2*2*2*2*2*2*2*2*2 it doesnt come out to 30. i have 5*5 and 3*3 but im stuck. how do i find this?
Firstly lets consider 9, 30 and 25. Smallest of them is 9, so GCF would be from 1 to 9. 30 and 25, however, do not divide by anything except 1 without a remainder, so GCF will include 1.
Next, we have x^2; x; and no x. Since third variable doesn't have x at all, you cannot factor x.
Next lets consider y: we have no y, y and y^2. Since first variable doesn't have y, it cannot be factored.
In conclusion, you can factor only 1: 1(9x^230xy+25y^2) factoring 1 doesn't do anything, but it is correct to say, that GCF of those 3 expressions is 1.
Next, we have x^2; x; and no x. Since third variable doesn't have x at all, you cannot factor x.
Next lets consider y: we have no y, y and y^2. Since first variable doesn't have y, it cannot be factored.
In conclusion, you can factor only 1: 1(9x^230xy+25y^2) factoring 1 doesn't do anything, but it is correct to say, that GCF of those 3 expressions is 1.
My homework assignment say to find the GCF but do not factor the polynomial so the first question is 33x^3 33x so would the answer just be 11 or 11x?
33x^3 is 11*3*x*x*x
33x is 11*3*x
There is a 11, 1 3, and an x in both of these.
So the greatest common factor would be 11*3*x which is
33x
I hope that helps make it click for you.
33x is 11*3*x
There is a 11, 1 3, and an x in both of these.
So the greatest common factor would be 11*3*x which is
33x
I hope that helps make it click for you.
how would you factor 10x^2y  5xy + y
From those 3 expressions, GCF is y, so you can factor y: y(10x^25x+1)
I'm not sure if you want to factor it all the way, since it is out of the scope of this particular video, but in case you want, it in a form (a+b)(a+b), you cannot factor *10x^25x+1* any further.
I'm not sure if you want to factor it all the way, since it is out of the scope of this particular video, but in case you want, it in a form (a+b)(a+b), you cannot factor *10x^25x+1* any further.
How am I suppose to find the trinomial in factored form as the product of two binomials?
use a calculator
How do you solve 7m^87 ?
Please help! D:
Please help! D:
The greatest common factor is 7, so "undistribute it".... 7(m^8  1). Now the part in parentheses is essentially a difference of squares so the entire expression could be written like 7((m^4)^2 1) = 7(m^4  1)(m^4 + 1). If this is unclear review the video http://www.khanacademy.org/math/algebra/polynomials/v/factoringdifferenceofsquares. You should continue to factor until you have no more difference of squares left, so the last expression becomes 7(m^2  1)(m^2 + 1)(m^4 + 1). There's one more step.
How would you do a problem like 15a25b+10 ?
Well, you have no "common" variables, so the greatest common factor must be a number. The biggest, and only, number that divides 15, 25 and 10 is 5. 15a/5=3a 25b/5=5b 10/5=2
so the expression could be written as: 5(3a5b+2).
so the expression could be written as: 5(3a5b+2).
I don't understand how he gets x to the third at 4:40. ??
x^3 is considered to be a factor of both x^4 and x^3 (x * x^3 = x^4 and 1 * x^3 = x^3). Not only is it a factor of them, but it is the largest of their factors. Just like 4 was the largest shared factor of 4 and 8, x^3 is the largest factor of the x variables. So Sal factors out x^3 like he did for 4 (and y).
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