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Factoring polynomials: common binomial factor
Sal factors n(n-1)+3(n-1) as (n+3)(n-1) by noticing that (n-1) is a common factor.
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- Why can't you continue simplify (n-1)(n+3) as n²+2n+3(3 votes)
- To do what you did, you multiplied the 2 binomials.
Factoring is the opposite of multiplication.
For example, if someone asks you for factors of 15, you would need to respond that the possible factors are: 1 x 15 and 3 x 5. You would not say that the factors are 15 are 15.
The problem in the video is asking for the factors of the polynomial which are: (n-1)(n+3)
Hope this helps.(7 votes)
- in my maths textbook i have two questions that i dont understand.
1 . (x - 3)^2 + 5(x-3)
2. a(a+1) - a(a+1)^2
please tell me how to solve these :) , i understand that i must group the common factors but what do i do the the brackets with the powers to 2(3 votes)
- Since you have (x-3)(x-3) + 5(x-3), you have x-3 in common, so (x-3)(x-3+5) or (x-3)(x+2)
Then you have a(a+1) in common, so (a(a+1))(1 - (a+1)) gives you (a)(a+1)(-a).(2 votes)
- I can't get it right. What am I doing wrong(2 votes)
- Sorry, without seeing what problem you are working on and what you did, I can't tell you what you are doing wrong. If you post with the exact problem and your work, then I can help you figure out what you are doing wrong.(4 votes)
- Why is (n-1) written first?(2 votes)
- Ever heard of the Commutative Property of Multiplication? It tells us the a*b = b*a. Or, with numbers: 2*3 = 3*2. What this means for your question is that the order of the factors does not matter.
You can write: (n-1) (n+3)
or (n+3) (n-1)(3 votes)
- how do we realize what to take common straight off by looking at the question?(2 votes)
- I have a question. How do we factor something that is squared?
For example: Factor 6(x-1)+(x-1)^2(1 vote)
- Use substitution. Swap out "(x-1)" and put is "a"
You would have:
6a + a^2
Hopefully you can see there is a common factor of "a".
This can factor into:
a [6 + a]
Now, swap out the "a" and bring back the "(x-1)":
Hope this helps.(2 votes)
- If I wanted to factor out the expression "6x^2+9x" as the product of two binomials, couldn't I write it as (3x+0)(2x+3) if I really wanted it in that form for some odd reason?(1 vote)
- They are equivalent, but you would then have an extraneous (not needed) number which is generally not included.(2 votes)
- I don't understand how you factor out the (n-1). Are you canceling it out? How is he allowed to factor out the (n-1)?(1 vote)
- Sal is using the distributive property.
Consider: nw + 3w
The common factor is "w". This can be factored out to create: w (n + 3)
Sal is doing the exact same thing. Except, in this case, the common factor is a binomial (n - 1).
Hope this helps.(2 votes)
- What if we made (n-1) cubed and squared, respectively?(1 vote)
- What is binomial factor of a^2+10a-24?(1 vote)
- 1) Watch the videos in this section: https://www.khanacademy.org/math/algebra/polynomial-factorization/factoring-quadratics-1/v/factoring-simple-quadratic-expression
2) You need to find 2 numbers that multiply to -24, that also add to +10
To create a -24, one number must be negative and one must be positive.
Factors of -24 are:
-1 and 24
-24 and 1
-2 and 12
-12 and 2
-3 and 8
-8 and 3
-4 and 6
-6 and 4
Now figure out which pair add to +10. The only pair is the -2 and 12.
The factors become (a - 2)(a + 12)(1 vote)
- Whereas to factor the polynomial below as the product of two binomials and we have n times n minus one plus 3 times n minus one. So I encourage you to pause this video and see if you can figure this out. Well, the key is to realizing that both of these terms have n minus one as a factor. Let me just rewrite the whole thing so we can work on it down here. So this is n times n minus one plus 3 times n minus one. And notice both of them have an n minus one, have an n minus one as a factor. So what we can do is factor out the n minus one or you can view it as undistributing the n minus one and if we do that, we're gonna factor out the n minus one and what are we going to have left over? Well if you take out the n minus one here if you undistribute it out, you're just going to be left with that n. So you're going to have an n there and then for this second term, you factor this n minus one out. You're just going to be left with this positive three plus three. And just like that we are done. We have factored the polynomial below as a product of two binomials. So this is the same thing as n minus one times n plus three. And once again you can check this. You can take this n minus one and distribute it. n minus one times n is this term right over here n times n minus one and then n minus one times three is this term right over here. n minus one times three or three times n minus one.