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## Algebra basics

### Course: Algebra basics>Unit 7

Lesson 4: Factoring polynomials by taking common factors

# 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.

## Want to join the conversation?

• Why can't you continue simplify (n-1)(n+3) as n²+2n+3
• 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.
• 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
• 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).
• I can't get it right. What am I doing wrong
• 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.
• Why is (n-1) written first?
• 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)
• how do we realize what to take common straight off by looking at the question?
• I'm so confused. Wouldn't the answer be 4n-3? How did he get (n-1)(n-3), and what does it mean?
• 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)": `(x-1)[6+(x-1)]` = `(x-1)(x+5)`
Hope this helps.
• 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.