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Algebra basics
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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Factoring by common factor review
The expression 6m+15 can be factored into 3(2m+5) using the distributive property. More complex expressions like 44k^5-66k^4 can be factored in much the same way. This article provides a couple of examples and gives you a chance to try it yourself.
Example 1
Factor.
Both terms share a common factor of start color #e07d10, 3, end color #e07d10, so we factor out the start color #e07d10, 3, end color #e07d10 using the distributive property:
Want a more in-depth explanation? Check out this video.
Example 2
Factor out the greatest common monomial.
The coefficients are 44, comma, 66, comma and 77, and their greatest common factor is start color #11accd, 11, end color #11accd.
The variables are k, start superscript, 5, end superscript, comma, k, start superscript, 4, end superscript, comma and k, cubed, and their greatest common factor is start color #11accd, k, cubed, end color #11accd.
Therefore, the greatest common monomial factor is start color #11accd, 11, k, cubed, end color #11accd.
Factoring, we get:
Want another example like this one? Check out this video.
Want to join the conversation?
- I heard there's a way that can solve all the polynomials speedy. That's cross-factoring, but anyone knows how to use cross-factoring, and could it really solve all kinds of polynomials speedy?(3 votes)
- I guess the term 'cross-factoring' is used when you're dividing a polynomial by a polynomial. There is a term 'cross out' when simplifying a polynomial. You just need to factor the denominator and numerator. Then, find the same factors and divide both numerator and denominator. We usually call this 'cross out'.
Hope this help! If you have any questions or need help, please ask! :)(3 votes)
- how do you factor
2x^3-7x^2+7x-5(2 votes)- You will start with a binomial, seeing 2 and 5 adding to 7, I would try (2x - 5), then there will be a trinomial, the first and last terms have to be 1x^2 and 1, so (2x-5)(x^2 - x + 1). Then check, 2x(x^2 - x + 1) - 5((x^2 - x + 1) = 2x^3 - 2x^2 + 2x - 5x^2 + 5x -5. Combining like terms shows it was correct.
If that did not work, you might have to try other combinations of 1 or 2 for the first term and 1, -1, 5, or -5 for the second term of the binomial.(3 votes)
- I don't get what it means by "Factor the polynomial as the product of two binomials". Does it mean to factor it out after the two are multiplied or does it mean to have the product be the factor?(2 votes)
- Wayne,
Factor the polynomial as the product of two binomials mean that you are asked to take an expression that looks like thisa^2+2ab+b^2
(a polynomial)
and algebraically manipulate the terms until the expression looks like this:(a+b)(a+b)
two binomial factors being multiplied(3 votes)
- I still am having a little bit of trouble but I think your video helped. If you can explain a little more I would rely think that it would be helpful?(2 votes)
- I think you don't have to rewrite the whole equation with the answer(2 votes)
- a^2b-ab^2
can you help me solve this?(2 votes) - i still dont know or understand how to do this(2 votes)
- my test is today and im still struggling(2 votes)
- I dont understand how 18b^7 when factored with 3b^4 can equal to 6b^3? How do you get 3 out of 4 and 7??(1 vote)
- b^7 = b*b*b*b*b*b*b
b^4 = b*b*b*b
Divide out 4 of the b's and you get b*b*b = b^3(2 votes)
- How come you don't have a lesson on factoring polynomials by grouping or factoring binomials?(1 vote)
- There are multiple sections on factoring. You just haven't got to those yet. This link is the entire section of factoring for Algebra I: https://www.khanacademy.org/math/algebra/polynomial-factorization
If you are in Algebra II, there are some additional techniques shown in that section.(1 vote)