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?

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! :)

how do you factor 2x^3-7x^2+7x-5

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.

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?

Wayne, _Factor the polynomial as the product of two binomials_ mean that you are asked to take an expression that looks like this `a^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_

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??

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

Main content

Algebra basics

Course: Algebra basics > Unit 7

Lesson 4: Factoring polynomials by taking common factors

Factoring by common factor review

Google Classroom

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.

6, m, plus, 15

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:

\begin{aligned} &6m+15\\\\ =&\goldD{3}(2m+5) \end{aligned}

Want a more in-depth explanation? Check out this video.

Example 2

Factor out the greatest common monomial.

44, k, start superscript, 5, end superscript, minus, 66, k, start superscript, 4, end superscript, plus, 77, k, cubed

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:

\begin{aligned} &44k^5-66k^4+77k^3\\\\ =&\blueD{11k^3}(4k^2)+\blueD{11k^3}(-6k)+\blueD{11k^3}(7)\\\\ =&\blueD{11k^3}(4k^2-6k+7) \end{aligned}

Want another example like this one? Check out this video.

Practice

Factor the polynomial below by its greatest common monomial factor.

3, b, start superscript, 5, end superscript, plus, 15, b, start superscript, 4, end superscript, minus, 18, b, start superscript, 7, end superscript, equals

Want more practice? Check out this exercise.

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💎Chυcκ Lørrε💎
Posted 6 years ago. Direct link to 💎Chυcκ Lørrε💎's post “I heard there's a way tha...”
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?
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(3 votes)
Answer
- David Lee
  Posted 6 years ago. Direct link to David Lee's post “I guess the term 'cross-f...”
  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! :)
  Comment on David Lee's post “I guess the term 'cross-f...”
  (3 votes)
Conrad
Posted 4 years ago. Direct link to Conrad's post “how do you factor 2x^3-7...”
how do you factor
2x^3-7x^2+7x-5
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(2 votes)
Answer
- David Severin
  Posted 4 years ago. Direct link to David Severin's post “You will start with a bin...”
  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.
  Button navigates to signup page
  (3 votes)
wayne
Posted 5 years ago. Direct link to wayne's post “I don't get what it means...”
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?
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(2 votes)
Answer
- AD Baker
  Posted 4 years ago. Direct link to AD Baker's post “Wayne, _Factor the polyn...”
  Wayne,
  Factor the polynomial as the product of two binomials mean that you are asked to take an expression that looks like this
  
  a^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
  Comment on AD Baker's post “Wayne, _Factor the polyn...”
  (3 votes)
Bhavya
Posted 5 years ago. Direct link to Bhavya's post “I still am having a littl...”
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?
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(2 votes)
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Makenna Petrusek
Posted 4 years ago. Direct link to Makenna Petrusek's post “I think you don't have to...”
I think you don't have to rewrite the whole equation with the answer
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(2 votes)
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emiliamolla10
Posted 5 years ago. Direct link to emiliamolla10's post “a^2b-ab^2 can you help me...”
a^2b-ab^2
can you help me solve this?
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pearl.herrera
Posted 5 years ago. Direct link to pearl.herrera's post “i still dont know or unde...”
i still dont know or understand how to do this
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(2 votes)
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diamond krigger
Posted 5 years ago. Direct link to diamond krigger's post “my test is today and im s...”
my test is today and im still struggling
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Narrycanoodles
Posted 5 years ago. Direct link to Narrycanoodles's post “I dont understand how 18b...”
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??
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(1 vote)
Answer
- Kim Seidel
  Posted 5 years ago. Direct link to Kim Seidel's post “b^7 = b*b*b*b*b*b*b b^4 =...”
  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
  Button navigates to signup page
  (2 votes)
homestuckian375
Posted 7 years ago. Direct link to homestuckian375's post “How come you don't have a...”
How come you don't have a lesson on factoring polynomials by grouping or factoring binomials?
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(1 vote)
Answer
- Kim Seidel
  Posted 7 years ago. Direct link to Kim Seidel's post “There are multiple sectio...”
  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.
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  (1 vote)