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### Course: Algebra basics>Unit 7

Lesson 4: Factoring polynomials by taking common factors

# 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.
$6m+15$
Both terms share a common factor of $3$, so we factor out the $3$ using the distributive property:
$\begin{array}{rl}& 6m+15\\ \\ =& 3\left(2m+5\right)\end{array}$
Want a more in-depth explanation? Check out this video.

### Example 2

Factor out the greatest common monomial.
$44{k}^{5}-66{k}^{4}+77{k}^{3}$
The coefficients are $44,66,$ and $77$, and their greatest common factor is $11$.
The variables are ${k}^{5},{k}^{4},$ and ${k}^{3}$, and their greatest common factor is ${k}^{3}$.
Therefore, the greatest common monomial factor is $11{k}^{3}$.
Factoring, we get:
$\begin{array}{rl}& 44{k}^{5}-66{k}^{4}+77{k}^{3}\\ \\ =& 11{k}^{3}\left(4{k}^{2}\right)+11{k}^{3}\left(-6k\right)+11{k}^{3}\left(7\right)\\ \\ =& 11{k}^{3}\left(4{k}^{2}-6k+7\right)\end{array}$
Want another example like this one? Check out this video.

## Practice

Factor the polynomial below by its greatest common monomial factor.
$3{b}^{5}+15{b}^{4}-18{b}^{7}=$

Want more practice? Check out this exercise.

## 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?
• 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! :)
• I think you don't have to rewrite the whole equation with the answer
• 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 still dont know or understand how to do this
• 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 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?