"Factor by grouping" is a technique generally used to factor polynomials of degree 3 or higher.
The technique requires the expression to contain 4 or more terms, and is based on the ability to find a common binomial factor between the groupings.
Factoring by grouping will be used more extensively under the unit Polynomials.

To Use Factoring by Grouping with Quadratics

In this unit, we are going to use "factoring by grouping", along with another technique,
to factor quadratic expressions and solve quadratic equations.

First, let's take a look at the essence of factoring by grouping,
and some simple situations where it applies.

Factor:  x(3x + 2) + 8(3x + 2) Instead of distributing. adding, and then factoring this problem, we will use the fact that each term has a common factor (3x + 2), and apply the Distributive Property in reverse. Factor out the common binomial (3x + 2). (3x + 2) • (x + 8) Done!

Factor: (3x + 8)(x -2) - (x - 3)(x -2) Factor out the common binomial (x - 2). (x - 2)[(3x + 8) - (x - 3)] factor out (x - 2) (x - 2)(3x + 8 - x + 3) distribute & combine (x - 2)(2x + 11) factor out common binomial Done!

Factor:   2x2 + 6x - 3x - 9 While there is no "common factor" for all terms, we can use parentheses to "group" terms. Then we can see a common factor in the first two terms, and then in the last two terms. (2x2 + 6x) + (- 3x - 9) group (beware signs) 2x(x + 3) - 3(x + 3) factor each group (x + 3) (2x - 3) factor out common binomial Done!

Factor:   6x2 - 20x - 16 To start, there is a common factor of 2. 2(3x2 - 10x - 8) But we now need another "middle term", as four terms are needed for "factoring by grouping". Let's change -10x into -12x + 2x. 2(3x2 - 12x + 2x - 8) 2[(3x2 - 12x) + (2x - 8)] group 2[3x(x - 4) + 2(x - 4)] factor each group 2(x - 4)(3x + 2) pull out common binomial Question: How did we know to turn the middle term of -10x into the two specific terms -12x + 2x ?

ANSWER: We needed 4 terms in the quadratic to utilize "factoring by grouping". But we cannot just add another random term. We need two terms equivalent to "-10" that will help us factor the problem. A new technique, called the "ac" Method, described in the next lesson, will help us find the two terms needed to replace the "-10" and allow us to "factor by grouping".

You can see from the last two examples above, that the process of "factoring by grouping" for quadratics will depend upon those two terms in the middle that we create, and the grouping that will lead to a common binomial factor.

Meet you on the next lesson page.