The idea behind all of these problems is to find the largest value that two terms have in common. The goal is to factor out this value (GCF), leaving a binomial. When factoring by grouping, you should end up with two binomials that are the same. When
that happens, you can rewrite the problem by placing both of the GCF's in one set of parentheses and the other set of parentheses will contain the binomial that remained after pulling out the GCF's.
For example: #2
1) x^3 and 11x both have an x in common, so we will pull an x out of each term (divide each term by x), leaving us with
x^2 and 11.
Since we pulled the same value out of each term, we will place the GCF outside the parentheses and the remainders inside the parentheses, like this:
x(x^2 + 11)
(you can check your work by multiplying each term by x and verifying that you get what you started with)
2) Do the same thing with the terms 4x^2 and 44. They both have a +4 in common. Divide each term by the GCF (+4) to get:
+4(x^2 + 11)
3) The problem should now look like this:
x(x^2 + 11) +4(x^2 + 11)
Notice how both set of parentheses are the same?
When this happens, we can rewrite the problem by grouping the GCF's (x and +4) in their own set of parentheses
(x + 4)
and multiplying them by the terms that are already in parentheses (x^2 + 11).
Giving us:
(x + 4)(x^2 + 11) which is the answer.
The most commonly asked question that I hear is: "Why do we only write the (x^2 + 11) one time when there are two of them in the problem?"
If you look at,
x(x^2 + 11) +4(x^2 + 11)
each term in (x^2 + 11) is being multiplied by x one time and by +4 one time. If we write (x^2 + 11) twice when we rewrite it, then we would effectively be multiplying each term by x twice and by +4 twice and we don't want to do that.
Nov 6

Kristina H.
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