**Factor the GCF out of this given expression:Please explain the steps on you got it so as i can practice with it for future similar problems**

**Factor by grouping**

(1) 3x(7x-2)+(7x-2)

(2) x^3+4x^2+11x+44

(3) x^2+3x-9x-27

(4) 5x^2-25x-12x+60

(5) 4x^2-48x+x-12

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Kristina H. | Credentialed Math Teacher and Much More.Credentialed Math Teacher and Much More.

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.

Well done Kristina!!

Hi Ma,

I took a little different approach on #1 than Brian.

The Greatest Common Factor is (7x-2) so it can be re-written as...

(7x-2)(3x+1) reversing the distributive law. You are done.

#2

x^{3} +4x^{2 }+11x +44...First set it = to zero and do trial and error to get one of the solutions....

Positive numbers I know won't work because 1...2...etc keep getting bigger positive. So I tried negative solutions and -4 will give you zero...(-64+64-44+44)=0 So we know x=-4 works.

Therefore one of the factors is (x+4)

Now check the remainder theorem lesson under the algebra section here in WyzAnt. Using Synthetic division I can now divide my original expression by (x+4) and get (x^{2
}+11)...

So my answer is (x+4)(x^{2 }+11)...

If you multiply this back using the FOIL method you will get the original expression.

#3 by grouping this time...

x^{2 }+ 3x -9x -27

1st Group x^{2 }+3x

Factor out the x and get x(x+3)

2nd group -9x-27

Factor out a -9 and get -9(x+3)

So my final answer is (x+3)(x-9)

Oops Ma...I got the same answer as Kristina to #2 using a different method but since you were supposed to factor by grouping her answer is much better than mine! It is good to note that there is more than one way to skin a cat.

3x(7x-2)+(7x-2)

first, lets factor 3x back in:

21x²-6x+(7x-2)

because we are adding (or multiplying the second term by +1) we can remove the parentheses:

21x²-6x+7x-2

wow we can join like terms:

12x²+x-2

and now you factor:

You should be looking for factors of 21, and -2 that differ by 1 when multiplied by each-other.

After Brian joined like terms here he meant

21x^{2 }+x -2.

I see Brian already caught his error.

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