5.+Factoring

Exercise 24.
__Step__ 1. from the expression 6x^3y^2-24xz^2, we can see that the two coefficient** s, **6 and 24, when factorized have the same G.F.C which i** s **6, and they also have a common exponent x.

__#|Step__ 2. We factorize the number 24 which is the same as 4 times 6 and the number 6 which is the same as 6 times 1.

Step 3. Factorize the exponents- the only common exponent is x, so we can we can place x out of the parenthesis which gives us 6x(x^2y^2-4z^2).


 * S's added by Cyndi. **

Exercise 26.
Because this exercise is a four polynomial terms, we will need to factor it by grouping. First, we need to find what the first two terms have in common: in order to solve this we need to find the GCF (Greates Common Factor) of 8 and 28 which is 4. So, what those terms have in common is 4s, Then you divide the two first terms by 4 and place the answers in a parenthesis.

4s(2x+7y)

Second, we do the same process with the two last terms: the GCF of 6 and 21 is 3. so, what those terms have in common is 3t

3t(2x+7y)

the process should be look like this:

4s(2x+7y) - 3t(2x+7y)

Now you have to still find and group what now those terms have in common, which is (2x+7y). The final answer should be:

(2x + 7y) ( 4s - 3t)


 * I deleted your -ing's in the sentence above. (Cyndi) **

Exercise 28.


2xy(4x-3)(4x+3).

The first step is to find out what the GCF of 32 and 18 is. It's 2. With this, we group together 2xy keeping the variables and open a parenthesis. Half (or 32/2) is equal to 16 so that is the first term we put in the parenthesis. We subtract 1 from the exponent 3 to square x. Then we keep the subtraction sign and divide 18 in half (18/2) to get 9. So it looks like this: 2xy (16x^2-9) Then you simplify or factor even further, keeping the 2xy and reducing because 16 and 9 are perfect squares (4 and 3, respectively). Since 16 is raised to the power of 2 you have to distribute it on each side. 9 reduces to 3. 2xy(4x-3)(4x+3)

Exercise 29.


Looking at the expression 6z^2+17z-3, we can see that the first coefficient is 6, the second coefficient is 17, and the last term is 3.

Now multiply the first coefficient 6 by the last term (3)(6)=18-1. //(We need two numbers which by multiplying will give us -18 and those same numbers which// by //adding will give us +17)//

Now the question is: what two whole numbers multiply to 18 ( the previous product) and add to the second coefficient 17.

To find these two numbers, we need to list all of the factors of 18 (the previous product)

Note: list the negative of each factor. this will allow us to find all possible combinations. The factors ha** ve ** to pair up and multiply to 18. But at the end ** it ** has to give you 17.

So the two numbers 6 and 3 both multiply to 18 and you subtract 1 witch give you 17.

Than you replace th** ese ** two numbers, ** and ** group them in two pairs.

Th** e **n factor out the GCF from the first group, than factor out the second group, and at the end all you do is combine like terms.

or more simpl** y ** you can do this...

First you have to find the factors of 6z^2 and -3.

Then you do a multiplication of those factors and the product has to be 17z... and at the end you cross multiply.

The answer would be (6z-1)(z+3).