Exercise 24.

Step 1. from the expression 6x^3y^2-24xz^2, we can see that the two coefficients, 6 and 24, when factorized have the same G.F.C which is 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 25.


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.


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


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 27.


Exercise 28.



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.

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 have 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 these two numbers, and group them in two pairs.

Then 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 simply 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).

Exercise 30.


Exercise 31.