Question

What is the complete factorization of 32-8z^2? A.-8(2+z)(2-z) B.8(2+z)(2-z) C. -8(2+z)^2 D.8(2-z)^2

Answer 1

What do you THINK we would do in this case to start the problem?

Answer 2

divide??

Answer 3

Divide what?

Answer 4

32 and 8

Answer 5

If you were to divide, that would change the value of the whole problem right?
For example, if i were to have a problem 12-6 , and i divide it by 6, it wouldnt be the same as 2-1 would it?

Answer 6

no it wouldnt...

Answer 7

Exactly, so here its asking to factor. But you had the right idea, 8 can go into 32 4 times.
The problem is 32-8z^2.

If I was to take that, and put it in the form, 8(4 - z^2)

would it be the same problem?

Answer 8

yes

Answer 9

Why?

Answer 10

Because when you divie you are going to get 4 and then you square it..

Answer 11

Take this example:

6(4+x)

and expand it. What do you get?

Answer 12

i dont know...?

Answer 13

24x?

Answer 14

Ok, let me explain factoring. When we get a problem like
6(4+x)

the x is a variable, meaning, the x is an unknown number. in this case, we're not trying to find x. We're just trying to rewrite that expression.

6(4+x) means 6*4 + 6*x
which means 24 + 6x.

You take the 6, and multiply it by 4, then, you multiply it by x

so the six multiplied by 4 gives you 24, and the 6 multiplied by x simply gives you 6x. You add those together, and you get 24+6x.

Let's try another, how can you expand:

5(3+z)