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

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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

Mathematics
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What do you THINK we would do in this case to start the problem?
divide??
Divide what?

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Other answers:

32 and 8
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?
no it wouldnt...
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?
yes
Why?
Because when you divie you are going to get 4 and then you square it..
Take this example: 6(4+x) and expand it. What do you get?
i dont know...?
24x?
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)

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