Question

Multiply and simplify. 4c^2/4c^2-32c+65*4c-16/2c please help me by showing the steps on how to do this. Thanks!

Answer 1

can you use some parentheses to clarify the question

Answer 2

Sure

Answer 3

\[(4c^2/4c^2-32c+64)\times(4c-16/2c)\]

Answer 4

\[\frac{4c^2}{4c^2-32c+64}*\frac{4c-16}{2c}\]

Answer 5

Yes exactly

Answer 6

I'd first factor a 4 out of the 1st fraction

Answer 7

So 1/-32c+64 ?

Answer 8

Now the 1st denominator is easy to factor

Answer 9

Oh I see. Let me see how far I get doing that.

Answer 10

i'm sorry actually the 4's would cancel, so you don't multiply the entire thing by 4

Answer 11

\[\frac{4c^2}{4(c^2-8c+16)}*\frac{4(c-4)}{2c}\]

Answer 12

can you work on it from there?

Answer 13

Let me try :)

Answer 14

Do I multiply straight across? 4c^2*4(c-4)?

Answer 15

not yet, factor the denominator of the 1st fraction first, you'll see why you shouldn't multiply yet

Answer 16

Oh! As if it is a quadratic?

Answer 17

yes

Answer 18

4(c-4)^2 ?

Answer 19

right

Answer 20

now you can start cancelling stuff

Answer 21

\[\frac{4c^2}{4(c^2-8c+16)}*\frac{4(c-4)}{2c}=\frac{\cancel{4}c^2}{\cancel{4}\cancel{(c-4)}(c-4)}*\frac{4\cancel{(c-4)}}{2c}\]

Answer 22

this can be further simplified

Answer 23

now you multiply across

Answer 24

you there moonie?

Answer 25

Yeah I'm trying to do the math. :) 4c^2/2c^2-4 ?

Answer 26

At this point we have this:
\[\frac{4c^2}{2c(c-4)}\]

Answer 27

Rob, thank you but the original equation was written incorrectly, Rickjbr is heping me get the original equation solved correctly.

Answer 28

Oh I see, I can't square the c's.

Answer 29

you can, but you would have to undo it for the next step, so it's pointless. Also,
2c(c-4)=2c^2-8c

Answer 30

how did you get 2c^2-4?

Answer 31

Bad math... I was trying to use the distributive law but I didn't follow all the way through with it.

Answer 32

ah ok

Answer 33

You always want to cancel out as many terms as you can before multiply anything.

Answer 34

Can you reduce the fraction we have at this point?

Answer 35

I think so...

Answer 36

2/-8c ?

Answer 37

no, walk me through how you got that

Answer 38

Be cause 2c^2 will go into 4c^2 one time with a remainder of 2? (God I'm horrible at this)

Answer 39

This is why i told you not to expand the fraction yet, because you can't reduce fractions with addition or subtraction signs in the middle

Answer 40

\[\frac{4c^2}{2c(c-4)}\]

Answer 41

This is what we're working with right now

Answer 42

OH!..... Okay let me try again.

Answer 43

Are you using this programs eqiation editor? (Just a side question)

Answer 44

yes

Answer 45

So would I expand the numerator since it is squared and then try to cancel from there?

Answer 46

the numerator can't be expanded

Answer 47

What is 4/2?

Answer 48

2

Answer 49

So coefficients first then variables and exponents?

Answer 50

Yes, but everything has to be in factored form for you to divide and cancel. It is in factored form right now, so you can do that.

Answer 51

rob, you should read the 5th post from the top :/

Answer 52

making any progress?

Answer 53

Um, maybe... 4c^2/2c(c-4) = 2c^2/c(c-4) Can I use the distributive law on the denominator now?

Answer 54

nope, there's some more cancelling to d o

Answer 55

Is my equation correct so far?

Answer 56

yes, it is equivalent

Answer 57

What is
\[\frac{c^2}{c}\]

Answer 58

Do I divide 2 by -4?

Answer 59

1/2?

Answer 60

no you can't do that

Answer 61

Properties of exponents, quotient rule:
\[\huge \frac{a^m}{a^n}=a^{m-n}\]

Answer 62

\[\huge \frac{c^2}{c^{1}}=c^{2-1}=c^1=c\]

Answer 63

Oh, I didn't know I could use 1 as an exponent

Answer 64

If that is the case then we are left with 2c/ (c-4) ?

Answer 65

You can't touch anything inside of the parentheses because there are addition and subtraction symbols in there

Answer 66

Yes, that's the final answer, no further factoring possible

Answer 67

Omg that was worse then pulling a tooth. Thank you so much for your patience!

Answer 68

you're welcome.

Answer 69

1 is an implied exponent, nobody writes it because it's always there

Answer 70

I understand. I just don't remember things like that. Okay well, I have four more problems to get figured out. Thank you again!

Answer 71

Refer to the solution attachment.