Question

\[\left(\begin{matrix}4c^2 \\4c^2-24c+36\end{matrix}\right)*\left(\begin{matrix}4c-12 \\ 2c\end{matrix}\right)\]

Answer 1

Multlipy and simplify

Answer 2

It's much simpler if you just use frac {num} {den} * frac {num} {den}

Answer 3

Oh ok sorry

Answer 4

Try this instead:
~[ \frac {4c^2} {4c^2-24c+36} \times \frac {4c-12} {2c} ~]

And change every ~ to a \

Answer 5

\[\frac {4c^2} {4c^2-24c+36} * \frac {4c-12} {2c }\]

Answer 6

It automatically formatted it for him.

Answer 7

There ya go. Now the first thing you'll want to do is factor the denominator of that first fraction.

Answer 8

Sorry everyone, thanks

Answer 9

2*1 times a 2*1
so such multiplication can be performed

Answer 10

no*

Answer 11

wait is he writing fractions or matrices?

Answer 12

it looks like matrices

Answer 13

They are fractions

Answer 14

disappointed..thought this was going to be some cool combinatoric problem ;)

Answer 15

lol, yes, just multiplication of fractions guys.

Answer 16

It's for algebra 2

Answer 17

oops earlier i thought he was adding matrices not fractions so i guess he got the wrong answer for that one question earlier

Answer 18

Sorry again everyone I'm a noob and horrible with math! :(

Answer 19

Don't put yourself down so easily there bud

Answer 20

A lot of people like to say they're bad at math, even if they aren't. I don't really understand the group-think behind that.

Answer 21

lol, thanks to clear the air I am a female, LMAO

Answer 22

So are you still working on this problem? Did you factor the denominator?

Answer 23

yes I got 4c^2+48c

Answer 24

females are good at math.

Answer 25

No, that's not quite right.

Answer 26

LOL, oh not this one!!!!

Answer 27

did I forget the 36

Answer 28

4c^2+48c+36 Is that right?

Answer 29

Bah, that's wrong actually.

It should be 4(c + a)(c+b)

So we need to factor a 4 from everything first.

Answer 30

Can you factor a 4 from each term?

Answer 31

yes

Answer 32

I think

Answer 33

I would divide them by 4 right?

Answer 34

i'm a female too
so if i can do this you can

Answer 35

Bringing the 4 out in front leaves 1 less factor of 4 in each, so yes, you divide by 4 to each of them to see what's left.

Answer 36

don't be sexist against your own kind lol

Answer 37

c^2-6c+9?

Answer 38

With a 4 out in front, and that part in parens, yes.

Answer 39

\[4c^2 - 24c + 36 = 4(c^2 - 6c + 9)\]

Answer 40

OH ok

Answer 41

Would I do the same for the top?

Answer 42

Myin, she can do this. Let her. =)

Answer 43

polpak i didn't do the whole thing lol
i was gonna let her finish the rest of it

Answer 44

Yeah, but this part is the hardest part. Don't steal the fun ;)

Answer 45

lol, thank you for the confidence everyone :)

Answer 46

ok so this is what I got for the top
16c^3-12?

Answer 47

wait huh?

Answer 48

The top of what?

Answer 49

The first fraction or the second?

Answer 50

Don't do the multiplication of the two fractions until after you simplify them, otherwise it'll mean more work later.

Answer 51

The fraction, the 4c^2 and 4c-12

Answer 52

ok now I am confused, I'm sorry

Answer 53

It's ok. You see that there are two fractions here being multiplied right?

Answer 54

Yes

Answer 55

Ok to clear this up for me, the part of the problem we just did was just the denominator for the first fraction right?

Answer 56

Yes exactly

Answer 57

We factored it. Now since there is a factor of 4 on top and bottom you can cancel it right?

Answer 58

\[\frac {4c^2}{4c^2-24c+36} * \frac{4c-12}{2c }\]\[=\frac {4c^2} {4(c^2-6c+9)} * \frac {4c-12} {2c }\]

Answer 59

This is what we've done so far.

Answer 60

Right, so would that cancel out the 4c^2 on top?

Answer 61

You cannot cancel the c^2 because we don't have a factor of c^2 on bottom. But we do have a factor of 4 on each, so the 4's (only) cancel.

Answer 62

Factors are things being multiplied to make a product. The c^2 you have on the bottom is one of the terms (somthing being added) to make a factor, but it is not a factor by itself.

Answer 63

ok, so then would it be \[\left(\begin{matrix}c^2 \\ c^2-62+9\end{matrix}\right)\]

Answer 64

Sorry I keep putting it in wrong

Answer 65

That's ok, you can use combinations for now, I'll know what you mean ;)

Answer 66

That 2 on the bottom should be a c right?

Answer 67

Thank you

Answer 68

lol, yes sorry

Answer 69

Ok, good. Now then, lets look at the other fraction:

\[=\frac {4c^2} {4(c^2-6c+9)} * \frac {4c-12} {2c }\]

We have a 2 on bottom, can we factor out a 2 from the top?

Answer 70

Whoops, forgot to cancel the 4's you got rid of:

\[=\frac {c^2} {c^2-6c+9} * \frac {4c-12} {2c }\]

Answer 71

So can you factor a 2 from the top of the right hand fraction?

Answer 72

So would the other fraction be \[2c-12\]

Answer 73

you have to factor a 2 from both terms in the numerator, you cannot cancel a 2 from just part of the factor.

Answer 74

4c-12 is a factor by itself, you have to extract from it a factor of 2. To do that (just like the denominator of the other fraction) we have to pull a 2 from each term, 4c and -12.

Answer 75

Then we can cancel the factor of 2 from the top and the bottom.

Answer 76

ok so 2c-6

Answer 77

That's the numerator after you cancel, what is left in the denominator?

Answer 78

c?

Answer 79

Also correct.

Answer 80

really?!

Answer 81

\[=\frac {c^2} {c^2-6c+9} * \frac {4c-12} {2c }\]\[=\frac {c^2} {c^2-6c+9} * \frac {c-6} {c }\]

Yes really. I told you that you weren't bad at math ;p

Answer 82

Thank you so so much!!!! I'm so excited!

Answer 83

So now lets combine the two, but just as factors, don't multiply them all out yet.

Answer 84

When you multiply two fractions, what's the proceedure?

Answer 85

Just describe it, don't try to give the result yet.

Answer 86

um, numerator by denominator (multiplying)

Answer 87

Right!, So lets just rewrite them like that without doing the whole distributing thing.. Just put them together with some parens.. What do you have on the top and what on the bottom?

Answer 88

On the top c^2(c)?
and the bottom c^2-6c+9(c-6)?

Answer 89

That doesn't look right to me

Answer 90

No, you flipped the right hand fraction for some reason.

Answer 91

\[=\frac {c^2} {c^2-6c+9} * \frac {2c-6} {c }\]

Answer 92

and you lost the 2 in front of the c on the c-6 also

Answer 93

Ok, so should it be, top c^2(c)
bottom 2c-6(c^2-6c+9)?

Answer 94

No. You still flipped them. Don't do that. Just multiply straight across.

Answer 95

Also put parens around the 2c-6, otherwise it looks like you just multipling the -6.

Answer 96

\[\frac{2}{5}\times \frac{3}{7} = \frac{2(3)}{5(7)}\]

Answer 97

c^2(2c-6)
c^2-6c+9(c)

Answer 98

That's right on top. On bottom you put parens around the wrong thing again. If the denominator has multiple terms, put parens around it when you multiply.