Question

@ganeshie8

Answer 1

So first I flip the right so that it's multiplication..

Answer 2

thats a good start !

Answer 3

keep going

Answer 4

And then I multiply them (multiplying one second..)

Answer 5

dont do that wait

Answer 6

try to factor and see if something cancel out

Answer 7

factor c^2-4c+4
and c^2-4

Answer 8

Mmk

Answer 9

so c^2 - 4c + 4 is (c - 2)^2
and c^2 - 4 is (c + 2)(c - 2)

Answer 10

so (c - 2) cancel out

Answer 11

perfect !

Answer 12

mess wid other two terms also..
factor them and see if u can cancel out anything

Answer 13

give it a good try, ping me if u get stuck :)

Answer 14

12c^3 + 30c is 6c(2c^2 + 5)
and 6c^4 + 15c^3 is 3c^2(2c^2 + 5c)?

Answer 15

look again, its 12c^3 + 30c^2

Answer 16

i mean in ur question, it is 12c^3 + 30c^2

Answer 17

so then 6c(2c^2 + 5c)?

Answer 18

yes !

Answer 19

was the other one right?

Answer 20

other one is right

Answer 21

so (2c^2 + 5c) can cancel out, (c - 2) can cancel out, and that's it I think?

Answer 22

do it in ur book... and tell me wat u get in the end

Answer 23

I got 3c^3 - 6c^2/6c^3 + 12c^2

Answer 24

you can still factor and cancel things, think a bit

Answer 25

before or after I multiply it out?

Answer 26

3c^3 - 6c^2/6c^3 + 12c^2

after this

Answer 27

mmk one sec

Answer 28

so I got 3c(c^2 - 2c)/6c(c^2 + 2c)

Answer 29

yeah, cancel..

Answer 30

canceled now I have 3c/6c do I reduce that to c/2c or leave it as it is?

Answer 31

\(\large \frac{c^2-4c+4}{12c^3+30c^2} \div \frac{c^2-4}{6c^4+15c^3} \)

\(\large \frac{c^2-4c+4}{12c^3+30c^2} \times \frac{6c^4+15c^3}{c^2-4} \)

\(\large \frac{(c-2)^2}{12c^3+30c^2} \times \frac{6c^4+15c^3}{(c-2)(c+2)} \)

\(\large \frac{c-2}{12c^3+30c^2} \times \frac{6c^4+15c^3}{c+2} \)

\(\large \frac{c-2}{6c^2(2c+5)} \times \frac{3c^3(2c+5)}{c+2} \)

\(\large \frac{c-2}{6c^2} \times \frac{3c^3}{c+2} \)

Answer 32

fine so far ?

Answer 33

Yes, I forgot to put the cubed parts there ><

Answer 34

\(\large \frac{c^2-4c+4}{12c^3+30c^2} \div \frac{c^2-4}{6c^4+15c^3} \)

\(\large \frac{c^2-4c+4}{12c^3+30c^2} \times \frac{6c^4+15c^3}{c^2-4} \)

\(\large \frac{(c-2)^2}{12c^3+30c^2} \times \frac{6c^4+15c^3}{(c-2)(c+2)} \)

\(\large \frac{c-2}{12c^3+30c^2} \times \frac{6c^4+15c^3}{c+2} \)

\(\large \frac{c-2}{6c^2(2c+5)} \times \frac{3c^3(2c+5)}{c+2} \)

\(\large \frac{c-2}{6c^2} \times \frac{3c^3}{c+2} \)

\(\large \frac{c-2}{2} \times \frac{c}{c+2} \)

\(\large \frac{c(c-2)}{2(c+2)} \)

Answer 35

Done. see if that makes some sense

Answer 36

Ohh okay, it makes sense now, I didn't complete cancel

Answer 37

good, gtg cya

Answer 38

Thank you for your help!

Answer 39

np :)