Question

HELP PLEASE! multiply k+3/4k-2 times (12k^2-3)

Answer 1

Hint:

Factor 4k-2 to get 2(2k-1)

and

factor 12k^2-3 to get 3(4k^2-1) which factors further to 3(2k+1)(2k-1)

Answer 2

You should see some common terms that will cancel

Answer 3

um i need a bit more help with this @jim_thompson5910

Answer 4

\[\Large \frac{k+3}{4k-2} \times (12k^2-3)\]


\[\Large \frac{k+3}{4k-2} \times \frac{12k^2-3}{1}\]


\[\Large \frac{k+3}{2(2k-1)} \times \frac{3(2k+1)(2k-1)}{1}\]


\[\Large \frac{k+3}{2\cancel{(2k-1)}} \times \frac{3(2k+1)\cancel{(2k-1)}}{1}\]


See how I'm getting all this?

Answer 5

yes i do

Answer 6

but 3(2k+1)/1 is not one of the multiple choice answers

Answer 7

one moment

Answer 8

ok

Answer 9

\[\Large \frac{k+3}{4k-2} \times (12k^2-3)\]

\[\Large \frac{k+3}{4k-2} \times \frac{12k^2-3}{1}\]

\[\Large \frac{k+3}{2(2k-1)} \times \frac{3(2k+1)(2k-1)}{1}\]

\[\Large \frac{k+3}{2\cancel{(2k-1)}} \times \frac{3(2k+1)\cancel{(2k-1)}}{1}\]

\[\Large \frac{k+3}{2} \times \frac{3(2k+1)}{1}\]

\[\Large \frac{(k+3)*3(2k+1)}{2*1}\]

\[\Large \frac{3(k+3)(2k+1)}{2}\]

\[\Large \frac{3(2k^2+k+6k+3)}{2}\]

\[\Large \frac{3(2k^2+7k+3)}{2}\]

\[\Large \frac{6k^2+21k+9}{2}\]

Answer 10

Hopefully that all makes sense

Answer 11

here are the multiple choice answers A.)3(k+3)(2k+1)/2 B) 3(k+3)(2k-1)/2 C) 3(k+3)/2 D) 3(k+3)(k-1)

Answer 12

ah so they left it factored, ok let me fix it

Answer 13

\[\Large \frac{k+3}{4k-2} \times (12k^2-3)\]

\[\Large \frac{k+3}{4k-2} \times \frac{12k^2-3}{1}\]

\[\Large \frac{k+3}{2(2k-1)} \times \frac{3(2k+1)(2k-1)}{1}\]

\[\Large \frac{k+3}{2\cancel{(2k-1)}} \times \frac{3(2k+1)\cancel{(2k-1)}}{1}\]

\[\Large \frac{k+3}{2} \times \frac{3(2k+1)}{1}\]

\[\Large \frac{(k+3)*3(2k+1)}{2*1}\]

\[\Large \frac{3(k+3)(2k+1)}{2}\]

Answer 14

So it looks like choice A

Answer 15

Ok ya that makes much more sense haha sorry i should have given u the choices to begin with

Answer 16

that's ok, I'm glad it makes more sense now

Answer 17

thanks again jim

Answer 18

np