Question

https://www.connexus.com/content/media/478541-3282011-25319-PM-1352166554.jpg

can some1 plz solve dis 4 me, i dont understand math dat good, thx

Answer 1

\[\frac{k+3}{4k-2}(12k^2+2k-4)\]
First, consider the denominator of the fraction 4k-2, what is the common factor of these 2 terms?

Answer 2

2?

Answer 3

Sorry, a typo there :|
Take the common factor out the group the rest => factorization.
So, 4k - 2 = 2 (2k-1)
Got it?

Answer 4

ok :)

Answer 5

Consider 12k^2+2k-4
What is the common factor of these 3 terms?

Answer 6

wait, do we end up with

3(k+3)(2k-1) divided by 2? lol, im probs way off, but im not good in math :/

Answer 7

2?

Answer 8

2 is the common factor of the terms. can you factorise 12k^2+2k-4 by taking the common factor 2 out?

Answer 9

yes, into 4

Answer 10

Hmm..?
12k^2+2k-4 = 2(6k^2) + 2(k) - 2(2)
Take common factor - 2 - out, what do you get?

Answer 11

im not sure..do u just take all da 2s out of da equation?

Answer 12

Hmm... kinda.
2(6k^2) + 2(k) - 2(2) = 2 (6k^2 + k - 2)
Got it?

Answer 13

ok :)

Answer 14

Consider (6k^2 + k - 2)
6k^2 + k - 2
= 6k^2 - 3k + 4k -2
Take out the common factor again,
= 3k (2k-1) + 2 ( 2k-1)
And again, take out the common factor (2k-1)
= (2k-1) ( 3k-2)
Got it so far?

Answer 15

yeah :)

Answer 16

So, now we get
2 (6k^2 + k - 2) = 2(2k-1) ( 3k-2)
\[\frac{k+3}{4k-2}(12k^2+2k-4) = \frac{k+3}{2(2k-1)}[2(2k-1) ( 3k-2)]\]\[=\frac{2(2k-1) ( 3k-2)(k+3)}{2(2k-1)}\]Can you cross out the common factors in numerator and denominator?

Answer 17

yeah..i think i got it from here, thx man :)

Answer 18

Welcome !~