Question

help guide me through this plzzz fan and medal


Find the lowest common denominator of the following

https://crookcountysd.owschools.com/media/g_alg01_2013/10/114.gif and
https://crookcountysd.owschools.com/media/g_alg01_2013/10/115.gif

Answer 1

Okay you are trying to find out the lowest common multiple of these polynomials:

\[\large{6y^2-6,~8y+8}\]

Answer 2

\[\large{6y^2 - 6 = 6(y^2-1)}\]

Now, \(\large{a^2-b^2 = (a-b)(a+b)}\)

Thus, \(\large{6(y^2-1^2) = 6(y-1)(y+1)}\)

Answer 3

And, \(\large{8y+8 = 8(y+1)}\)

Answer 4

Now can you find out the least common multiple of \(\large{6(y-1)(y+1)}\) and \(\large{8(y+1)}\) ?

Answer 5

@sara45 ?

Answer 6

oh my bad off line

Answer 7

Its okay :)

Answer 8

\[24(y^2-1)(y+1)\]

Answer 9

You don't need \(\large{(y^2-1)}\) because it contains both (y-1),(y+1). So either use (y-1) in place of it or don't write (y+1)

Answer 10

o so \[24y^2\]

Answer 11

my bad whith y+1

Answer 12

No no no it would be :

\[\large{24(y-1)(y+1)}\]

Answer 13

i dont think so

Answer 14

Why ?

Answer 15

\[y^2\]

Answer 16

How ?

Answer 17

24y^2

24(y^2 - 1)

24(y^2 - 1)(y + 1)

Answer 18

\[24y^2 = 24(y^2 - 1) =24(y^2 - 1)(y + 1)\]

Answer 19

See 2y^2 is not the same as 24(y^2-1) and it is not at all the same as 24(y^2-1)(y+1)

Answer 20

so is it \[24(y^2 - 1)(y + 1)\]

Answer 21

No no see (y^2-1) already is (y-1)(y+1)

This will mean that you are using the same factor (y+1) twice but this is not allowed

Answer 22

o so it is\[24(y^2 - 1)\]

Answer 23

Yes

Answer 24

ok ty

Answer 25

yw