Question

i have kind of a messy one but its my last subject. i have((2x^3-16)/6x^2+6x-36)) multiplied by ((9x+18)/3x^2+6x+12)) where do i begin?

Answer 1

\[((2x^3-16)(9x+18))\div((6x^2+6x-36)(3x^2+6x+12))\]

Answer 2

Then take the greatest common divisor out of (3x^2+6x+12) Hint(its 3)

Answer 3

now you will have \[(2x^3−16)(9x+18))\div(3(x^2+2x+4)(6x^2+6x−36))\]

Answer 4

ok so youd foil the numerator, what would u do to the denominator?

Answer 5

factor a 6 out of (6x^2 +6x-36)

Answer 6

factor (x^2+x-6) to get (x+3)(x-2)

Answer 7

oh i see

Answer 8

did you get the answer

Answer 9

if not ill keep going

Answer 10

i got the denominator factored out im just figuring out the top now

Answer 11

you should now have (9(2x^3-16)(x+2))/(6*3(x+3)(x+2)(x^2+2x+4) now factor a 2 out of (2x^3-16) to get (x^3-8) which is equal to (x^3-2^3)

Answer 12

now take x^3-2^3 and use difference of two cubes to get (x-2)(x^2+2x+2^2)

Answer 13

now it should look like \[\frac{ 2*9(x-2)(x^2+2x+2^2)(x+2) }{ 3*6(x+3)(x+2)(x^2+2x+4)}\]

Answer 14

ok I think i got it!!!! so the answer should be x+2/x+3?

Answer 15

yup

Answer 16

\[\frac{ x+2 }{ x+3 }\] thank you so much!