Question

solve for all values of x: 2x^3-3x^2-8x+12=0 are

Answer 1

I'll give you the first step and you will do it, OK?

Answer 2

ok thanks!

Answer 3

2x^3-3x^2-8x+12=0 write it in a different order
2x^3-8x-3x^2+12=0 factor
2x(x^2-4)-3(x^2-4)=0
(2x-3)(x^2-4)=0

Go

Answer 4

(HINT: one of the parenthesis has to be equal zero)

Answer 5

So (2x-3)=0 and (x^2-4)=0

2x-3=0 x^2-4=0
x=? x=?

Solve for x in each one....

Answer 6

thank you! appreciate the assistance.

Answer 7

Anytime!

Answer 8

I don't know what x equals on the 2x-3=0 side.. confused

Answer 9

2x-3=0 x^2-4=0
2x-3 + 3 = 0 + 3 x^2-4 + 4 =0 + 4
2x = 3 x^2 = 4
x= 3/2 x= 2, -2

Answer 10

tnx for the medal brah!

Answer 11

thanks 4 your help. I understand now. and your welcome!!

Answer 12

Anytime!

Answer 13

I know what it feels like when you don't get your questions answered and explained.
I had this in Literature, and that's why i helped you.