Calculus
posted by Z32 .
Find the inflection points for f(x) = 12x^5 + 45x^4  360x^3 + 7

f'(x) = 60x^4 + 180x^3  1080x^2
f''(x) = 240x^3 + 540x^2  2160x
= 0 for points of inflection
240x^3 + 540x^2  2160x = 0
divide by 60
4x^3 + 9x^2  36x = 0
x(4x^2 + 9x  36) = 0
Can you take it from there ?
There are 3 solutions for x 
I got 0 for one of my inflections since a 0 in the place of the "x"(4x^2 + 9x  36) would make it zero. But I'm not sure what would need to go into the other x's to make it 0. I tried 4/9 9/36 and 36/9 but I couldn't get it.

Oh, okay. I found out I had to use the quadratic formula. Thanks for the help!