Solve the equation 12xcube-8xsquare+kx+18=0 and find the value of k given that one root is negative of other

Answer 1

12x^3-8x^2+kx+18 = 0
If one root is a and the other is -a, then we have (with the third root being p)
12(x-a)(x+a)(x-p) = 12(x^2-a^2)(x-p)
= 12x^3-12px^2-12a^2x+12a^2p

So, we need
-12p = -8
p = 2/3

12a^2p = 18
a^2 = 9/4
a = 3/2

k = -12a^2 = -27

check: (3x-2)(2x-3)(2x+3) = 12x^3-8x^2-27x+18

Answer 2

check