Question

find all real zeroes of the function:
f(x)= x^3 - 6x^2 -3x + 18.

Answer 1

x^2(x-6) -3(x-6)
(x^2-3)(x-6)=0
x=6, \[\pm \sqrt{3}\]

Answer 2

wow, i see where i went wrong now. on (x^2 -3) I put x^2 - 6 and got x + square root of x and x = plus or minus square root of 6.

Answer 3

thank you