<PRE><B>Question 982209</B>
Try factoring and substituting for x^3.


{{{y=-2(x^6+16x^3+59)}}}


Let v=x^3 and .....


{{{y=-2(v^2+16v+59)}}}


-----
CHANGE IN THOUGHT


IF you are in a Calculus 1 course, try taking derivative, and set equal to zero.  You should be able to find something near or at x=2.  This question would be better in a Calculus 1 course than a College Algebra course.</PRE>