Question 465494
{{{f(x) = 2x^4 - 18x^2}}}
If x = 0 (up to multiplicity) is an obvious root of the polynomial, then factor out the highest power of x from the expression, and then apply the relevant theorem.  Hence {{{f(x) = 2x^2(x^2 - 9)}}}, and apply simply look at {{{x^2 - 9}}}.
a. There is only one variation of sign among the terms of the polynomial, and so there is one positive real root.  If we substitute -x for x in the polynomial, we get the same function, and so this tells us that there is also one negative root.

b.  From the rational roots theorem, the possible rational roots of {{{x^2 - 9}}} are simply the divisors of 9, namely -3, -1,1, and 3.

c. Using {{{x^2 - 9}}}, {{{(-1)^2 - 9 = 1-9 = -8 < 0}}}, while {{{(-6)^2 - 9 = 36-9 = 27 >0}}}, hence by the intermediate value theorem, there is r between (-6, -1) such that f(r) = 0.

d.  {{{f(x) = 2x^4 - 18x^2 = 2x^2(x^2 - 9) = 2x^2(x-3)(x+3)}}}

==> x = 0,0,-3,3 are the roots.