Question 554217
POSSIBLE RATIONAL ZEROS
The rational zeros will be rational numbers (fractions) with denominators that are divisors/factors of the leading coefficient, and numerators that are divisors/factors of the independent term.
For {{{f(x)=x^4+2x^2-24}}} the leading coefficient is the invisible {{{1}}} in front of {{{x^4}}}, and the independent term is {{{-24}}}.
The only positive divisors/factor of 1 is 1. 
The positive divisors/factors of 24 are:
1, 2, 3, 4, 6, 8, 12, and 24.
The possible rational zeros are 1, -1, 2, -2, 3, -3, etc.
HOW TO FIND THE POSSIBLE RATIONAL ZEROS
My method to find the factors, was to start with 1, and check integers to see if they would divide 24 evenly. Once I got to a factor that squared was equal or greater than 24, I used another strategy.
Factors come in pairs that multiply to give you 24:
1 x 24 = 24,
2 x 12 = 24
3 x 8 = 24
4 x 6 = 24
Once I got to 6, and saw that {{{6^2=36>24}}},
I found the larger factors from the smaller factors I found before, by dividing as in
{{{24/3=8}}}, and {{{24/2=12}}}.
Another method to find the factors is to work form the prime factorization:
{{{24=2^3*3}}} so the factors/divisors will all be
{{{2^a*3^b}}}, with {{{0<=a<=3}}} and {{{0<=b<=1}}}.
There are 4 possible values for {{{a}}} and 2 for {{{b}}}, so I knew there would be 8 factors.
For a number like 24, calculating them as {{{2^a*3^b}}} would have been a pain, so I used the other method.
NOTE:
To find the zeros of {{{f(x)=x^4+2x^2-24}}}, I would solve the equation
{{{x^4+2x^2-24=0}}} by changing variables to {{{y=x^2}}}, so that the equation would transform into
{{{y^2+2y-24=0}}} ---> {{{(y-4)(y+6)=0}}}
Back to the original function, I would write it as
{{{f(x)=x^4+2x^2-24=(x^2-4)(x^2+6)=(x-2)(x+2)(x^2+6)}}}
Then I would know that the only real zeros are 2, and -2, which were 2 of the 24 possible rational zeros.
The factor {{{(x^2+6)}}} is not zero for any real value of {{{x}}}.