Question 715742
Rational roots would be fractions (negative and positive)
whose numerator is a factor of the independent term, the constant {{{12}}} ,
and whose denominator is a factor of the leading coefficient, the invisible {{{1}}} in front of {{{x^4}}}.
The denominator can only be {{{1}}}, the only factor of {{{1}}},
but {{{12}}} has 6 factors:
1, 2, 3, 4, 6, and 12.
So the possible rational roots are:
-12, -6, -4, -3, -2, -1, 1, 2, 3, 4, 6, and 12.
 
Finding the roots is easier.
No need to try any of those rational roots.
If we change variables using {{{y=x^2}}} ,
the equation becomes
{{{y^2-7y+12=0}}} --> {{{(y-3)(y-4)=0}}} and that factoring tells us that
{{{y=3}}} and {{{y=4}}} are solutions of {{{y^2-7y+12=0}}}
and will lead us to solutions of {{{x^4-7x^2+12=0}}}
{{{x^2=3}}} leads us to {{{highlight(x=-sqrt(3))}}} and {{{highlight(x=sqrt(3))}}} (not rational roots< but at least real roots)
{{{x^2=4}}} leads us to
{{{x=-sqrt(4)}}} --> {{{highlight(x=-2)}}} and
{{{x=sqrt(4)}}} --> {{{highlight(x=2)}}} (two of the 12 possible rational roots).