#-2x^6+2x^5-4x^4-44x^3+120x^2#
#=-2x^2(x^4-x^3+2x^2+22x-60)#
The #-2x^2# factor means that #color(red)(x=0)# is a zero of the polynomial.
Using the rational root theorem, rational roots of
#x^4-x^3+2x^2+22x-60=0#
must be factors of #60#
So possible roots to try are #+-1#, #+-2#, #+-3#, #+-4#, #+-5#, #+-6#, #+-10#, #+-15#, #+-20#, #+-30#, #+-60#.
It's easy to see that #+-1# are not roots. How about #2#?
If #x=2#
#x^4-x^3+2x^2+22x-60#
#= 2^4-2^3+2*2^2+22*2-60#
#=16-8+8+44-60 = 0#
So #color(red)(x=2)# is another zero of the polynomial.
So #(x-2)# is a factor of #x^4-x^3+2x^2+22x-60#
#x^4-x^3+2x^2+22x-60#
#= (x-2)(x^3+x^2+4x+30)#
Rational roots of #x^3+x^2+4x+30 = 0# must be factors of #30# and negative, so #-1#, #-2#, #-3#, #-5#, #-6#, #-10#, #-15# or #-30#.
We have already eliminated #-1# as a possibility.
Trying #-2# does not work.
How about #-3#?
#(-3)^3+(-3)^2+4(-3)+30#
#=-27+9-12+30 = 0#
So #color(red)(x=-3)# is a zero and #(x+3)# is a factor.
#x^3+x^2+4x+30 = (x+3)(x^2-2x+10)#
The discriminant of #x^2-2x+10# is:
#Delta = (-2)^2-(4xx1xx10) = 4-40 = -36 < 0#
So #x^2-2x+10=0# has no real roots.
So the only real roots of the original polynomial are #x=0#, #x=2# and #x=-3#.
