-2x^6+2x^5-4x^4-44x^3+120x^2−2x6+2x5−4x4−44x3+120x2
=-2x^2(x^4-x^3+2x^2+22x-60)=−2x2(x4−x3+2x2+22x−60)
The -2x^2−2x2 factor means that color(red)(x=0)x=0 is a zero of the polynomial.
Using the rational root theorem, rational roots of
x^4-x^3+2x^2+22x-60=0x4−x3+2x2+22x−60=0
must be factors of 6060
So possible roots to try are +-1±1, +-2±2, +-3±3, +-4±4, +-5±5, +-6±6, +-10±10, +-15±15, +-20±20, +-30±30, +-60±60.
It's easy to see that +-1±1 are not roots. How about 22?
If x=2x=2
x^4-x^3+2x^2+22x-60x4−x3+2x2+22x−60
= 2^4-2^3+2*2^2+22*2-60=24−23+2⋅22+22⋅2−60
=16-8+8+44-60 = 0=16−8+8+44−60=0
So color(red)(x=2)x=2 is another zero of the polynomial.
So (x-2)(x−2) is a factor of x^4-x^3+2x^2+22x-60x4−x3+2x2+22x−60
x^4-x^3+2x^2+22x-60x4−x3+2x2+22x−60
= (x-2)(x^3+x^2+4x+30)=(x−2)(x3+x2+4x+30)
Rational roots of x^3+x^2+4x+30 = 0x3+x2+4x+30=0 must be factors of 3030 and negative, so -1−1, -2−2, -3−3, -5−5, -6−6, -10−10, -15−15 or -30−30.
We have already eliminated -1−1 as a possibility.
Trying -2−2 does not work.
How about -3−3?
(-3)^3+(-3)^2+4(-3)+30(−3)3+(−3)2+4(−3)+30
=-27+9-12+30 = 0=−27+9−12+30=0
So color(red)(x=-3)x=−3 is a zero and (x+3)(x+3) is a factor.
x^3+x^2+4x+30 = (x+3)(x^2-2x+10)x3+x2+4x+30=(x+3)(x2−2x+10)
The discriminant of x^2-2x+10x2−2x+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.
