Brainliest award will be given!with math ive been asking for hours! ! 1) solve the following equation by identifying all of its roots including all real and complex numbers. in your final answer, include the necessary steps and calculations. hint: use your knowledge of factoring polynomials. (x^2 + 1)(x^3 + 2x)(x^2 - 64) = 0 2) solve for the roots of x in each of the equations below. x^4 - 81 = 0 x^4 + 10x^2 + 25 = 0 x^4 - x^2 - 6 = 0 3) solve for the roots in the equation below. x^4 + 3x^2 - 4 = 0

1. (x^2+1)*(x^3+2*x)*(x^2-64)
=(x^2+1)*x*(x^2+2)*(x+8)(x-8)
Solving for each factor in turn, for example, 
x^2+1=0 => x^2=-1 => x=+i, x=-i
x=0 => x=0
x^2+2=0 => x^2=2 => x=+sqrt(2)i, -sqrt(2)i
x+8=0 => x=-8
x-8=0 => x=+8
we have solution set
S, whereS={+i, -i, 0, +sqrt(2)i, -sqrt(2)i, -8, +8)

2. A.
x^4-81=0 => x^4=81 => x^2=+9 or x^2=-9
x^2=+9  => x=+3, -3
x^2=-9  => x=+3i, -3i
S={+3i, -3i, +3, -3}

B.
x^4+10x^2+25=0 => (x^2+5)^2=0 => ± (x^2+5)=0 => x^2=-5
=> x=+sqrt(5)i (multiplicity 2 and x=-sqrt(5)i (multiplicity 2)
S={+sqrt(5)i (multiplicity 2) -sqrt(5)i (multiplicity 2)}

C.
x^4-x^2-6=0 => (x^2-3)(x^2+2)=0 => x^2=3 or x^2=-2
S={+sqrt(2)i,-sqrt(2)i, +sqrt(3), -sqrt(3) }

3.
x^4+3x^2-4=0 = (x^2-1)(x^2+4) => x^2=1 or x^2=-4
S={+2i, -2i, +1, -1}