BRAINLIEST AWARD WILL BE GIVEN! PLEASE HELP WITH MATH IVE BEEN ASKING FOR HOURS! PLEASE ANYONE!!!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) = 02) Solve for the roots of x in each of the equations below.x^4 - 81 = 0x^4 + 10x^2 + 25 = 0x^4 - x^2 - 6 = 03)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}