How to find the real-number solutions of these polynomials

i'm trying to factor these polynomials to get the x intercepts, but i suck at factoring, so could someone please factor them for me and get me the solutions while showing their work? thanks! 10 points for the best answer!

3x^4 + 15x^2 - 72 = 0

x^3 + 2x^2 - x = 2

x^4 + 7x^3 - 8x -56 = 0

3(x^4 + 5x^2 - 24) = 3(x^2 + 8)(x^2 - 3) = 0
x^2 + 8 = 0 or x^2 - 3 = 0
no real solutions.........x^2 = 3
.....................................x = ± √3

x^3 + 2x^2 - x = 2
x^3 + 2x^2 - x - 2 = 0
x^2(x + 2) - 1(x + 2) = 0
(x + 2)(x^2 - 1) = 0
x + 2 = 0 or x^2 - 1 = 0
x = -2.............x^2 = 1
........................x = ± 1

x^4 + 7x^3 - 8x - 56 = 0
x^3(x + 7) - 8(x + 7) = 0
(x + 7)(x^3 - 8) = 0
x + 7 = 0 or x^3 - 8 = 0
x = -7.............x^3 = 8
.......................x = 2
....................(other solutions are complex)

3x^4+15x^2-72=0
3(x^4+5x^2-24)=0
3(x^2+8)(x^2-3)=0
x^2+8=0 or x^2-3=0
x^2=0-8 or x^2=0+3
x^2=-8 or x^2=3
√x^2=√-8 or √x^2=√3
x=2i√2 or x=√3
Roots: √3 and 2i√2

x^3+2x^2-x=2
x^3+2x^2-x-2=0
(x^2-1)(x+2)=0
(x+1)(x-1)(x+2)=0
x+1=0 or x-1=0 or x+2=0
x=0-1 or x=0+1 or x=0-2
x=-1 or x=1 or x=-2
Roots: -1, 1, and -2

x^4+7x^3-8x-56=0
(x^3-8)(x+7)=0
x^3-8=0 or x+7=0
x^3=0+8 or x=0-7
x^3=8 or x=-7
x=2 or x=-2 or x=-7
Roots: 2, -2, and -7

Maybe you should ask your teacher to tutor before or after school. You don't want answers. At test time, that won't help!