I have two math problems... factor completely?

Question

3x(to the fifth power) - 3x


And

3x(to the third) y squared - 6x(to th fourth) y -12x squared y

Answer 1

3x^5 - 3x =
3x(x^4 -1) =
3x(x^2 -1)(x^2 +1) =
3x(x -1)(x +1)(x^2 +1)

3x^3y^2 -6x^4y - 12x^2y =
3x^2y(xy - 2x^2 - 4)

Answer 2

3x^5-3x
=3x(x^4-1)
=3x(x^2+1)(x^2-1)
=3x(x^2+1)(x+1)(x-1)
For usual smaller grade classes this would be the end. But if you are into advanced classes and if you are aware of the complex numbers then there would be one more step it can factorized to

=3x(x+i1)(x-i1)(x+1)(x-1)

This is the final answer

3x^3y^2-6x^4y-12x^2y
=3x^2y(xy-2x^2-4)

thats all.. I cannot reduce the second one though

Answer 3

I won't answer it, but this is how you should of phrased it:

3X^5 - 3X


3X^3*Y^2 - 6X^4*Y-12X^2*Y

Answer 4

3x(to the fifth power) - 3x
3x is common in this eq
=(3x) x(fourth power) - 1
= (3x) (x-1)whole square (x+1)whole square
= (3x)(x-1) (x+1) (x+1)whole square
or
(3x) (x-1)(x+1) (x+1)(x+1)

Answer 5

3X^5-3X
= 3X (X^4 -1)
= 3X (X^2 + 1)(X^2 - 1)
= 3X (X^2 + 1)(X + 1)(X - 1)

and

3 x^3 y^2 - 6 x^4 y - 12 x^2 y
= 3 x^2 y ( xy - 2x^2 - 4)

Answer 6

3x^5 - 3x =
3x(x^4 - 1) =
3x[(x^2 + 1)(x^2 - 1)] =
3x(x^2 + 1)(x + 1)(x - 1).

3(x^3)(y^2) - 6(x^4)(y) - 12(x^2)(y) =
[3(x^2)(y)] [xy - 2(x^2) - 4].