4(x^2 + x + 3)
(2x+3)(2x+4)
No
Re=Write as 16x^(3) + 12x^(2) + 4x Take out the common factor of '4' Hence 4(4x^(3) + 3x^)2) + x) Take out the common factor of 'x' Hence 4x( 4x^(2) + 3x + 1) The answer!!!!!
2(x^2 + 2)(x + 3)
4(x+y)^2
No solution in integers.
(2x-1)(2x-1) = 4x^2 -4x + 1
2(x - 1)(2x + y)
(4x + 1)(2x + 3)
There is no rational factorisation.
I assume you wanted this factored: (x+6)(x-2)
(2x - 1)(6x + 5)