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4(x^2 + x + 3)
(2x+3)(2x+4)
No
4x(4x^2 + 3x + 1)
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 - 3)(2x - 3)