If we assume that it equals zero and you wish to solve for x, then the answer is:
-x2 + 2x - 3 = 0
x2 - 2x + 3 = 0
x2 -2x + 1 = -2
(x - 1)2 = -2
x - 1 = ± √(-2)
x = 1 ± i√2
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(x4 - 2x3 + 2x2 + x + 4) / (x2 + x + 1)You can work this out using long division:x2 - 3x + 4___________________________x2 + x + 1 ) x4 - 2x3 + 2x2 + x + 4x4 + x3 + x2-3x3 + x2 + x-3x3 - 3x2 - 3x4x2 + 4x + 44x2 + 4x + 40R∴ x4 - 2x3 + 2x2 + x + 4 = (x2 + x + 1)(x2 - 3x + 4)
2x3 + 16 = 2(x3 + 8) = 2(x3 + 23) = 2(x + 2)(x2 - 2x + 22) = 2(x + 2)(x2 - 2x + 4)
Do you want its factorisation? 2x3 + 2x2 - 12x = 2x(x2 + x - 6) = 2x(x + 3)(x - 2).
-2x3 + 2x2 + 12x = -2x(x2 - x - 6) = -2x(x2 + 2x - 3x - 6) = -2x[ x(x + 2) - 3(x + 2) ] = -2x(x - 3)(x + 2)
(-2x3 - 2x2 + 12x) = -2x (x2 + x - 6) = -2x (x + 3) (x - 2)