-3
3-2 x 3-3 = 3-5 or 1/243
x2_12x-3 = 0 Using the formula for solving quadratics: x = 12 plus/minus sqrt (144 - (4 x 1 x -3))/2 x = 12 plus/minus (sqrt 156)/2 x = 12 + (sqrt 156)/2 = 12.25 or x = 12 - (sqrt 156)/2 = -0.25
3x2 when x = 2 is the answer 3 X 4 = 12 or is the answer (3 X 2)2 = 62 = 36
x^3 - x^2 + x - 1 = (x^3 - x^2) + (x - 1) = x^2(x - 1) + (x - 1) = (x -1)(x^2 + 1)
so, if 2 minus Ln times 3 minus x equals 0, then 2 minus Ln times 3 equals x, therefore 2 minus Ln equals x divided by three, so Ln + X/3 = 2 therefore, (Ln + [X/3]) = 1
-3
x5
3-2 x 3-3 = 3-5 or 1/243
x over x is one, so the problem would be 1-2-3/2=-5/3
(x - 2)(x^2 + x + 3)
x >2 x2 - x - 6/(x2 - 4) (x + 2)(x - 3)/(x + 2)(x - 2) cross out x + 2 (x - 3)/(x - 2) this is the most simplified this can be
It is either 2|x| - 3 or 2|x-3| depending on if the minus three is part of the absolute value.
3(x - 2)(x + 2)
(x - 2)(2x + 3)
It is: x^2 -9
x³ - x² + x - 3 (x² + 1)(x - 1) - 2