Those aren't functions. If you mean f(x) = x + 1 and f(x) = 3x - 2, the domain is the the set of all real numbers, R. I, of course, am assuming you haven't been introduced to the complex number plane, so I'm ignoring those.
if you rearrange it, it becomes 4y + 8 = 3x2 -2x +1 4y = 3x2 -2x -7 y= (3x2 -2x -7)/4 which is a parabola
3x2 + 10x + 3 = (x + 3)(3x + 1).
19
3x2 + 2x + 3 + x2 + x + 1 = 4x 2+ 3x + 4
x3 + 3x2 - 9x + 5 = 0 has roots of -5,1 and 1. CHECK : x3 + 3x2 - 9x + 5 = (x + 5)(x - 1)(x - 1)
3x2 + 22x + 7 = 3x2 +21x + x + 7 = 3x(x + 7) + (x + 7) = (x + 7)(3x + 1)
3x2 + 12x = - 1 ie 3x2 + 12x + 1 = 0 has no rational roots. The irrational roots are [-12 +/- sqrt(132)]/6 = -3.915 and -0.085
3x2 + 5x + 2 is a quadratic expression that can be factored as follows: 3x2 + 5x + 2 = 3x2 + 3x + 2x + 2 = 3x(x + 1) + 2(x + 1) = (3x + 2)(x + 1)
x2 + x2 + x2 = (1 + 1 + 1)x2 = 3x2
== == Suppose f(x) = x3 + 3x2 - 2x + 7 divisor is x + 1 = x - (-1); so rem = f(-1) = 11
(x + 1)(3x + 5)
(3x2 + 2x + 1)