-8x + 3x2 - 3
In general, no.
x2+8x = -3x2+5 4x2+8x-5 = 0 (2x+5)(2x-1) = 0 x = -5/2 or x = 1/2
(x + 1)(3x + 5)
3x^2 + 8x + 4 = (x+2) (3x+2) x = -2 x = -2/3 So there are no complex roots, they are real. You can test this by b^2 - 4ac if greater than 0, it is real if equal, there will be 2 identical roots. if less than 0 you get imaginary roots.
If you mean: 3x2+8+5 = 0 Then it crosses the x axis at points -1 and -5/3
Twice. Between negative two and negative one.
2x^2 + 8x + 3 = 0
The sum 6 (3x^2) + 8x -3 plus 3 (3x^2) - 7x + 2 = 27 x^2 + x -1
= (3x + 5)(x + 1) so x = -5/3 or -1
I do not believe that there are any rational roots.
It is 2x*(4x^2 - 3x - 8). The quadratic cannot be factorised: it has no rational roots.