(x + 1)(3x + 5)
In general, no.
two negative
x2+8x = -3x2+5 4x2+8x-5 = 0 (2x+5)(2x-1) = 0 x = -5/2 or x = 1/2
3x2 + 8x - 8 doesn't factor neatly. Applying the quadratic formula, we find two real solutions: (-4 plus or minus 2 times the square root of 10) divided by three.x = 0.7748517734455863x = -3.441518440112253
5 - 8x + 3x2 = 3x2 - 8x + 5 (since the sign of the second term is negative, then 5 can be factored as (-1)(-5)) = (3x - )(x - ) (try to put 1 or 5 to the empty places, in order to obtain -8x) = (3x - 5)(x - 1)
2x^2 + 8x + 3 = 0
The sum 6 (3x^2) + 8x -3 plus 3 (3x^2) - 7x + 2 = 27 x^2 + x -1
If you notice, 8x + 4 = 4(2x + 1), but there is an odd coefficient for x2, so there is guaranteed to be some remainder (that is 8x + 4 is not a factor of the polynomial): (24x3 - 5x2 - 48x - 8) ÷ (8x + 4) = 3x2 - 2x - 5 - (x2 - 12)/(8x + 4)
= (3x + 5)(x + 1) so x = -5/3 or -1
That does not factor neatly.
If you mean: 3x2+8+5 = 0 Then it crosses the x axis at points -1 and -5/3