x2+8x = -3x2+5 4x2+8x-5 = 0 (2x+5)(2x-1) = 0 x = -5/2 or x = 1/2
(x + 1)(3x + 5)
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)
In algebra and mathematics the simple above equation can be written and simplified as follows . (5x2 8x)-(3x2-x) = 10 + 8x -6 +x = 4 +7x.
If you mean: 3x2+8+5 = 0 Then it crosses the x axis at points -1 and -5/3
-8x + 3x2 - 3
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
(3x + 2)(x + 2) so x = -2/3 or -2
(x + 1)(3x + 5)
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)
In algebra and mathematics the simple above equation can be written and simplified as follows . (5x2 8x)-(3x2-x) = 10 + 8x -6 +x = 4 +7x.
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
If you mean: 3x2+8+5 = 0 Then it crosses the x axis at points -1 and -5/3