x3 - x2 - 2x +2 you can divide this into two parts for now x3 - x2 and -2x +2 factor out everything you can from both x2(x-1) and -2(x+1) since the insides of the parentheses are the same, we can put them together, then multiply it by the sum of the two parts outside the parentheses. (x-1)(x2-2)
3x2 + 2x + 3 + x2 + x + 1 = 4x 2+ 3x + 4
14x2 + 15x + 4 = (7x + 4)(2x +1)
4X + 2x = 1. Where x = 0.166666666666666666666666666666666666666 recurring.
2x(1-5)
It is a quadratic expression and when factored is: (x+1)(x+1)
x2 + 1 can't be factored in the real numbers. If complex numbers are acceptable, it can be factored as (x + i)(x - i).
4x2 + 8x + 4 = 4 (x2 + 2x + 1) = 4 (x + 1)2
2x2+5x+3 = (2x+3)(x+1) when factored
x3 - x2 - 2x +2 you can divide this into two parts for now x3 - x2 and -2x +2 factor out everything you can from both x2(x-1) and -2(x+1) since the insides of the parentheses are the same, we can put them together, then multiply it by the sum of the two parts outside the parentheses. (x-1)(x2-2)
If you mean (1+2x)2, that is already factored.
x2 + 2x -6 = 0 x2 + 2x + 1 = 7 (x + 1)2 = 7 x = -1 ± √7
If x2 + 5 = -2x, then x2 + 2x + 5 = 0 This equation cannot be readily factored so using the quadratic formula :- x= {-2 ± √[(-2)2 - 4x5]} ÷ 2 = {-2 ±√-16} ÷ 2 = -1 ± 2i Therefore x = -1 + 2i or x = -1 - 2i
(2x+5)(2x+1)
(2x+1)(2x+3) when factored
That expression can not be factored.
Not neatly. Applying the quadratic formula, we find two real solutions: -1 plus or minus the square root of three.x = 0.7320508075688772x = -2.7320508075688772