Assuming the missing signs are pluses, that factors to (2x + 5)(2x + 5)
362+60x+25 (6x+5)(6x+5)
(x-5i)(x+5i)
(6x + 5)(6x + 5) or (6x + 5)2
x2 + 25 doesn't factor neatly. Applying the quadratic formula, we find two imaginary solutions: Zero plus or minus 5i times the square root of 1.x = 5ix = -5iwhere i is the square root of negative one.
(2x - 5)(2x - 5)
(2x - 5)(2x - 5) or (2x - 5) squared
c = 25 4x2 + 20x + 25 (2x + 5)(2x + 5)
(2x+5)(2x+5)
The discriminant is 0.
4x2-20+25 = (2x-5)(2x-5) when factored with the help of the quadratic equation formula.
16x2 + 40x + 25 = 16x2 + 20x + 20x + 25 = 4x(4x + 5) + 5(4x + 5) = (4x + 5)(4x + 5) = (4x + 5)2
(2x + 5)(4x2 - 10x + 25)
Assuming the missing signs are pluses, that factors to (2x + 5)(2x + 5)
(2x + 5)(6x - 5)
12x2 + 20x - 25 IS a polynomial that factors into (2x + 5)(6x - 5)
(2x + 5)(2x + 5)