Q/A...

1) X² + 5X + 5 = 0
a=1
b=5
c=5

-b ± ( √ b²-4ac )
¯¯¯¯¯2a¯¯¯¯
x = -5 ± ( √ 5²-4(1)(5) )
¯¯¯¯¯2(1)¯¯¯¯¯¯¯¯

x = -5 ± ( √ 25-20 )
¯¯¯¯¯2¯¯¯¯¯¯¯

(x = -5 ± ( √ 5 )
¯¯¯¯¯2¯¯¯¯

2) X³ - 6X² + 11x - 6 = 0

x³ - 6x² + 11x - 6 = 0
Based on the factors of 6, factors will be ±1, ±2, and/or ±3:
Assume (x - 1):
(x³ - 6x² + 11x - 6)/(x - 1) = x² - 5x + 6 (bingo!)
So, (x - 1) is the first factor. Now factor (x² - 5x + 6):
x² - 5x + 6 = (x - 2)(x - 3)
Thus, the three factors are (x - 1), (x - 2), and (x - 3).


3) X^4 + 16 = 0
* maybe i copy wrong question..
x^4-16
(x^2-4)(x^2+4)
(x-2)(x+2)(x^2+4)