a(x^2-8x+16)+bx-4b+c
ax^2 - 8ax + 16a + bx - 4b + c
ax^2 - (8a+b)x + 16a - 4b + c
so a = 5
-(8 x 5) = b = -37
so b = 3
15(5) - 4(3) +c = 68
so c = 0
Hence, 5(x-4)^2+ 3(x-4) + 0
(x-4) (5(x-4) + 3)
= (x-4)(5x-17)
Q2. b^2 - 4ac
k^2 + 6k + 9 + 4k >= 0
k^2 + 10k + 9 >= 0
so it is real...