maths
posted by Ann .
x+2a is a factor of (x^3+2(a2b)x^2b(8a3b)x+6ab^2)

I assume you want to find a and b so it divides evenly?
x^3 + 2(a2b)x^2  b(8a3b)x + 6ab^2

x+2a
= x^2  4bx + 3b^2 remainder 6ab(1b)
so, to divide evenly, either a=0 or b=1
check:
a=0: x^3  4bx^2 + 3b^2x + 0 = x(x^24bx+3b^2) for any b
b=1: x^3 + 2(a2)x^2  (8a+3)x + 6a
= (x+2a)(x^2  4x  3) for any a