If (x-2) and (x+2) are factors of 6x^3+ax^2+bx+16, determine a and b, and any remaining factors.?

Let f(x) = 6x³ + ax² + bx + 16

(x - 2) is a factor of f(x). Then, f(2) = 0
6(2)³ + a(2)² + b(2) + 16 = 0
2a + b = -32 …… [1]

(x + 2) is a factor of f(x). Then, f(-2) = 0
6(-2)³ + a(-2)² + b(-2) + 16 = 0
2a - b = 16 …… [2]

[1] + [2]:
4a = -16
a = -4

[1] - [2]:
2b = -48
b = -24

f(x) = 6x³ - 4x² - 24x + 16
= 6x³ + 12x² - 16x² - 32x + 8x + 16
= (6x³ + 12x²) - (16x² + 32x) + (8x + 16)
= 6x²(x + 2) - 16x(x + 2) + 8(x + 2)
= (x + 2)(6x² - 16x + 8)
= 2(x + 2)(3x² - 8x + 4)
= 2(x + 2)(x - 2)(3x - 2)

The answers:
a = -4 and b = -24
Remaining factors: 2 and (3x - 2)

(x - 2) * (x + 2) = x^2 - 4

(x^2 - 4) * (mx + n) = 6x^3 + ax^2 + bx + 16
mx^3 + nx^2 - 4mx - 4n = 6x^3 + ax^2 + bx + 16

mx^3 = 6x^3
m = 6

nx^2 = ax^2
n = a

-4mx = bx
-4m = b
-4 * 6 = b
-24 = b

-4n = 16
n = -4
a = -4

a = -4 , b = -24

Put f(x) = 6x^3+ax^2+bx+16;
Given : (x-2) & (x+2) are both factors of f(x);
Therefore (x-2)*(x+2) = (x^2-4) is a factor of f(x);
Then there exists a linear factor g(x) = (px + q) such that (x^2-4)(px +q) = f(x). Now
(x^2-4)(px+q) = px^3+qx^2-4px-4q = 6x^3+ax^2+bx+16. Equating coefficients of like
powers of x gives p = 6, q = a, -4p = b, -4q = 16. Then q = a = -4 and b = -24. Then
g(x) = (6x-4), f(x) = 6x^3 -4x^2 -24x +16 = (6x-4)(x^2-4) = 2(x+2)(3x-2)(x-2). I have
deliberately lined up the factors containing the zeros of f(x) in ascending order.  which are -2, 2/3, 2. 

6x^3+ax^2+bx+16=0. x=2 is aroot, Put it 6*8+4a+2b+16=0
4a+2b=-48-16=-64.
Now put x=-2
-48+4a-2b+16=0. This gives 4a-2b=32
2a+b=-32, 2a-b=16, Add the two. 4a=-16, a=-4,
For b use the first equation. -8+b=-32. b=-24. Please check the result.