Answer:
b) x + 1
Explanation:
you can either
1) take (6x³ + 6) and divide by all the choices to see which one gives you a factor. You will realize that if you divide this by option b, you will be able to factorize the equation as follows:
(6x³ + 6) = 6(x+1)(x²−x+1)
Hence option b is a factor
or
2) (my preferred method), utilize the properties of functions and roots.
Let function f(x) = 6x³ + 6
any value of a which gives f(a) = 0 is a root , i.e (x-a) is a factor.
In this case, lets consider option b
let x + 1 = 0 -------> or x = -1
substitute this into the function f(x)
f(-1) = 6 (-1)³ + 6
f(-1) = -6 + 6 = 0
hence x = -1 is a root , or (x+1) is a factor.
as a sanity check, lets try choice a) x -1
let x - 1 = 0 -------> or x = +1
substitute this into the function f(x)
f(1) = 6 (1)³ + 6
f(1) = 6 + 6 = 12 ≠0
hence x = 1 is NOT a root , or (x-1) is NOT a factor.
You can do the same for c and d and find that they too are NOT factors.