Which of the following is a factor of 6x^3 + 6?
a) x - 1
b) x + 1
c) x² - 2x + 1
d) x² + x + 1

B

Step-by-step explanation:

Given

6x³ + 6 ← factor out 6 from each term

= 6(x³ + 1)

x³ + 1 is a sum of cubes and factors as

x³ + 1 = (x + 1)(x² - x + 1)

Hence

6x³ + 6 = 6(x + 1)(x² - x + 1)

With factor (x + 1) → B

b) x + 1

Step-by-step 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.

thats not a system of equations, that is a solution

where is abdul's work? don't see that anywhere posted : /

step-by-step explanation: