How would you factor the following: 54x^4 + 2x

Since the real numbers have as common factor the numbers 1
and 2, we'll factorize the expression by 2, for the
beggining:

54x^4 + 2x = 1*2*(27x^4 +
x)

The terms inside the brackets have as common factor only
the variable x.

We'll factorize again by
x:

1*2*(27x^4 + x) = 2*x(x^3 +
1)

We can re-write the sum of cubes, applying the
formula:

a^3 + b^3 = (a+b)(a^2 - ab +
b^2)

x^3 + 1 = (x+1)(x^2 - x +
1)

2*x(x^3 + 1) = 2x(x+1)(x^2 - x +
1)

We'll determine the roots of the quadratic x^2 - x +
1;

 x^2 - x + 1 = 0

x1 =
[1+sqrt(1 - 4)]/2

x1 =
(1+isqrt3)/2

x2 =
(1-isqrt3)/2

x^2 - x + 1 = [x - (1+isqrt3)/2][x +
(1-isqrt3)/2]

The completely factorized expression will
become:

54x^4 + 2x = 2x(x+1)[x -
(1+isqrt3)/2][x + (1-isqrt3)/2]