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

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We factorise 54x^4+2x .

54x^4 = 2x*27x^3 = 2x*(3x)63.

Therefore 5x^4+2x = 2x(27x^3+1)

54x^4+2x = 2x{(3x)^3 +1^3}. We use a^3+b^3 = (a+b)(a^2-ab+b^2), where a = 3x and b = 1.

54x^4+2x = 2x{3x+1){(3x)^2-3x*1+1^2}.

54x^4+2x = 2x(3x+1)(9x^2-3x+1).

You have not provided an equation but a mathematical expression as 54x^4+2x.

To factorize this:

54x^4 + 2x

separate the common factors which are 2 and x

=> 2x*27*x^3 + 2x

=> 2x( 27x^3 + 1)

Now we can factor 27x^3 + 1, using the formula a^3 + b^3 = (a+b)*(a^2 - ab+ 1) as (3x + 1)*( 9x^2 - 3x + 1)

**Therefore the factors of ****54x^4+2x are 2x, ( 3x+ 1) and 9x^2 -3x + 1.**

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]**

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