Question

How do you factor `3x^8+45x^5+129x^2`?

Answer 1

`3x^8+45x^5+129x^2`

`= 3x^2(x+root(3)(15/2-sqrt(53)/2))(x^2-(root(3)(15/2-sqrt(53)/2))x+root(3)(139/2-(15sqrt(53))/2))(x+root(3)(15/2+sqrt(53)/2))(x^2-(root(3)(15/2+sqrt(53)/2))x+root(3)(139/2+(15sqrt(53))/2))`

Explanation:

First note that all of the terms are divisible by `3x^2`, so we can separate that out as a factor:

`3x^8+45x^5+129x^2 = 3x^2(x^6+15x^3+43)`

We can treat the remaining sextic as a quadratic in `x^3` to factor it:

`x^6+15x^3+43 = (x^3+15/2)^2-(15/2)^2+43`

`color(white)(x^6+15x^3+43) = (x^3+15/2)^2-(225-172)/4`

`color(white)(x^6+15x^3+43) = (x^3+15/2)^2-53/4`

`color(white)(x^6+15x^3+43) = (x^3+15/2)^2-(sqrt(53)/2)^2`

`color(white)(x^6+15x^3+43) = (x^3+15/2-sqrt(53/2))(x^3+15/2+sqrt(53)/2)`

The sum of cubes identity can be written:

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

Note that:

`(15/2-sqrt(53)/2)^2 = 225/4-(15sqrt(53))/2+53/4 = 139/2-(15sqrt(53))/2`

`(15/2+sqrt(53)/2)^2 = 225/4+(15sqrt(53))/2+53/4 = 139/2+(15sqrt(53))/2`

Hence we find:

`x^3+15/2-sqrt(53)/2 = x^3+(root(3)(15/2-sqrt(53)/2))^3`

`color(white)(x^3+15/2-sqrt(53)/2) = (x+root(3)(15/2-sqrt(53)/2))(x^2-(root(3)(15/2-sqrt(53)/2))x+root(3)(139/2-(15sqrt(53))/2))`

`x^3+15/2+sqrt(53)/2 = x^3+(root(3)(15/2+sqrt(53)/2))^3`

`color(white)(x^3+15/2+sqrt(53)/2) = (x+root(3)(15/2+sqrt(53)/2))(x^2-(root(3)(15/2+sqrt(53)/2))x+root(3)(139/2+(15sqrt(53))/2))`

So putting it all together, we have:

`3x^8+45x^5+129x^2`

`= 3x^2(x+root(3)(15/2-sqrt(53)/2))(x^2-(root(3)(15/2-sqrt(53)/2))x+root(3)(139/2-(15sqrt(53))/2))(x+root(3)(15/2+sqrt(53)/2))(x^2-(root(3)(15/2+sqrt(53)/2))x+root(3)(139/2+(15sqrt(53))/2))`