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

1 Answer
Jan 16, 2017

Answer:

#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))#