How do you factor # 2x^4+2x^3-2x^2-x#?

1 Answer
Oct 27, 2016

Answer:

#2x^4+2x^3-2x^2-x = x(2x^3+2x^2-2x-1)#

#color(white)(2x^4+2x^3-2x^2-x) = 2x(x-x_0)(x-x_1)(x-x_2)#

where:

#x_n = 1/3(4 cos(1/3 cos^(-1)(5/32) + (2npi)/3) - 1)" "#

Explanation:

All of the terms are divisible by #x#, so we can separate that out as a factor first:

#2x^4+2x^3-2x^2-x = x(2x^3+2x^2-2x-1)#

Let:

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

This cubic is not easy to factor, so I have posted it as a Precalculus question, with solution https://socratic.org/s/az7KUZgd

#f(x)# has zeros:

#x_n = 1/3(4 cos(1/3 cos^(-1)(5/32) + (2npi)/3) - 1)" "# for #n = 0, 1, 2#

and hence factorises as:

#f(x) = 2 (x-x_0)(x-x_1)(x-x_2)#

So putting it all together, we find:

#2x^4+2x^3-2x^2-x = 2x(x-x_0)(x-x_1)(x-x_2)#

where:

#x_n = 1/3(4 cos(1/3 cos^(-1)(5/32) + (2npi)/3) - 1)" "#