How do you factor 2x^4+2x^3-2x^2-x2x4+2x3−2x2−x?
1 Answer
where:
x_n = 1/3(4 cos(1/3 cos^(-1)(5/32) + (2npi)/3) - 1)" "xn=13(4cos(13cos−1(532)+2nπ3)−1)
Explanation:
All of the terms are divisible by
2x^4+2x^3-2x^2-x = x(2x^3+2x^2-2x-1)2x4+2x3−2x2−x=x(2x3+2x2−2x−1)
Let:
f(x) = 2x^3+2x^2-2x-1f(x)=2x3+2x2−2x−1
This cubic is not easy to factor, so I have posted it as a Precalculus question, with solution https://socratic.org/s/az7KUZgd
x_n = 1/3(4 cos(1/3 cos^(-1)(5/32) + (2npi)/3) - 1)" "xn=13(4cos(13cos−1(532)+2nπ3)−1) forn = 0, 1, 2n=0,1,2
and hence factorises as:
f(x) = 2 (x-x_0)(x-x_1)(x-x_2)f(x)=2(x−x0)(x−x1)(x−x2)
So putting it all together, we find:
2x^4+2x^3-2x^2-x = 2x(x-x_0)(x-x_1)(x-x_2)2x4+2x3−2x2−x=2x(x−x0)(x−x1)(x−x2)
where:
x_n = 1/3(4 cos(1/3 cos^(-1)(5/32) + (2npi)/3) - 1)" "xn=13(4cos(13cos−1(532)+2nπ3)−1)
