it's a sum of cubes, so
(x+4)(x^2-4x+16)
Factor
x^4+64
3 answers
Could you elaborate on how you got that answer
Good question! I misread it as x^3+64. How about this?
There is no factoring over real numbers. If you must factor it completely, then we have a difference of squares, and we get
(x^2+8i)(x^2-8i)
Now, we have to factor those.
x^2-8i = (2+2i)^2
and so on,
so we finally get values for x of ±(2±2i)
(x-(2+2i))(x+(2+2i))(x-(2-2i))(x+(2-2i))
There is no factoring over real numbers. If you must factor it completely, then we have a difference of squares, and we get
(x^2+8i)(x^2-8i)
Now, we have to factor those.
x^2-8i = (2+2i)^2
and so on,
so we finally get values for x of ±(2±2i)
(x-(2+2i))(x+(2+2i))(x-(2-2i))(x+(2-2i))
0 answers