Factor
x^4+64
it's a sum of cubes, so
(x+4)(x^2-4x+16)
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))
To factor the expression x^4 + 64, we can use a special factoring formula called the difference of squares. The difference of squares formula states that a^2 - b^2 can be factored as (a + b)(a - b).
In order to apply this formula to our expression, we first need to rewrite 64 as a perfect square. 64 can be expressed as 8^2.
Therefore, we can rewrite the original expression as x^4 + 8^2.
Now, we notice that x^4 can be expressed as (x^2)^2.
Applying the difference of squares formula, we have:
x^4 + 8^2 = (x^2)^2 + 8^2
Now, we can see that we have a difference of squares pattern:
a^2 - b^2 = (a + b)(a - b)
In our case, a = (x^2) and b = 8. Applying the formula, we get:
(x^2)^2 + 8^2 = (x^2 + 8)(x^2 - 8)
Therefore, the factored form of x^4 + 64 is (x^2 + 8)(x^2 - 8).