Question

why is it that you can factor a sum of two cubes, but not a sum of two square?

Answer 1

@zepdrix

Answer 2

Here is an explanation to consider.\[\Large\bf\sf x^2+4=0\]Subtracting 4 from each side,\[\Large\bf\sf x^2=-4\]Taking the square root of each side,\[\Large\bf\sf x=\pm \sqrt{-4}\]Uh oh, we can't take the square root of a negative number! :(
So no `real` solutions exist.

Answer 3

If you include complex/imaginary solutions though,
Taking the root gives us,\[\Large\bf\sf x=\pm 2i\]

That's what happens when we factor the `sum of squares`.
We get complex factors, not real ones.\[\Large\bf\sf x^2+4=(x+2i)(x-2i)\]

Answer 4

If you haven't been introduced to imaginary numbers yet, then I apologize.
This might seem a little confusing in that case hehe.

Answer 5

lol thanks for your help this is one of the question on my quiz i have to write why ?

Answer 6

hmm

Answer 7

I'll give you a little more info to think about.
Word your answer the way that makes sense to you. It's your test! XD\[\Large\bf\sf x^3+8=0\]Again, solving for x,\[\Large\bf\sf x^3=-8\]We `CAN` take the cube root of a negative number.
Doing so gives us,\[\Large\bf\sf x=-2\]

Answer 8

http://en.wikipedia.org/wiki/Intermediate_value_theorem