How do you factor: x^10-x^5-2?

Question

So i got this far...
x^10-x^5-2
(x^5-1)(x^2+2) ..... Now when i foil it....
x^10+2x^5-x^2-2.....now i am stuck how do i subtract that 2x^5-x^2.....??
Thank you

Answer 1

You need to treat the x^10 like a square. If you like you can substitute y = x^5 and then you have
y^2 - y - 2 = 0
This factors as
(y-2)(y+1)
or
(x^5-2)(x^5+1)
Looks like you're learning about quadratic forms, that is, expressions that aren't quadratic but they can, with a change in variables, be made to behave like quadratics. Look for patterns like this, ways you can change a variable and get a quadratic expression, in all of your exercises of this type.
Remember, when you multiply, you add exponents, so when you fiol the expression you have above, x^5 times x^2 = x^7, not x^10. So you need to do a substitution like I've shown.
For more info search on quadratic form substitution tutorial
You'll turn up references like this...
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut20_quadform_ans.htm

Answer 2

This is a quadratic, even though it doesn't look like it.
let y = x^5
Then x^10 = x^5*x^5 = y^2

Our equation becomes:
y^2 - y - 2
Which factors into:
(y - 2)(y + 1)

or, in terms of x:
(x^5 - 2)(x^5 +1)

remember: when multiplying terms with exponents, you add the exponents; x^5*x^2 = x^7 not x^10

Answer 3

x^10 -x^5-2
let y = x^5 and sub into orginal eqn
y^2 - y - 2 then factor as
(y - 2) (y + 1) and sub back in x^5
(x^5 - 2) (x^5 + 1)
x = -1 because (-1)^5 = -1, the root of (x^5 +1)
x = 2^(1/5) because (2^(1/5))^5 = 2, the root of (x^5 - 2)

Answer 4

(x^10 - 1) - x^5 - 1 =

(x^5-1)(x^5+1) - (x^5+1)=

(x^5+1)(x^5-1-1)=

(x^5-2)*(x^5+1)...I guess that's all you can do, unless you know the formula for factoring a^n + b^n...