SOLUTION: How do you factor completely x^8-y^20?
Algebra.Com
Question 357583: How do you factor completely x^8-y^20?
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Think "difference of squares"...
A^2-B^2 = (A-B)(A+B)
And, "sum of squares"...
AČ + BČ = (A + B + √(2AB)) (A + B − √(2AB))
.
x^8-y^20
Can be rewritten as:
(x^4)^2 - (y^10)^2
which can be factored as
(x^4-y^10)(x^4+y^10)
.
rewriting the above:
((x^2)^2-(y^5)^2)((x^2)^2+(y^5)^2)
(x^2-y^5)(x^2+y^5)(x^2+y^5+√(2x^2y^5))(x^2+y^5-√(2x^2y^5))
(x^2-y^5)(x^2+y^5)(x^2+y^5+xy^2√(2y))(x^2+y^5-xy^2√(2y))
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