So here is the question. We're reviewing factoring right now.
"If you had the choice of factoring a polynomial as the difference of two squares or as the difference of two cubes, which do you do first? Why?"
I'm not sure I fully understand the question. Is there even a polynomial that can be factored as both a difference of two squares and differences of two cubes? Can you provide me with an example please? Or is the question simply open-ended and an opinion?
Thanks so much :)
"If you had the choice of factoring a polynomial as the difference of two squares or as the difference of two cubes, which do you do first? Why?"
I'm not sure I fully understand the question. Is there even a polynomial that can be factored as both a difference of two squares and differences of two cubes? Can you provide me with an example please? Or is the question simply open-ended and an opinion?
Thanks so much :)
-
x^6 - 1 is one example.
You can re-write it as
(1): (x^2)^3 -1
or
(2): (x3)^2 -1
(1): (x^2)^3 -1 =(x^2-1)(x^4+x^2+1) = (x-1)(x+1)(x^4+x^2+1) =
(x-1)(x+1)(x^4 +2x^2 +1 - x^2) = (x-1)(x+1)[(x^2+1)^2-x^2]
=(x-1)(x+1)(x^2+x+1)(x^2-x+1)
(2): (x3)^2 -1 =(x^3-1)(x^3+1)= (x-1)(x^2+x+1)(x+1)(x^2-x+1)
I guess the second one is easier to do.
You can re-write it as
(1): (x^2)^3 -1
or
(2): (x3)^2 -1
(1): (x^2)^3 -1 =(x^2-1)(x^4+x^2+1) = (x-1)(x+1)(x^4+x^2+1) =
(x-1)(x+1)(x^4 +2x^2 +1 - x^2) = (x-1)(x+1)[(x^2+1)^2-x^2]
=(x-1)(x+1)(x^2+x+1)(x^2-x+1)
(2): (x3)^2 -1 =(x^3-1)(x^3+1)= (x-1)(x^2+x+1)(x+1)(x^2-x+1)
I guess the second one is easier to do.
-
Sorry, I meant "I love how you provided an example."
Report Abuse
-
factorize using difference of 2 squares, its easier to manage than difference of 2 cubes