SOLUTION: factoring polynomials [Sum of cubes]: x^3 + y^3 = (x + y)(x^2 &#8722; xy + y^2) [Difference of cubes]: x^3 - y^3 = (x-y)(x^2 + xy + y^2) problem: 250x^4-54x

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Question 552011: factoring polynomials
[Sum of cubes]: x^3 + y^3 = (x + y)(x^2 − xy + y^2)
[Difference of cubes]: x^3 - y^3 = (x-y)(x^2 + xy + y^2)
problem:
250x^4-54x
Found 2 solutions by Earlsdon, solver91311:
Answer by Earlsdon(6294) (Show Source):
You can put this solution on YOUR website!
Factor:
Factor 2x.
Now factor the difference of cubes.

Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

Factor out leaving you with:

And since and , just follow the pattern for the difference of two cubes.

BTW, the cubes factorizations are much easier to remember as one statement rather than two as you gave them:

And the key to remembering the way the signs go is to remember San Diego Padres -- SDP -- Same, Different, Positive.

John

My calculator said it, I believe it, that settles it

SOLUTION: factoring polynomials [Sum of cubes]: x^3 + y^3 = (x + y)(x^2 &#8722; xy + y^2) [Difference of cubes]: x^3 - y^3 = (x-y)(x^2 + xy + y^2) problem: 250x^4-54x

SOLUTION: factoring polynomials [Sum of cubes]: x^3 + y^3 = (x + y)(x^2 − xy + y^2) [Difference of cubes]: x^3 - y^3 = (x-y)(x^2 + xy + y^2) problem: 250x^4-54x