How do I factorise this cubic expression?

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Homework Help Overview

The discussion revolves around the factorization of a cubic expression, specifically the polynomial x^3 + 3x^2y + 2xy^2 + 6y^3. The original poster expresses familiarity with quadratics but indicates difficulty with cubic expressions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants suggest various strategies for factorization, including testing for rational roots and grouping terms. There is a mention of attempting to factor by grouping and factoring out common terms.

Discussion Status

The discussion is active, with participants providing different approaches and suggestions for tackling the factorization problem. There is no explicit consensus on a single method, but several lines of reasoning are being explored.

Contextual Notes

The original poster indicates a lack of experience with cubic expressions, which may influence the approaches discussed. There is also an implication that the problem may involve rational roots, though this is not confirmed.

physicsfun_12
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Homework Statement


Hello there, hope you'r well.

I am having trouble factorising this cubic. I am comfortable with quadratics however I have never had to factorise anything to the power 3 before.

x^3+3x^2y+2xy^2+6y^3

Thanks in advance for any input.

Mike


Homework Equations


n/a


The Attempt at a Solution


I have had ago but didn't get very far!

(3x+2y)(xy) that does the bit in the middle, excluding the two terms to the power 3.

Thanks again for any help with this.

Mike
 
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physicsfun_12 said:
x^3+3x^2y+2xy^2+6y^3

Hi Mike! :smile:

(try using the X2 tag just above the Reply box :wink:)

Assuming it has rational roots, they must be x/y = ± 1 or 2 or 3 or 6, so just try all of them. :biggrin:

(to save time, do the "odd" terms together, and the "even" terms together :wink:)
 
Try grouping the first two and last two terms together and factor each group. Then...
 
Factor x2 from the first two terms and 2y2 from the last two terms.
 

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