SOLUTION: Factor this problem x^3 y+2x^2 y^2+xy^3

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Question 203514: Factor this problem
x^3 y+2x^2 y^2+xy^3
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
When factoring, always start by factoring out the Greatest Common Factor (GCF) (unless the GCF is a 1). Then try factoring using any other techniques you may know:Factoring by patterns
  • Difference of squares: a%5E2+-+b%5E2+=+%28a+%2B+b%29%28a-b%29
  • Difference of Cubes: a%5E3+-+b%5E3+=+%28a-b%29%28a%5E2+%2B+ab+%2B+b%5E2%29
  • Sum of Cubes: a%5E3+%2B+b%5E3+=+%28a%2Bb%29%28a%5E2+-+ab+%2B+b%5E2%29
  • Perfect square trinomials:
    • a%5E2+%2B+2ab+%2B+b%5E2+=+%28a%2Bb%29%5E2+=+%28a%2Bb%29%28a%2Bb%29
    • a%5E2+-+2ab+%2B+b%5E2+=+%28a-b%29%5E2+=+%28a-b%29%28a-b%29
  • Factoring trinomials
  • Factoring by grouping
  • Factoring by trial and error of the possible rational roots

    • And you keep factoring, using and reusing any of the above until you cannot factor any further.

    Now let's try to use this on your expression.
    x%5E3y%2B2x%5E2y%5E2%2Bxy%5E3
    Start with the GCF. We are looking for the greatest/longest set of factors whic are factors of every term. The terms are: x%5E3y, 2x%5E2y%5E2 and xy%5E3. Until you develop an eye for find the GCF it may help to find the GCF in the following way. I'll factor each term fully, numbers into prime factors and variables with exponents into the product of the appropriate number of that variable. Note how spacing is used so that common factors can be lined up:
    x^3*y    =   x*x*x*y
2x^2*y^2 = 2*x*x*  y*y
x*y^3    =   x*    y*y*y

    Looking at the above it should be clear that the only factors that are in all three terms are an x and a y. So the GCF is the their product: x*y. When you factor out the GCF you write the GCF and then, in parentheses, you write each term of the polynomial with the GCF removed:


    Now we try other factoring techniques. This is pretty easy at this point because it should be pretty clear how (x^2 + 2xy + y^2) matches the first perfect square trinomial pattern. So we can factor it using the pattern:


    This will not factor any further so we are done.