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Easy factoring problem

  1. Jan 25, 2007 #1
    how to factor (x+3)^3

    maybe foil after factoring?

    thanks for any help.
  2. jcsd
  3. Jan 25, 2007 #2


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    Huh? It's already factored.
  4. Jan 25, 2007 #3
    hmm, maybe i used the wrong terminolgy, i want to take it apart, is that possible?

    is (x+3)^3 the same as (x+3)(x+3)^2
  5. Jan 25, 2007 #4


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    Ah, you want to expand it. (There are other synonyms too)

    Yes, those two are the same; that's essentially the definition of raising something to the third power.

    Factoring is the process where, for example, you start with

    x^2 + 4x + 3

    and end up with

    Last edited: Jan 25, 2007
  6. Jan 26, 2007 #5
    okay so to fully expand (x+3)^3

    could i just take the foil of (x+3)^2 and then multiply by (x+3) again?

    if so, i'm unsure how to multiply the x+3 and the foiled polynomial
  7. Jan 26, 2007 #6
    (a+b)^2 = a^2 + 2ab + b^2
    (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

    There's a really easy trick to finding these. Expanding expressions of the form (x+y)^n can be done as follows:

    First write x^n. The coefficient of the next term is equal to the current coefficient multiplied by the current exponent of x, and divided by the term number. For the coefficient of the second term, this is 1 * n / 1 = n. Decrease the exponent of x by 1, and increase the exponent of y by 1. Repeat this process until you get to y^n.

    If they are of the form (x-y)^n, the signs just alternate. If you're unsure, use the property that a - b = a + (-b).

    Hope that helps!
  8. Jan 31, 2007 #7


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    "Foil" is just a way of remembering that you must multiply each term in one factor by each term in the other.

    Using "foil" on (x+3)(x+3) give x2+ 3x+ 3x+ 9= x2+ 6x+ 9. Now to multiply that by x+3 again, first multiply each part by x: x(x2+ 6x+ 9)= x3+ 6x2+ 9x, then multiply each part by 3: 3(x2+ 6x+ 9)= 3x2+ 18x+ 27, and finally add: x3+ 6x2+ 9x+ 3x2+ 18x+ 27= x3+ 9x2+ 27x+ 27.

    This is not "abstract algebra"- I'm moving it to "General Math".
    Last edited by a moderator: Jan 31, 2007
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