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Can't figure out a simple factorization

  1. Jan 2, 2008 #1
    I'm doing some basic factoring now to refresh my skills.

    On one problem, it is asking to factor: [tex](y^2 + 8)^2 - 36 y^2[/tex]

    I was able to get to where I factor the trinomials: (y-4)(y-2)(y+4)(y+2)

    When it says to give the summary of factorization, it gives: [tex](y^2 + 8)^2 - (6y)^2[/tex]

    I'm going through a tutorial and this last step is the one I can't figure out.

    For some reason, I thought that (y-4)(y-2)(y+4)(y+2) = [tex](y^2 + 8)^2[/tex]

    but instead: (y-4)(y-2)(y+4)(y+2) = [tex](y^2 + 8)^2 - (6y)^2[/tex]

    My question is how the extra [tex](6y)^2[/tex] comes about.
     
    Last edited: Jan 2, 2008
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  3. Jan 2, 2008 #2

    arildno

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    "(y-4)(y-2)(y+4)(y+2) = (y^2 +8)^2

    Why on earth should this be correct??

    Whereas the left hand side has 4 zeroes, the righthand side has no zeroes.
    Therefore, they cannot be identical.
    Do you understand that?
     
  4. Jan 2, 2008 #3
    What do you mean by 4 zeros? (I've been out of college for 7 years so I've forgotten quite a bit).

    Does it mean that constants (like -4, -2) are zeros?

    I guess I'm having trouble with reverse factoring. I just see this as a pair of squares.
     
  5. Jan 2, 2008 #4

    dst

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    From the start - compare:

    (y^2 +8)^2 - 36y^2 --- (1)

    And the book also gives:

    (y^2 +8)^2 - (6y)^2 --- (2)

    Since you know these have the same factors multiplied out then you know (1) = (2).

    In any case, note that (6y)^2 = (6y)(6y) = 6*6*y*y = 6^2y^2 = 36y^2 as seen at the start.

    The 6y comes straightaway when you multiply the first 2 brackets (y-4)(y-2), and it should eventually become (6y)^2.
     
  6. Jan 2, 2008 #5
    I ocassionally have problems with factorizations. Maybe I could tell you how would I do that one (it is easy), if it is of any help.

    I would try to 'unwrap', or 'flatten', the whole expression, to get it into the form of a polynomial, ax^4 + bx^3 + cx^2 + dx + e; that is, get rid of the parenthesis, so that you can group similar exponents together. (This way at least you would get a different feeling of the expression, what its degree is, or whatever.) In your case, once you group similar exponents the answer should be pretty straightforward.

    Other factorization techniques that could help you are
    • Checking for two numbers that multiplied give C and added give B, when you have an expression like x^2 + Bx + C. If those two numbers are p and q, then you know the factorization is (x + p)(x + q). The numbers p and q can be positive or negative.
    • "Completing the square"; not always leading to a complete factorization, but very handy. (You can google for that term.)
    • Multiplying/dividing the whole expression by some constant, for example if you need to get rid of one of the coefficients; then you may apply other techniques, and then restore back the factor that you removed (if you divided by 3 and the did some trick, you must multiply by 3 at the end to restore the original problem).

    What Arildno was saying above about some zeroes, is that (y-4)(y-2)(y+4)(y+2) becomes 0 when y = 4, 2, -4 or -2; while the right side (y^2 +8)^2 is always positive no matter the value of y; so they cannot be equal.

    Hope this helps.
     
    Last edited: Jan 2, 2008
  7. Jan 2, 2008 #6

    HallsofIvy

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    He means that y= 4, y= -4, y= 2 and y= -2 all make one of the factors of (y-4)(y-2)(y+4)(y+2) equal to 0 and so make the entire polynomial 0. On the other hand y^2+ 8 cannot be smaller than 9 and so (y^2+ 8)^2 is never 0 for any value of y. They can't possibly be the same!

    You said you see [itex](y^2+ 8)^2- 36y^2[/itex] as "pair of squares". I assume you are talking about the "difference of square" formula: [itex]a^2- b^2= (a+ b)(a- b)[/itex]
    Okay, [itex](y^2+ 8)^2- 36y^2[/itex] is of that form with [itex]a= x^2+ 8[/itex] and [itex]b= 36y^2[/itex]. That means [itex]a= y^2+ 8[/itex] and [itex]b= 6y[/itex]
    [itex](y^2+ 8)^2- 36y^2= (y^2+ 8+ 6y)(y^2+ 8- 6y)[/itex]. You can then factor [itex]y^2+ 6y+ 8= (y+4)(y+2)[/itex] and [itex]y^2- 6y+ 8= (y- 4)(y- 2)[/itex].
     
  8. Jan 2, 2008 #7

    symbolipoint

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    You want to understand factoring trinomials AND recognize the difference of squares.

    [tex] \[
    \begin{array}{l}
    ((y^2 + 8) + 6y)((y^2 + 8) - 6y) \\
    (y^2 + 6y + 8)(y^2 - 6y + 8) \\
    (y + 2)(y + 4)(y^2 - 6y + 8) \\
    \end{array}
    \]
    [/tex]
     
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