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Math Help - Help factoring something. What's the process/rule?

  1. #1
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    Help factoring something. What's the process/rule?

    I'm having trouble figuring out how to factor this.

    81y^4-72y^2+16

    I know 16 is 2^4, so I rewrite as

    81y^4-72y^2+2^4

    I don't know where to go from here.

    Thanks in advance.
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  2. #2
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    Re: Help factoring something. What's the process/rule?

    Use a substitution. (Turn the problem into a factorable quadratic)
    let a = y^2

    Then, 81a^2-72a+16 is the new equation using a (since a^2 = y^4)

    Now to factor, you need two numbers whose product is (81x16=1296) and whose sum is -72 (-36 and -36)
    Factor by grouping:

    81a^2 - 36a - 36a + 16
    9a(9a - 4) - 4(9a - 4)
    (9a - 4)(9a - 4)

    Resubstiture a = y^2

    (9y^2-4)(9y^2-4)
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  3. #3
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    Re: Help factoring something. What's the process/rule?

    Thanks, I didn't know I could do that.

    But the answer in the book is, (3y+2)^2(3y-2)^2.
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  4. #4
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    Re: Help factoring something. What's the process/rule?

    Quote Originally Posted by StudentMCCS View Post
    Thanks, I didn't know I could do that.

    But the answer in the book is, (3y+2)^2(3y-2)^2.
    I suppose you can factor some more:
    (9y^2-4) = (3y+2)(3y-2) (difference of squares)
    So, (9y^2-4)(9y^2-4) = (3y+2)(3y-2)(3y+2)(3y-2) = (3y+2)^2*(3y-2)^2
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  5. #5
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    Re: Help factoring something. What's the process/rule?

    Thanks!
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  6. #6
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    Re: Help factoring something. What's the process/rule?

    Hi again.

    I've been wondering if there is an easy way to recognize the answer to this solution based on the relationships of its terms.

    What I came up with is that.

    Given AX^4-BX^2+C.
    If (rootA)(rootB)(2)=B
    Then AX^4-BX^2+C=, sorry I don't know how to properly write this out.
    [(4th root A)X+(4th root C)]^2*[(4th root A)X-(4th root C)]^2

    Is this really the case, I only checked for 2 equations?
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  7. #7
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    Re: Help factoring something. What's the process/rule?

    Quote Originally Posted by StudentMCCS View Post
    Hi again.

    I've been wondering if there is an easy way to recognize the answer to this solution based on the relationships of its terms.

    What I came up with is that.

    Given AX^4-BX^2+C.
    If (rootA)(rootB)(2)=B
    Then AX^4-BX^2+C=, sorry I don't know how to properly write this out.
    [(4th root A)X+(4th root C)]^2*[(4th root A)X-(4th root C)]^2

    Is this really the case, I only checked for 2 equations?
    There's lots of "if's" to a generalization to this. If B < 0, for example, then simplest factored form is not written using 4th roots. Example:

    If the original equation was

    81y^4+72y^2+16 (B = -72)

    Then it would factor to (9y^2 + 4)(9y^2 + 4)

    Since both of the above factors have imaginary solutions, you would stop factoring here.

    In short, you are correct, but occasionally taking this shortcut will yield a factored expression using imaginary factors.

    Also, I would replace "If (rootA)(rootB)(2)=B" with "If (B/2)^2 = AC"
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  8. #8
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    Re: Help factoring something. What's the process/rule?

    Wait sorry, I meant (rootA)(RootC)(2)=B
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  9. #9
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    Re: Help factoring something. What's the process/rule?

    Is this the exercice:
    Given ax^4+bx^2+c (1)
    If 2\sqrt{a}\cdot \sqrt{b}=b (2)
    then ax^4+bx^2+c=...?
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