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Thread: Factoring...3

  1. #1
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    Factoring...3

    What is the first step in factoring the following problem?

    (5a^2 - 11a + 10)^2 - (4a^2 -15a + 6)^2
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    Re: Factoring...3

    Quote Originally Posted by mathdad1965 View Post
    What is the first step in factoring the following problem?

    (5a^2 - 11a + 10)^2 - (4a^2 -15a + 6)^2
    It's the difference of two squares.

    Recall that $x^2 - y^2 = (x - y)(x + y)$.


    $(5a^2 - 11a + 10)^2 - (4a^2 -15a + 6)^2 = $

    $[(5a^2 - 11a + 10) - (4a^2 - 15a + 6)][(5a^2 - 11a + 10) + (4a^2 - 15a + 6)] =$
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    Re: Factoring...3

    Quote Originally Posted by greg1313 View Post
    It's the difference of two squares.

    Recall that $x^2 - y^2 = (x - y)(x + y)$.


    $(5a^2 - 11a + 10)^2 - (4a^2 -15a + 6)^2 = $

    $[(5a^2 - 11a + 10) - (4a^2 - 15a + 6)][(5a^2 - 11a + 10) + (4a^2 - 15a + 6)] =$
    I will post my work and solution this weekend.
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    Re: Factoring...3

    (a^2 - 26a + 16)(a^2 - 26a + 16)

    a^4 - 26a^3 + 16a^2 - 26a^3 + 676a^2 - 416a + 16a^2 -

    416a + 256 = a^4 - 52a^3 + 740a^2 - 832a + 256.

    Is this correct?
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    Re: Factoring...3

    Quote Originally Posted by greg1313 View Post

    $(5a^2 - 11a + 10)^2 - (4a^2 -15a + 6)^2 = $

    $[(5a^2 - 11a + 10) - (4a^2 - 15a + 6)][(5a^2 - 11a + 10) + (4a^2 - 15a + 6)] =$
    Quote Originally Posted by mathdad1965 View Post
    (a^2 - 26a + 16)(a^2 - 26a + 16)

    a^4 - 26a^3 + 16a^2 - 26a^3 + 676a^2 - 416a + 16a^2 -

    416a + 256 = a^4 - 52a^3 + 740a^2 - 832a + 256.

    Is this correct?
    No.

    $(5a^2 - 11a + 10) - (4a^2 - 15a + 6) = $

    $5a^2 - 11a + 10 - 4a^2 + 15a - 6 = $

    $5a^2 - 4a^2 - 11a + 15a + 10 - 6 = $

    $a^2 + 4a + 4$


    and


    $(5a^2 - 11a + 10) + (4a^2 - 15a + 6) = $

    $5a^2 - 11a + 10 + 4a^2 - 15a + 6 = $

    $5a^2 + 4a^2 - 11a - 15a + 10 + 6 = $

    $9a^2 - 26a + 16$


    Substitute them back in:


    $(a^2 + 4a + 4)(9a^2 - 26a + 16) = $


    Do not multiply them back together. Try to factor each trinomial further into a pair of binomial factors.
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    Re: Factoring...3

    Quote Originally Posted by greg1313 View Post
    No.

    $(5a^2 - 11a + 10) - (4a^2 - 15a + 6) = $

    $5a^2 - 11a + 10 - 4a^2 + 15a - 6 = $

    $5a^2 - 4a^2 - 11a + 15a + 10 - 6 = $

    $a^2 + 4a + 4$


    and


    $(5a^2 - 11a + 10) + (4a^2 - 15a + 6) = $

    $5a^2 - 11a + 10 + 4a^2 - 15a + 6 = $

    $5a^2 + 4a^2 - 11a - 15a + 10 + 6 = $

    $9a^2 - 26a + 16$


    Substitute them back in:


    $(a^2 + 4a + 4)(9a^2 - 26a + 16) = $


    Do not multiply them back together. Try to factor each trinomial further into a pair of binomial factors.
    I get it now. I can take it from here. I made simple errors in my quick reply. I love this stuff. Thank God for math!
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    Re: Factoring...3

    Quote Originally Posted by greg1313 View Post
    No.

    $(5a^2 - 11a + 10) - (4a^2 - 15a + 6) = $

    $5a^2 - 11a + 10 - 4a^2 + 15a - 6 = $

    $5a^2 - 4a^2 - 11a + 15a + 10 - 6 = $

    $a^2 + 4a + 4$


    and


    $(5a^2 - 11a + 10) + (4a^2 - 15a + 6) = $

    $5a^2 - 11a + 10 + 4a^2 - 15a + 6 = $

    $5a^2 + 4a^2 - 11a - 15a + 10 + 6 = $

    $9a^2 - 26a + 16$


    Substitute them back in:


    $(a^2 + 4a + 4)(9a^2 - 26a + 16) = $


    Do not multiply them back together. Try to factor each trinomial further into a pair of binomial factors.
    a^2 + 4a + 4 factors out to be (a + 2)(a + 2).

    9a^2 - 26a + 16 factors out to be (3a - 2)(3a - 8).

    Answer:

    (a + 2)(a + 2)(3a - 2)(3a - 8).
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    Re: Factoring...3

    Quote Originally Posted by mathdad1965 View Post
    a^2 + 4a + 4 factors out to be (a + 2)(a + 2).

    9a^2 - 26a + 16 factors out to be (3a - 2)(3a - 8). No, that should be (a - 2)(9a - 8).

    Answer:

    (a + 2)(a + 2)(3a - 2)(3a - 8). So, that makes this answer also incorrect.
    Answer: (a + 2)(a + 2)(a - 2)(9a - 8).
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    Re: Factoring...3

    Quote Originally Posted by greg1313 View Post
    Answer: (a + 2)(a + 2)(a - 2)(9a - 8).
    I can see that my factoring needs polishing.
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