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Math Help - help with specific trinomials.

  1. #1
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    Exclamation help with specific trinomials.

    right, so i know how to do trinomials in a normal case with only one pronumeral.

    a^2 + 2a + 1
    (a+1)(a+1)

    and i can do non-monic trinomials no problem too, problem for me is when i discover a trinomial which has 2 pronumerals.

    a^2 + 2ab + b^2
    now i have no idea how i'm meant to work it out.
    i know that this simplifies to (a+b)(a+b) because i can do the expansion for it.
    but i have no idea how to simplify it and if it were slightly more complex
    eg. k^2 + 8kp + p^2 i would not know what to do in the exam.

    help please! could someone please show me how to work out these expressions a^2 + 2ab + b^2

    thankyou very much in advance.
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  2. #2
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    Quote Originally Posted by Zellus View Post
    right, so i know how to do trinomials in a normal case with only one pronumeral.

    a^2 + 2a + 1
    (a+1)(a+1)

    and i can do non-monic trinomials no problem too, problem for me is when i discover a trinomial which has 2 pronumerals.

    a^2 + 2ab + b^2
    now i have no idea how i'm meant to work it out.
    i know that this simplifies to (a+b)(a+b) because i can do the expansion for it.
    but i have no idea how to simplify it and if it were slightly more complex
    eg. k^2 + 8kp + p^2 i would not know what to do in the exam.

    help please! could someone please show me how to work out these expressions a^2 + 2ab + b^2

    thankyou very much in advance.
    Complete the square:

    k^2 + 8p + p^2 = (k + 4p)^2 - 15p^2 = (k + 4p)^2 - (\sqrt{15} p)^2.

    Now use the difference of two squares formula to get k^2 + 8p + p^2 = (k - [\sqrt{15} - 4] p) (k + [ \sqrt{15} + 4] p).
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Complete the square:

    k^2 + 8p + p^2 = (k + 4p)^2 - 15p^2 = (k + 4p)^2 - (\sqrt{15} p)^2.
    how did you know that k^2 + 8p + p^2 when has it's square completed = (k + 4p)^2 - 15p^2 without having to go through a slow, laborious process of trial and error?
    Last edited by mr fantastic; April 26th 2009 at 03:53 AM. Reason: Closed the quote
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  4. #4
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    Quote Originally Posted by Zellus View Post
    how did you know that k^2 + 8p + p^2 when has it's square completed = (k + 4p)^2 - 15p^2 without having to go through a slow, laborious process of trial and error?
    The key is to know that {\color{red}A^2 + 2AB} + B^2 = (A + B)^2. So {\color{red}k^2 + 8p} + 16p^2 = (k + 4p)^2.

    Hence k^2 + 8p + p^2 = k^2 + 8p + 16p^2 - 15p^2 = (k + 4p)^2 - 15p^2.
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