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Math Help - Question on algebra

  1. #1
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    Question on algebra

    Hi guys,i had posted a few threads on algebra just now.

    There are many solution but i find the one which i could understand was:

    Question: Factorise 121 - (p+2)Square

    Answer/Solution:
    Quote Originally Posted by skeeter View Post
    11^2 - (p+2)^2 = [11 - (p+2)][11 + (p+2)] = (9-p)(13+p)
    But, what i don't understand is this solution:

    Question: Factorise 121 - (p+2)Square

    Answer/Solution: 121 - (p)Square + 2(p)(2) + 2Square
    = 121 - (p)Square + 4p + 4
    = 117 - (p)Square + 4p
    = (9-p)(13+p)
    I mean, how do you know if it's "(9-p)(13+p)" ?

    Thank you so much for your time
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  2. #2
    MHF Contributor Unknown008's Avatar
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    You can expand as I showed you earlier. Expand both expressions and you'll get the same final expression.
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  3. #3
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    Quote Originally Posted by Unknown008 View Post
    You can expand as I showed you earlier. Expand both expressions and you'll get the same final expression.
    Hi, but how can you expand from "117 - (p)Square + 4p" onwards?
    Can you show it to me?
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  4. #4
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    \displaystyle = -(p^2 - 4p - 117).

    Now what numbers multiply to give \displaystyle -117 and add to give \displaystyle -4?
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  5. #5
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    Quote Originally Posted by Prove It View Post
    \displaystyle = -(p^2 - 4p - 117).

    Now what numbers multiply to give \displaystyle -117 and add to give \displaystyle -4?
    OOOOOO, so thats how you do it, i was searching for this.
    So,
    -13 x 9 = -117
    9 - 13 = -4

    Thanks so much !
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Well, what I initially meant is:

    11^2 - (p+2)^2 = 121 - p^2 - 4p - 4 = 117 - p^2 - 4p

    And taking the other,

    (9-p)(13+p) = 117 - p^2 - 13p + 9p = 117 - p^2 - 4p

    Which is the same as before. Hence, both are equivalent, meaning that the answer (9-p)(13+p) is correct.
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