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Math Help - [SOLVED] Multiplying Polynomials?

  1. #1
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    [SOLVED] Multiplying Polynomials?

    This is the problem:

    (x + 3)(3x - 4)

    My math teacher taught us this:
    F - first
    O - outside
    I - inside
    L - last

    So would I go something like this?
    (x * 3x) + (3 * 3x) + (x * - 4) + (3 * 4)

    I'm really confused.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Zach View Post
    This is the problem:

    (x + 3)(3x - 4)

    My math teacher taught us this:
    F - first
    O - outside
    I - inside
    L - last

    So would I go something like this?
    (x * 3x) + (3 * 3x) + (x * - 4) + (3 * {\color{red}4})

    I'm really confused.
    no. the red 4 should be -4. you got it (except you did FIOL instead of FOIL )

    just multiply out and simplify

    you do know that "first" means "the first terms in each bracket" and analogous meanings for the others?


    anyway, what FOIL basically boils down to, is taking each term in one set of brackets and multiplying every term in the other set
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  3. #3
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    So:

    F - first terms in each bracket
    O - ??
    I - ??
    L - ??
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  4. #4
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    I simplified and got this:

    (x+3)(3x-4)
    (3x^2)+(9x)+(-4x)+(12)
    3x^2+5x+12
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Zach View Post
    So:

    F - first terms in each bracket
    O - the outside terms, that is, the terms on the ends
    I - the inside terms, the terms that are closest together
    L - the last terms in each bracket, that is, the right-most terms
    so, if we are talking about (a + b)(c + d) for example, we have:

    F: a and c, so we do a*c

    O: a and d, so we do a*d

    I: b and c, so we do b*c

    L: b and d, so we do b*d
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Zach View Post
    I simplified and got this:

    (x+3)(3x-4)
    (3x^2)+(9x)+(-4x)+(12)
    3x^2+5x+12
    you made the same mistake i told you about last time. it is a minus 4. so you must do (3 * -4) = -12
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    so, if we are talking about (a + b)(c + d) for example, we have:

    F: a and c, so we do a*c

    O: a and d, so we do a*d

    I: b and c, so we do b*c

    L: b and d, so we do b*d
    Thank you so much I finally understand!
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  8. #8
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    Another problem:

    (2x+7)(2x-7)

    My solution:

    (2x+7)(2x-7)
    (4x^2)+(-14x)+(14x)+(-49)
    4x^2-49
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Zach View Post
    Another problem:

    (2x+7)(2x-7)

    My solution:

    (2x+7)(2x-7)
    (4x^2)+(-14x)+(14x)+(-49)
    4x^2-49
    you are correct

    you could have also realized that this was the form of the difference of two squares and get the answer in one step (if you don't know what "the difference of two squares" is referring to, forget about it, you will soon enough. it seems that practicing FOIL is your objective, and it will always work for multiplying two binomials)
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  10. #10
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    Quote Originally Posted by Jhevon View Post
    you are correct

    you could have also realized that this was the form of the difference of two squares and get the answer in one step (if you don't know what "the difference of two squares" is referring to, forget about it, you will soon enough. it seems that practicing FOIL is your objective, and it will always work for multiplying two binomials)
    Yeah I'm in 8th grade Saxon Algebra 1.

    Thanks so much you were of great help.
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Zach View Post
    Yeah I'm in 8th grade Saxon Algebra 1.

    Thanks so much you were of great help.
    you're welcome . you were a great helpee!

    "666" appears in my post count again!
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