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Math Help - Multiplying trinomials

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    Senior Member vaironxxrd's Avatar
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    Multiplying trinomials

    I have to solve for : (n+3)(n-2)(n+3)

    My answer: n^3+4n^2-3n-18
    I went through a tedious solving form because of lack of knowledge.
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  2. #2
    MHF Contributor
    Prove It's Avatar
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    Re: Multiplying trinomials

    Quote Originally Posted by vaironxxrd View Post
    I have to solve for : (n+3)(n-2)(n+3)

    My answer: n^3+4n^2-3n-18
    I went through a tedious solving form because of lack of knowledge.
    First of all, this is the product of three BINOMIALS. Trinomials have three terms inside each bracket.

    Your answer is correct.
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    Senior Member vaironxxrd's Avatar
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    Re: Multiplying trinomials

    Quote Originally Posted by Prove It View Post
    First of all, this is the product of three BINOMIALS. Trinomials have three terms inside each bracket.

    Your answer is correct.
    So the long process is needed?
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    MHF Contributor Siron's Avatar
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    Re: Multiplying trinomials

    Quote Originally Posted by vaironxxrd View Post
    So the long process is needed?
    What do you mean with the 'long process'?
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    Re: Multiplying trinomials

    Quote Originally Posted by vaironxxrd View Post
    So the long process is needed?
    I'm not sure what you mean by the long process but the easiest way is to multiply two of them and then their product by the third. In your example:

    (n+3)(n-2)(n+3) = (n+3)^2(n-2) = (n^2+6n+9)(n-2) = (n^3 +6n^2+9n) + (-2n^2-12n-18) = n^3+4n^2-3n-18
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    Senior Member vaironxxrd's Avatar
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    Re: Multiplying trinomials

    Quote Originally Posted by Siron View Post
    What do you mean with the 'long process'?
    Quote Originally Posted by e^(i*pi) View Post
    I'm not sure what you mean by the long process but the easiest way is to multiply two of them and then their product by the third. In your example:

    (n+3)(n-2)(n+3) = (n+3)^2(n-2) = (n^2+6n+9)(n-2) = (n^3 +6n^2+9n) + (-2n^2-12n-18) = n^3+4n^2-3n-18
    well I multiplied the first two terms and then the product of them by the last term.
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    Re: Multiplying trinomials

    Quote Originally Posted by vaironxxrd View Post
    well I multiplied the first two terms and then the product of them by the last term.
    That's pretty much the fastest way
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