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Math Help - factorising

  1. #1
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    factorising

    the question is: a polynomial is given by P(x) = x^3 + 3x^2 - 10x - 24. given that (x + 2) is a factor factorise P(x) completely

    now i actually know a way of doing this question but it involves a bit of trial and error so i was wondering if there was a quicker way. i would start by using the idea that 2ab = -24 leaving ab = -12. but then i'd have to go through 1, 2, 3, 4, 6, 12 and their minuses to find the right one. could someone tell me if there's a quicker way or this it?

    thanks, Mark
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  2. #2
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    Quote Originally Posted by mark View Post
    the question is: a polynomial is given by P(x) = x^3 + 3x^2 - 10x - 24. given that (x + 2) is a factor factorise P(x) completely

    now i actually know a way of doing this question but it involves a bit of trial and error so i was wondering if there was a quicker way. i would start by using the idea that 2ab = -24 leaving ab = -12. but then i'd have to go through 1, 2, 3, 4, 6, 12 and their minuses to find the right one. could someone tell me if there's a quicker way or this it?

    thanks, Mark
    Hi


    Do the long division ,then you will get a quadratic equation , further factorise it if possible . I think this is the fastest for such questions .
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  3. #3
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    Quote Originally Posted by mark View Post
    the question is: a polynomial is given by P(x) = x^3 + 3x^2 - 10x - 24. given that (x + 2) is a factor factorise P(x) completely

    now i actually know a way of doing this question but it involves a bit of trial and error so i was wondering if there was a quicker way. i would start by using the idea that 2ab = -24 leaving ab = -12. but then i'd have to go through 1, 2, 3, 4, 6, 12 and their minuses to find the right one. could someone tell me if there's a quicker way or this it?

    thanks, Mark
    synthetic division ...

    Code:
    -2]........1......3......-10......-24
    .................-2......-2........24
    -------------------------------------
    ..........1.......1.....-12........0
    note the depressed coefficients 1, 1, -12

    (x+2)(x^2 + x - 12)

    now you can factor the quadratic factor rather easily ...

    (x+2)(x+4)(x-3)
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  4. #4
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    P(x) = x^3 + 3x^2 - 10x - 24
    Assume that P(x) = (x+2)(Ax^2 + Bx + C)

    Equating the coefficient of x^3, x^2, x, you will get
    A = 1
    2A+ B = 3 => B = 1
    C + 2B = -10 => C = -12

    p(x) = (x+2)(x^2 + x -12) = (x+2)(x+4)(x-3)

    Therefore P(x) = (x+2)(x+4)(x-3)


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