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Math Help - Polynomial

  1. #1
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    Polynomial

    Question>> the polynomial x^4+5x+a is denoted by p(x). It is given that x^2-x+3 is a factor of p(x).
    part (1) Find the value of a in p(x).

    Can somebody please reply to this quickly. Thank you
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Polynomial

    (x^2 - x + 3)(a_2x^2 + a_1x + a_0) = x^4 + 5x + a so [tex] we get  a_2x^4 + a_1x^3 + a_0x^2 - a_2x^3 - a_1 x^2 - a_0x + a_2*3*x^2 + 3*a_1x + 3a_0= a_2x^4 + (a_1 - a_2)x^3 + (a_0 - a_1 + 3a_2)x^2 + (3a_1 - a_0)x + 3a_0 = x^4 + 5x + a which means a_2 = 1. So a_2x^4 + (a_1 - 1)x^3 + (a_0 - a_1 + 3)x^2 + (3a_1 - a_0)x + 3a_0 and since  a_1 - 1 = 0  a_1 = 1 . So now you got a_2x^4 + (a_0 + 2)x^2 + (3 - a_0)x + 3a_0 and since  a_0 + 2 = 0  a_0 = -2 so again x^4 + (5)x -6 so a = -6.
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  3. #3
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    Re: Polynomial

    Code:
    ................. x^2  +  x  - 2
    .................-----------------------------
    x^2 - x + 3 | x^4 + 0x^3 + 0x^2 + 5x + a
    ............  x^4 -  x^3  + 3x^2
    ..............-----------------------------
    ....................x^3  - 3x^2 + 5x + a
    ....................x^3  -  x^2 + 3x
    ....................---------------------
    ........................- 2x^2 + 2x + a
    ........................- 2x^2 + 2x - 6
    ........................----------------
    ......................................a+6
    remainder, a+6 = 0

    a = -6
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  4. #4
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    Re: Polynomial

    thanks jakncoke. Thankyou skeeter, i was also using the same method but my remainder included the variable 'x' so i thought it was wrong. Found my mistake though, thanks once again.
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