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Math Help - [SOLVED] write an equation of a polynomial

  1. #1
    Junior Member moonman's Avatar
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    [SOLVED] write an equation of a polynomial

    Write an equation of a polynomial of degree 4 and integer coefficients with roots -2i, 1/3 and 5.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by moonman View Post
    Write an equation of a polynomial of degree 4 and integer coefficients with roots -2i, 1/3 and 5.
    If a is a root of a polynomial, then x-a divides this polynomial.
    So you can say that (x-(-2i))(x-1/3)(x-5) divides the polynomial.

    What about the 4th factor ? Since the coefficients are integers, if a complex number (like -2i) is a root, then its conjugate is also a root.

    The conjugate of a complex number a+ib is a-ib. Here, a=0 and b=-2. Can you do it ?
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  3. #3
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    Quote Originally Posted by moonman View Post
    Write an equation of a polynomial of degree 4 and integer coefficients with roots -2i, 1/3 and 5.
    1. I assume that this polynomial has +2i as a root too.

    2. p(x)=a\left((x^2+4)(x-5)\left(x-\frac13\right)\right)

    3. Expand the brackets:

    p(x)=a\left(x^4-\frac{16}3 x^3 + \frac{17}3 x^2 - \frac{64}3 x + \frac{20}3  \right)

    4. Choose a = 3 and you'll get:

    p(x)=3x^4-16x^3+17x^2-64x+20

    -----------------------------------------------------------

    EDIT: Too fast for me, Moo
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  4. #4
    Junior Member moonman's Avatar
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    Quote Originally Posted by earboth View Post
    1. I assume that this polynomial has +2i as a root too.

    2. p(x)=a\left((x^2+4)(x-5)\left(x-\frac13\right)\right)

    3. Expand the brackets:

    p(x)=a\left(x^4-\frac{16}3 x^3 + \frac{17}3 x^2 - \frac{64}3 x + \frac{20}3 \right)

    4. Choose a = 3 and you'll get:

    p(x)=3x^4-16x^3+17x^2-64x+20

    -----------------------------------------------------------

    EDIT: Too fast for me, Moo

    In step 2; can you show me how you got (x^2 + 4)
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  5. #5
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    Quote Originally Posted by moonman View Post
    In step 2; can you show me how you got (x^2 + 4)
    If the roots of the polynomial are x = -2i~\vee~x=+2i then you have the product:

    (x-2i)(x+2i)=x^2-4i^2=x^2-4\cdot (-1)=x^2+4
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  6. #6
    Junior Member moonman's Avatar
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    Quote Originally Posted by earboth View Post
    If the roots of the polynomial are x = -2i~\vee~x=+2i then you have the product:

    (x-2i)(x+2i)=x^2-4i^2=x^2-4\cdot (-1)=x^2+4

    awesome
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