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algebra 3

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Find a polynomial of lowest degree with only real coefficients and having the given zeros.

-2+i, -2-i, 3, -3

  • algebra 3 - ,

    Zeros are also "x = ..." statements, so x = -2 + i, x = -2 - i, etc...

    Since you need REAL coefficients in the polynomial, change the x = -2 + i and x = -2 - i statements to "clean" polynomials, like this:

    x = -2 + i
    x + 2 = i (square each side to get rid of the i)
    (x+2)^2 = i^2
    x^2 + 4x + 4 = -1
    ==> x^2 + 4x + 5 = 0

    x = -2 - i
    (x+2)^2 = (-i)^2
    I'm not going to go further with this because you get the same exact equation as the one before. So now when you multiply all the zeroes together, you get this:

    0 = (x^2 + 4x + 5)(x-3)(x+3)

    Multiply it out and you get:

    x^4 + 4x^3 - 4x^2 - 36x - 45 = 0


    Does this make sense?

  • algebra 3 - ,

    The polynomial would have to have 4 zeros, meaning it would have to be a polynomial of the 4th degree. The general form for a polynomial of the 4th degree with zeros a, b, c and d would be:

    f*(x-a)*(x-b)*(x-c)*(x-d)

    where f is a random real number (lets take this to be 1 in this case).
    So, if we fill in the zeros you were given we get that:

    (x-(-2+i))*(x-(-2-i))*(x-3)*(x+3) =
    (x+2-i))*(x+2+i))*(x-3)*(x+3) =

    When we multiply the first two factors and the last two, we get:

    (x^2 + 2x + ix + 2x + 4 + 2i -ix - 2i +1) * (x^2 - 9) =

    (x^2 + 4x + 5)*(x^2 - 9) =
    (x^4 - 9x^2 + 4x^3 - 36x + 5x^2 -45) =
    x^4 + 4x^3 - 4x^2 -36x - 45

    this is a polynomial of the 4th degree which has the given values as its zeros

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