pre-calculus

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form a polynomial f(x) with real coefficients having the given degree and zeros. degree 5; zeros -7; -i;-9+i

enter the polynomial.

f(x)=a(?)

  • pre-calculus -

    Complex numbers always appear as conjugate pairs, so if you have -i, then you also have +i
    and if you have -9+i, there will also be -9 - i

    so we know we have factors of (x+7) , (x^2 + 1) and two more

    I will use the sum and product rule to find the other
    sum of -9+i and -9 - i = -18
    product of the above is 81 - i^2 = 81 + 1 = 82
    resulting in the quadratic factor
    x^2 + 18x + 82

    so f(x) = (x+7)(x^2 + 1)(x^2 + 18x + 82)

    notice, if expanded this will give you a 5th degree polynomial. If you have to expand it, do it very carefully and patiently.

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