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7 September, 18:47

What is the polynomial function of lowest degree with lead coefficient 1 and roots 1 and 1 + i? f (x) = x2 - 2x + 2 f (x) = x3 - x2 + 4x - 2 f (x) = x3 - 3x2 + 4x - 2 f (x) = x2 - x + 2

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  1. 7 September, 19:04
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    If a+bi is a root in a polynomial with real coefients, then a-bi is also a root

    since 1+i is a root, 1-i is alos a root

    a function that has the roots r1, r2 can be factored into form

    f (x) = a (x-r1) (x-r2)

    a is leading coefient

    a=1 in this equation

    r1=1+i

    r2=1-i

    r3=1

    f (x) = (x - (1+i)) (x - (1-i)) (x-1)

    f (x) = (x-1-i) (x-1+i) (x-1)

    do math stuff

    remember that i²=-1

    f (x) = x³-3x²+4x-2
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