FInd a cubic equation with intergral coeficients that has roots 3 and 7 - i(assume the leading coeficient is 1?

Show all work and thank you soooooooo much!!! I reaaallly appreciated it!

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  • 9 years ago
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               x³ + bx² + cx + d = 0

    Complex roots occur in pairs: 7 – i and 7 + i

    then all three roots shown as product terms is:

         (x – 3) • (x – 7 + i) • (x – 7 – i) = 0

     ... Expanding these terms ...

          (x – 3) • (x² – 14x + 50) = 0 ... intermediate step

         x³ – 17x² + 92x – 150 = 0

    so the initial coefficient = a = 1

     [b = -17] [c = 92] and [d = -150]

  • Anonymous
    9 years ago

    Ok so here is your answer.

    x^3-17 x^2+92 x-150

    You get this by multiplying (x-3)(x-7-i)(x-7+i)

    The only tricky part to this is finding the conjugates. Its very easy, all you have to do is change the - sign in 7-i to 7+i. If you know that and how to set up factors, you just FOIL the (x-7-i)(x-7+i) first. And then do the same thing again with your answer and (x-3)

    Source(s): HS Algebra tutor
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