anonymous
  • anonymous
How would I find a cubic polynomial with roots 1 and i? Thanks!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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y2o2
  • y2o2
(x-i)(x-1)² or (x-1)(x-i)²
phi
  • phi
You need to know that complex roots come in complex conjugate pairs also, a cubic (3rd power) has 3 roots. for a quadratic with roots 2 and 3, the poylnomial is (x-2)(x-3)
phi
  • phi
y2, not repeated i, but complex conjugate pairs.

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anonymous
  • anonymous
I am honestly trying to understand, but I am not sure how I would use that information to start the question. Is there a general process for this type of question?
phi
  • phi
First, can you identify the 3 roots? Do you know what complex conjugate pair is?
anonymous
  • anonymous
I'm not sure how to find the third root, but I realize that two of the roots are (1+0i) and (0+1i).
phi
  • phi
no. The roots are 1, +i , and -i (the complex conjugate of a+bi is a-bi, here a=0, b=1)
anonymous
  • anonymous
Oh, I see. Okay.
phi
  • phi
Now we know the terms (x-i)(x- -i)(x-1)= p(x) or (x-i)(x+i)(x-1) = p(x) multiply out to get the full expression
y2o2
  • y2o2
Oh, i thought he wanted the roots to be only +i and 1 i didn't know that -i is also included
phi
  • phi
You always get imaginary roots in conjugate pairs (unless your polynomial has complex coeffs, which we typically do not have)
anonymous
  • anonymous
Okay, thank you very much! I understand it now. :)
y2o2
  • y2o2
Hmm, Ok, thanks for the information

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