anonymous
  • anonymous
Find the polynomial function with roots 11 and 2i.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ganeshie8
  • ganeshie8
if 2i is one root, then -2i is also a root
ganeshie8
  • ganeshie8
cuz comples roots come in pairs always
ganeshie8
  • ganeshie8
so the roots are 11, 2i, and -2i the polynomial function wud be... ?

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More answers

anonymous
  • anonymous
(x-11)(x+2)(x-2) ?
ganeshie8
  • ganeshie8
done forget the i :)
anonymous
  • anonymous
(x-11i)(x+2i)(x-2i)
ganeshie8
  • ganeshie8
i is oly for 2i
ganeshie8
  • ganeshie8
11 is a real simple root
ganeshie8
  • ganeshie8
it wud be :- (x-11)(x+2i)(x-2i)
anonymous
  • anonymous
sooo (x-11)(x+2i)(x-2i) (x²-11x+24i)(x-2i)
ganeshie8
  • ganeshie8
we can do that, but let me show u how to expand easily :)
ganeshie8
  • ganeshie8
(x-11)(x+2i)(x-2i) first expand the last two factors
ganeshie8
  • ganeshie8
\((x-11)(x^2+4)\)
ganeshie8
  • ganeshie8
cuz, \((a+ib)(a-ib) = a^2 + b^2\)
ganeshie8
  • ganeshie8
\((x-11)(x^2+4) \) now you can simplify easily ?
anonymous
  • anonymous
x^3-11x²+4x-44
ganeshie8
  • ganeshie8
very good work :) wolfram says the same : http://www.wolframalpha.com/input/?i=%28x-11%29%28x%2B2i%29%28x-2i%29
ganeshie8
  • ganeshie8
you can test ur answers in that site
anonymous
  • anonymous
Thankyouu (:>

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