SOLUTION: Find a cubic polynomial in standard form with real coefficients, having the zeros 2 and 6i. Let the leading coefficient be 1

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Question 999717: Find a cubic polynomial in standard form with real coefficients, having the zeros 2 and 6i. Let the leading coefficient be 1
Found 3 solutions by Fombitz, ikleyn, Alan3354:
Answer by Fombitz(32388)   (Show Source):
You can put this solution on YOUR website!
Real coefficients means complex roots come in complex conjugate pairs.

Answer by ikleyn(52803)   (Show Source):
You can put this solution on YOUR website!
.
Find a cubic polynomial in standard form with real coefficients, having the zeros 2 and 6i. Let the leading coefficient be 1
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(x-2)*(x-6i)*(x-(-6i)) = .

1. If a polynomial with real coefficients has a complex root, it has the conjugate complex number as a root, too.

2. If a polynomial f(x) of a degree n with leading coefficient 1 has n roots , , . . . , then the polynomial is the product

      f(x) = .

Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website!
Find a cubic polynomial in standard form with real coefficients, having the zeros 2 and 6i. Let the leading coefficient be 1
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If 6i is a zero then -6i is too.
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--> (x-2)*(x-6i)*(x+6i)
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