document.write( "Question 1059919: Find a polynomial of degree 2 with real-number coefficients and zero 3-2i \n" ); document.write( "

Algebra.Com's Answer #674960 by Boreal(15235) $\"\"$ $\"About$

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The zeros are 3 +/- 2i, since complex roots come as conjugate
\n" ); document.write( "This is the result of -b+/-sqrt(b^2-4ac) all divided by 2a
\n" ); document.write( "Therefore, b=-6, because it will be divided by 2 at the end
\n" ); document.write( "b^2-4ac=36-4*c, and that has to equal -16. The sqrt (-16) is +/-4i, and dividing that by 2 will make 2i
\n" ); document.write( "Therefore, 4c=52, and c=13.
\n" ); document.write( "x^2-6x+13=0
\n" ); document.write( " $\"graph%28300%2C300%2C-10%2C10%2C-10%2C10%2Cx%5E2-6x%2B13%29\"$ \n" ); document.write( "

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