document.write( "Question 1006537: Find a polynomial function with real coefficients that has the given zeros. (There are many correct answers.) Thanks!\r
\n" ); document.write( "\n" ); document.write( "6, 7 + i
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Algebra.Com's Answer #622658 by ikleyn(52797)\"\" \"About 
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\n" ); document.write( "p(x) = (x-6)*(x-(7+i))*(x-(7-i)) = (x-6)*(x^2 - 14x + 50).\r
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\n" ); document.write( "\n" ); document.write( "It is based on two theorems.\r
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\n" ); document.write( "\n" ); document.write( "First theorem says:
\n" ); document.write( "\"If a polynomial with real coefficients has a complex root (the root which is complex number) \r
\n" ); document.write( "\n" ); document.write( "then it has the root which is complex conjugate to the first one\".\r
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\n" ); document.write( "\n" ); document.write( "Second theorem says:
\n" ); document.write( "\"If a polynomial p(x) of degree n with the leading coefficient 1 at \"x%5En\" has n roots \"x%5B1%5D\", \"x%5B2%5D\", . . . , \"x%5Bn%5D\", \r
\n" ); document.write( "\n" ); document.write( "then the polynomial is p(x) = \"%28x+-+x%5B1%5D%29\"*\"%28x+-+x%5B2%5D%29\"* . . . *\"%28x-x%5Bn%5D%29\".\r
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