document.write( "Question 1113491: A polynomial f(x) with real coefficients and leading coefficient 1 has the given zeros and degree. Express f(x) as a product of linear and/or quadratic polynomials with real coefficients that are irreducible over the set of real numbers.
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document.write( "7 + 5i, −1 + i; degree 4 \n" );
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Algebra.Com's Answer #728570 by ikleyn(52814)![]() ![]() You can put this solution on YOUR website! . \n" ); document.write( " \r\n" ); document.write( "At given condition (polynomial with real coefficients) each complex (non-real) root goes in pair with its conjugate.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "So, if 7+5i is the root, then 7-5i is the root, too.\r\n" ); document.write( "\r\n" ); document.write( "Also, if -1+i is the root, then -1-i is the root, too.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "Thus we have 4 roots 7+5i, 7-5i, -1+i, -1-i.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "Hence, our polynomial is the product\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "p(x) = (x-(7+5i))*(x-(7-5i))*(x-(-1+i))*(x-(-1-i)) = \r\n" ); document.write( "\r\n" ); document.write( " = ((x-7)-5i)*((x-7)+5i)*((x+1)-i)*((x+1)+i) = \r\n" ); document.write( "\r\n" ); document.write( " = ((x-7)^2 - (5i)^2)*((x+1)^2 - i^2) =\r\n" ); document.write( "\r\n" ); document.write( " = (x^2 - 14x + 49 + 25)*(x^2 + 2x + 1 + 1) = \r\n" ); document.write( "\r\n" ); document.write( " = (x^2 - 14x + 74)*(x^2 + 2x + 2).\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "It is the required expression.\r\n" ); document.write( "\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |