document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=942775>Question 942775</A>: <I><SPAN STYLE=\"word-break: break-all;\"> P(x)= x^3+x^2-4x+6\r<BR>\n" );
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document.write( "Express P(x) as a product of irreducible factors over the set of real numbers given that 1+i is a root of the polynomial.\r<BR>\n" );
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document.write( "Thanks!!  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #574789 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=574789\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1+i is a root of P(x) --> 1-i is also a root of P(x)\r<BR>\n" );
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document.write( "The sum of the roots of P(x) is -1, so (1+i) + (1-i) + r = -1 --> r = -3. So -3 is also a root of P(x) (you can check this).\r<BR>\n" );
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document.write( "Then P(x) = (x - (1+i))(x - (1-i))(x+3). </SPAN>\n" );
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