document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1057226>Question 1057226</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Use the given root to help in writing the given equation as a product of linear and quadratic factors with real coefficients.  X^4+3x^3-5x^2-29x-30=0 ; -2-I  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #672281 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=672281\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If -2-i is one root, then -2+i is another.<BR>\n" );
document.write( "That works when a quadratic equation is of the form of x^2+4x+c=0<BR>\n" );
document.write( "and x=(1/2)(-4+/-2i)<BR>\n" );
document.write( "Therefore, b^2-4ac must be -4, because the sqrt (-4) is +/-2i<BR>\n" );
document.write( "b^2-4ac=16-4c=-4<BR>\n" );
document.write( "-4c=-20<BR>\n" );
document.write( "c=5<BR>\n" );
document.write( "x^2+4x+5 is one factor<BR>\n" );
document.write( "If this is divided into x^4+3x^3-5x^2-29x-30, using polynomial division, the result is (x^2-x-6), which factors into (x-3)(x+2)<BR>\n" );
document.write( "(x^2+4x+5)(x-3)(x+2)<BR>\n" );
document.write( "roots are -2+i,-2-i,3,-2<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );