document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1181501>Question 1181501</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Check that (2 + i) is a root of (z^4) + (2(z^3)) - (9(z^2)) - 10z + 50 = 0. What are the remaining 3 roots? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #811424 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=811424\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "In the hopes of making the problem easier, we can, as the other tutor did, assume (\"hope\") that the polynomial has real coefficients.<br><BR>\n" );
document.write( "In that case, follow the solution given by the other tutor:<BR>\n" );
document.write( "if (2+i) is a root then (2-i) is a root;<BR>\n" );
document.write( "determine the quadratic factor that comes from those two roots;<BR>\n" );
document.write( "divide the given function by that quadratic and observe that the result is another quadratic with no remainder -- confirming that our assumption of real coefficients was okay; and<BR>\n" );
document.write( "use the quadratic formula to find the two roots corresponding to that second quadratic factor.<br><BR>\n" );
document.write( "I would like to add a bit to her solution, showing different ways that parts of the work can be done.<br><BR>\n" );
document.write( "She finds the quadratic corresponding to the two root (2+i) and (2-i) by doing the multiplication<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28z-%282%2Bi%29%29%28z-%282-i%29%29 ALIGN=MIDDLE  ALT=\"%28z-%282%2Bi%29%29%28z-%282-i%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "Note that she makes that multiplication easier by writing it as<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%28z-2%29-i%29%28%28z-2%29%2Bi%29 ALIGN=MIDDLE  ALT=\"%28%28z-2%29-i%29%28%28z-2%29%2Bi%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "So that the product can be found as a difference of squares.<br><BR>\n" );
document.write( "Here is another way to form the quadratic corresponding to those two roots.<br><BR>\n" );
document.write( "The linear term of the quadratic is the opposite of the sum of the two roots: -((2+i)+(2-i)) = -4<br><BR>\n" );
document.write( "The constant term of the quadratic is the product of the two roots: (2+i)(2-i)=4+5 = 5<br><BR>\n" );
document.write( "So the quadratic factor having the roots (2+i) and (2-i) is z^2-4x+5.<br><BR>\n" );
document.write( "In the next step, she shows dividing the given polynomial by z^2-4x+5 to get z^2+6x+10 with no remainder.<br><BR>\n" );
document.write( "She doesn't show how she got that.  Synthetic division only works for dividing by linear polynomials, and long division of polynomials is awkward.<br><BR>\n" );
document.write( "So here is the process I like to use to divide a 4th degree polynomial by a quadratic.<br><BR>\n" );
document.write( "We are looking for a quadratic polynomial <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=az%5E2%2Bbz%2Bc ALIGN=MIDDLE  ALT=\"az%5E2%2Bbz%2Bc\"   DISABLED_style= \"max-width:90%; height:auto;\" > for which<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28az%5E2%2Bbz%2Bc%29%28z%5E2-4z%2B5%29=z%5E4%2B2z%5E3-9z%5E2-10z%2B50 ALIGN=MIDDLE  ALT=\"%28az%5E2%2Bbz%2Bc%29%28z%5E2-4z%2B5%29=z%5E4%2B2z%5E3-9z%5E2-10z%2B50\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "By looking at the leading coefficient and the constant term of the product, we can immediately determine that a=1 and c=10.  So the multiplication of the two quadratics is<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28z%5E2%2Bbz%2B10%29%28z%5E2-4z%2B5%29=z%5E4%2B2z%5E3-9z%5E2-10z%2B50 ALIGN=MIDDLE  ALT=\"%28z%5E2%2Bbz%2B10%29%28z%5E2-4z%2B5%29=z%5E4%2B2z%5E3-9z%5E2-10z%2B50\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "We can now determine b by looking at where the z^3 term of the product comes from: bz times z^2, plus z^2 times (-4z)<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28b%29%281%29%2B%281%29%28-4%29+=+2 ALIGN=MIDDLE  ALT=\"%28b%29%281%29%2B%281%29%28-4%29+=+2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b-4=2 ALIGN=MIDDLE  ALT=\"b-4=2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b=6 ALIGN=MIDDLE  ALT=\"b=6\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "So our second quadratic factor is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=z%5E2%2B6z%2B10 ALIGN=MIDDLE  ALT=\"z%5E2%2B6z%2B10\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "That process for finding the second quadratic factor is easier for me than long division of polynomials.<br>\r<BR>\n" );
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document.write( " </SPAN>\n" );
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