document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=935110>Question 935110</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I have been asked to show that (x+2) and (x-5) are factors of x^4+px^2+qx-40\r<BR>\n" );
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document.write( "Using synthetic division, I identified that for x+2; \r<BR>\n" );
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document.write( "Equation 1: 4p+16-2q-40\r<BR>\n" );
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document.write( "Equation 2: -25p+5q+585\r<BR>\n" );
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document.write( "However, if I were to subtract Equation 1 from 2; I do not have either P or Q fully nullified, there is a remainder left over.\r<BR>\n" );
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document.write( "For example, 4p-25p=-21p and 5q-2q=3q\r<BR>\n" );
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document.write( "However, in the case of 3p+27-q and -p-1+q\r<BR>\n" );
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document.write( "The -q and +q cancel each other out; leaving us only with the unknown P variable term. How would I remedy/rectify such a situation?\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #568244 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=568244\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1) using synthetic division, we divide x^4+px^2+qx-40 by (x+2), note that the equation becomes x^4+0x^3+px^2+qx-40, then we have<BR>\n" );
document.write( "-2 |  1  0  p q -40  |<BR>\n" );
document.write( "we get the following cubic x^3-x^2+(4+p)x-8-2p+q where -8-2p+q is the constant<BR>\n" );
document.write( "we also know that 16+4p-2q-40 = 0, therefore 4p-2q = 24, 2p-q=12<BR>\n" );
document.write( "2) now use the cubic result and divide it by synthetic division<BR>\n" );
document.write( "5 | 1  -2  4+p  -8-2p+q<BR>\n" );
document.write( "we get the following quadratic x^2+3x+19+p where 19+p is the constant<BR>\n" );
document.write( "we also know that 95+5p-8-2p+q=0, 3p+q=-87<BR>\n" );
document.write( "our two equations are<BR>\n" );
document.write( "2p-q=12<BR>\n" );
document.write( "3p+q=-87<BR>\n" );
document.write( "solve first equation for p<BR>\n" );
document.write( "p = (q/2)+6<BR>\n" );
document.write( "substitute for p in second equation and solve for q, then p<BR>\n" );
document.write( "3((q/2)+6)+q = -87<BR>\n" );
document.write( "multiply both sides of = by 2<BR>\n" );
document.write( "3q+36+2q=-174<BR>\n" );
document.write( "5q = -210<BR>\n" );
document.write( "q = -42<BR>\n" );
document.write( "p = -15<BR>\n" );
document.write( "our original equation is<BR>\n" );
document.write( "x^4+px^2+qx-40<BR>\n" );
document.write( "substitute for p and q<BR>\n" );
document.write( "x^4-15x^2-42x-40<BR>\n" );
document.write( "zeros for this equation using its graph<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+graph%28+300%2C+200%2C+-5%2C+6%2C+-300%2C+600%2C+x%5E4-15x%5E2-42x-40+%29+ ALIGN=MIDDLE  ALT=\"+graph%28+300%2C+200%2C+-5%2C+6%2C+-300%2C+600%2C+x%5E4-15x%5E2-42x-40+%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" > <BR>\n" );
document.write( "we see the graph crosses the x-axis at -2 and 5, these are the only zeros\r<BR>\n" );
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document.write( "  \r<BR>\n" );
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