document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1081688>Question 1081688</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Given the cubic function g(x)=x^3 + 6x +20. Show that g(x) can be written as g(x)=(x+2)Q(x), where Q(x) is a real quadratic. Hence,solve the equation g(x)=0 completely. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #695735 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=695735\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> see the following worksheet.\r<BR>\n" );
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document.write( "step 1 copies the original equation that will be divided by x+2.\r<BR>\n" );
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document.write( "step 2 inserts the missing order of exponents, so they're all present.\r<BR>\n" );
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document.write( "what was missing was the x^2 term, which was inserted as 0 * x^2.\r<BR>\n" );
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document.write( "step 3 justtakes the coefficients and divides they by -2 using the concept of synthetic division.\r<BR>\n" );
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document.write( "you are dividing by x + 2.<BR>\n" );
document.write( "set x + 2 = 0 and solve for x.<BR>\n" );
document.write( "you will get x = -2<BR>\n" );
document.write( "that's what you divide by.\r<BR>\n" );
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document.write( "you bring down the first coefficient.<BR>\n" );
document.write( "you then multiply that by -2 (not divide).<BR>\n" );
document.write( "the result is then added to the next coefficient.<BR>\n" );
document.write( "in this case the 1 was multiplied by -2 and then added to 0 to get -2.\r<BR>\n" );
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document.write( "you repeat that down the line until you get to the last term.<BR>\n" );
document.write( "if that term is 0, then x+2 is a factor of x^3 + 6x + 20.<BR>\n" );
document.write( "it is.\r<BR>\n" );
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document.write( "step 4 then adds the variable back in, but the order is one less than you started with.\r<BR>\n" );
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document.write( "that gets you x^2  -2x + 10.\r<BR>\n" );
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document.write( "step 5 takes (x+2) and multiplies it by (x^2 - 2x + 10)\r<BR>\n" );
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document.write( "this is to confirm the solution is correct.\r<BR>\n" );
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document.write( "step 6 completes the two parts of the multiplicsation and adds them together.\r<BR>\n" );
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document.write( "the result is what you see in step 7.\r<BR>\n" );
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document.write( "this is the original equation.\r<BR>\n" );
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document.write( "this tells you that the synthetic division was successful.\r<BR>\n" );
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document.write( "becauswe the last result in step 3 was 0, that told you that (x+2) divides evenly into x^3 + 6x + 20, which tells you that (x+2) is a factor of x^3 + 6x + 20.\r<BR>\n" );
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document.write( "here's a reference on synthetic division.\r<BR>\n" );
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document.write( "<a  rel=nofollow HREF= \"http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut37_syndiv.htm\" target = \"_blank\">http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut37_syndiv.htm</a>\r<BR>\n" );
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document.write( "you could also have used long division and gotten the same result.\r<BR>\n" );
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document.write( "if the remainder was 0, then x+2 could be considered a factor of x^3 + 6x + 20\r<BR>\n" );
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document.write( "here's a reference on long division.\r<BR>\n" );
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document.write( "<a  rel=nofollow HREF= \"http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut35_div.htm\" target = \"_blank\">http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut35_div.htm</a>\r<BR>\n" );
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document.write( "if the remainder was 0, then x+2 could be considered a factor of x^3 + 6x + 20\r<BR>\n" );
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document.write( " </SPAN>\n" );
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