document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=970091>Question 970091</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A function is shown below:\r<BR>\n" );
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document.write( "f(x) = x3 + 3x2 - x - 3\r<BR>\n" );
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document.write( "Part A: What are the factors of f(x)? Show your work.\r<BR>\n" );
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document.write( "Part B: What are the zeros of f(x)? Show your work.\r<BR>\n" );
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document.write( "Part C: What are the steps you would follow to graph f(x)? Describe the end behavior of the graph of f(x). </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #592762 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=592762\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The potential factors of x^3-3x^2-x-3  are +/- 3  and +/- 1; I need 3 roots or 2 roots with one a square.\r<BR>\n" );
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document.write( "Trying -3 (factor x+3) \r<BR>\n" );
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document.write( "-3   1 ; 3  ;-1  ;;-3<BR>\n" );
document.write( ";;;  1; -3;;  0 ;;  3 (+3)  (difficult to format synthetic division)<BR>\n" );
document.write( ";;;;;;;;0;; -1;;0<BR>\n" );
document.write( "This works with no remainder, so  (x+3) and (x^2-1) are factors.\r<BR>\n" );
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document.write( "Difference of squares noted:<BR>\n" );
document.write( "Factors are (x+3) (x+1) (x-1)\r<BR>\n" );
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document.write( "Each of those =0 will give me zeros.\r<BR>\n" );
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document.write( "-3, -1, 1\r<BR>\n" );
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document.write( "For -3:   -27  +27  +3   -3  =0\r<BR>\n" );
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document.write( "For 1:    1+3-1-3=0\r<BR>\n" );
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document.write( "For -1:   -1+3 +1 -3=0\r<BR>\n" );
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document.write( "To graph it <BR>\n" );
document.write( "y=a (x+3) (x-1)(x+1)<BR>\n" );
document.write( "point (0,-3) is on the graph (the constant).\r<BR>\n" );
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document.write( "-3= a (3)(-1) (1)<BR>\n" );
document.write( "a=1<BR>\n" );
document.write( "y=(x+3)(x+1)(x-1)<BR>\n" );
document.write( "To the left of minus 3, the graph goes negative, driven by the first negative cubic term.\r<BR>\n" );
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document.write( "It rises through the root -1 and drops to 0,-3, when x=0.  It then rises for good, passing through (1,0) and becoming driven by the positive cube.\r<BR>\n" );
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document.write( "     </SPAN>\n" );
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