document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=118563>Question 118563</A>: <I><SPAN STYLE=\"word-break: break-all;\"> CAN YOU HELP ME. I REALLY DONT UNDERSTAND. I HAVE TO VERIFY THE GIVEN FACTORS OF THE FUNCTION, FIND THE REMAINING FACTORS OF f, USE MY RESULTS TO WRITE THE COMPLETE FACTORIZATION OF f, LIST ALL REAL 0'S OF f, AND CONFRIM MY RESULTS BY USING A GRAPHING ULTILITY TO GRAPH THE FUNCTION.\r<BR>\n" );
document.write( "\n" );
document.write( "f(x)=2x(to the 3rd power)+x(to the 2nd power)-5x+2      Factors<BR>\n" );
document.write( "                                                      (x+2) (x-1) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #86693 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=86693\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You need to use long division or synthetic division to verify that the given factors are indeed factors of the function.  I'll try to walk you through the steps for one of them:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%282x%5E3%2Bx%5E2-5x%2B2%29%2F%28x%2B2%29 ALIGN=MIDDLE  ALT=\"%282x%5E3%2Bx%5E2-5x%2B2%29%2F%28x%2B2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The first term of the denominator, x, goes into the first term of the numerator, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2x%5E3 ALIGN=MIDDLE  ALT=\"2x%5E3\"   DISABLED_style= \"max-width:90%; height:auto;\" >, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2x%5E2 ALIGN=MIDDLE  ALT=\"2x%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > times, so that becomes the first term of your quotient.  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2x%5E2%2A%28x%2B2%29=2x%5E3%2B4x%5E2 ALIGN=MIDDLE  ALT=\"2x%5E2%2A%28x%2B2%29=2x%5E3%2B4x%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > so subtract these two terms from the first two terms of the numerator expression yielding <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-3x%5E2 ALIGN=MIDDLE  ALT=\"-3x%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  Bring down the next numerator term and divide the resulting binomial, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-3x%5E2-5x ALIGN=MIDDLE  ALT=\"-3x%5E2-5x\"   DISABLED_style= \"max-width:90%; height:auto;\" >, by <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%2B2 ALIGN=MIDDLE  ALT=\"x%2B2\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  The first term of the divisor, x, goes into the first term <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-3x ALIGN=MIDDLE  ALT=\"-3x\"   DISABLED_style= \"max-width:90%; height:auto;\" > times, so that becomes the second term of your quotient.  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-3x%2A%28x%2B2%29=3x%5E2-6x ALIGN=MIDDLE  ALT=\"-3x%2A%28x%2B2%29=3x%5E2-6x\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  Subtract these two terms from the (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-3x%5E2-5x ALIGN=MIDDLE  ALT=\"-3x%5E2-5x\"   DISABLED_style= \"max-width:90%; height:auto;\" > dividend part resulting in <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x ALIGN=MIDDLE  ALT=\"x\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  Bring down the <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%2B2 ALIGN=MIDDLE  ALT=\"%2B2\"   DISABLED_style= \"max-width:90%; height:auto;\" > resulting in <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%2B2 ALIGN=MIDDLE  ALT=\"x%2B2\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  Divide the <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%2B2 ALIGN=MIDDLE  ALT=\"x%2B2\"   DISABLED_style= \"max-width:90%; height:auto;\" > by <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%2B2 ALIGN=MIDDLE  ALT=\"x%2B2\"   DISABLED_style= \"max-width:90%; height:auto;\" > resulting in 1, which becomes the third term of your quotient.  The quotient is now <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2x%5E2-3x%2B1 ALIGN=MIDDLE  ALT=\"2x%5E2-3x%2B1\"   DISABLED_style= \"max-width:90%; height:auto;\" > and there is no remainder.  Since there is no remainder, you have verified that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%2B2 ALIGN=MIDDLE  ALT=\"x%2B2\"   DISABLED_style= \"max-width:90%; height:auto;\" > is a factor of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%282x%5E3%2Bx%5E2-5x%2B2%29 ALIGN=MIDDLE  ALT=\"%282x%5E3%2Bx%5E2-5x%2B2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Repeat the process using the quotient you just derived, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2x%5E2-3x%2B1 ALIGN=MIDDLE  ALT=\"2x%5E2-3x%2B1\"   DISABLED_style= \"max-width:90%; height:auto;\" >, and dividing that by the other given factor, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x-1 ALIGN=MIDDLE  ALT=\"x-1\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  Since the process is the same as described above, I'll leave the details to you, but you should get a quotient of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2x-1 ALIGN=MIDDLE  ALT=\"2x-1\"   DISABLED_style= \"max-width:90%; height:auto;\" > with no remainder.  That tells you that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x-1 ALIGN=MIDDLE  ALT=\"x-1\"   DISABLED_style= \"max-width:90%; height:auto;\" > is a factor and that the third factor is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2x-1 ALIGN=MIDDLE  ALT=\"2x-1\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Now you can say that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%28x%29=2x%5E3%2Bx%5E2-5x%2B2=%28x%2B2%29%28x-1%29%282x-1%29 ALIGN=MIDDLE  ALT=\"f%28x%29=2x%5E3%2Bx%5E2-5x%2B2=%28x%2B2%29%28x-1%29%282x-1%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  The zeros of f are those values of x for which <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%28x%29+=+0 ALIGN=MIDDLE  ALT=\"f%28x%29+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  \r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=f%28x%29+=+0 ALIGN=MIDDLE  ALT=\"f%28x%29+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" > if and only if <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%2B2=0 ALIGN=MIDDLE  ALT=\"x%2B2=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x-1=0 ALIGN=MIDDLE  ALT=\"x-1=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2x-1=0 ALIGN=MIDDLE  ALT=\"2x-1=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Solving each of these equations yields:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5B1%5D=-2 ALIGN=MIDDLE  ALT=\"x%5B1%5D=-2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5B2%5D=1 ALIGN=MIDDLE  ALT=\"x%5B2%5D=1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5B3%5D=1%2F2 ALIGN=MIDDLE  ALT=\"x%5B3%5D=1%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28600%2C600%2C-8%2C8%2C-8%2C8%2C2x%5E3%2Bx%5E2-5x%2B2%29 ALIGN=MIDDLE  ALT=\"graph%28600%2C600%2C-8%2C8%2C-8%2C8%2C2x%5E3%2Bx%5E2-5x%2B2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Notice that the graph crosses the x-axis at the three zeros of f.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Notice that this graph has a peak and a valley.  It's beyond the scope of this discussion why, but the values of x at the top of the peak and the bottom of the valley are given by solving <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6x%5E2%2B2x-5=0 ALIGN=MIDDLE  ALT=\"6x%5E2%2B2x-5=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >, yielding <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-1%2B-sqrt%2831%29%29%2F6 ALIGN=MIDDLE  ALT=\"%28-1%2B-sqrt%2831%29%29%2F6\"   DISABLED_style= \"max-width:90%; height:auto;\" >, very roughly speaking -1 and 3/4, as you can see on the graph.  The y values for these points would obviously be the original function evaluated at these two x values.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Hope that helps.<BR>\n" );
document.write( "John </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );