document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=148921>Question 148921</A>: <I><SPAN STYLE=\"word-break: break-all;\"> how do I solve (x-2)(x+1)&#8805;0<BR>\n" );
document.write( "related equation: (x-2) (x+1)=0<BR>\n" );
document.write( "solutions for the equation: (2,-1)<BR>\n" );
document.write( "intervals<BR>\n" );
document.write( "(-&#8734;,-1)(-1,2)(2,&#8734;) After this I am confused ?<BR>\n" );
document.write( "f(x)=(x-4)(x+2) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #109249 by </B><A HREF=/tutors/aboutme.mpl?userid=tangitehewagen><B>tangitehewagen(7)</B></A><img src=\"/images/stars/stars-05.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=tangitehewagen><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=109249\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> It looks like you are on the right track although I do not know where f(x) = (x-4)(x+2) came from so lets ignore that and look at the rest of your work. \r<BR>\n" );
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document.write( "Lets summarize what you have done so far, okay?\r<BR>\n" );
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document.write( "You decided that you needed to solve the equation (x-2)(x+1) = 0 to get your \"cut points\" which divide up the number line into intervals. \r<BR>\n" );
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document.write( "You found the cut points to be x = -1 and x = 2. \r<BR>\n" );
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document.write( "Next you used these cut points to divide the interval (-&#8734;,&#8734;) into the subintervals (-&#8734;,-1)(-1,2)(2,&#8734;).\r<BR>\n" );
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document.write( "Now you are in the home stretch. This is the point at which you remind yourself of the original problem (x-2)(x+1) &#8805; 0.\r<BR>\n" );
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document.write( "In other words, you want to know where the product (x-2)(x+1) is either 0 or greater than 0 (Positive), right?\r<BR>\n" );
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document.write( "Well, you found that the product will be 0 at x = -1 and x = 2. We will come back to this in a moment. \r<BR>\n" );
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document.write( "Lets determine where the product is positive so we go back to the intervals you found earlier. We know that in each of these intervals (x+1)(x-2) < 0 or (x+1)(x-2) > 0 but not both (why?).\r<BR>\n" );
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document.write( "To determine the sign of (x+1)(x-2) over each of the intervals, we take a number in each intervals and evaluate (x-2)(x+1) at the number.\r<BR>\n" );
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document.write( "So for the interval (-&#8734;,-1), lets use x = -2. <BR>\n" );
document.write( "(-2 + 1)(-2 - 2) = (-1)(-4) = 4 which is positive so (x+1)(x-2) > 0 on the interval (-&#8734;,-1).\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28+400%2C+100%2C+-10%2C+10%2C+-10%2C+10%2C++%0D%0A+line%282.5%2C-4%2C2.5%2C0%29%2C%0D%0A+locate%28+2.3%2C+3.2%2C+0+%29%2C%0D%0A+line%28-2%2C-4%2C-2%2C0%29%2C%0D%0A+locate%28+-2.2%2C+3.2%2C+0+%29%2C%0D%0A+line%28-5.5%2C.3%2C-5.5%2C2.3%29%2C%0D%0A+line%28-5.8%2C1.3%2C-5.2%2C1.3%29%2C%0D%0A+locate%28+-8%2C+6%2C+number_line%28+300%2C+-5%2C+5%2Cx+=+-1%2Cx+=+2+%29+%29%0D%0A+%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "For the interval (-1,2), lets use x = 0.<BR>\n" );
document.write( "(0 + 1)(0 - 2) = 1(-2) = -2 which is negative so (x+1)(x-2) < 0 on the interval (-1,2).\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28+400%2C+100%2C+-10%2C+10%2C+-10%2C+10%2C++%0D%0A+line%282.5%2C-4%2C2.5%2C0%29%2C%0D%0A+locate%28+2.3%2C+3.2%2C+0+%29%2C%0D%0A+line%28-2%2C-4%2C-2%2C0%29%2C%0D%0A+locate%28+-2.2%2C+3.2%2C+0+%29%2C%0D%0A+line%28-5.5%2C.3%2C-5.5%2C2.3%29%2C%0D%0A+line%28-5.8%2C1.3%2C-5.2%2C1.3%29%2C%0D%0A+line%280.0%2C1.3%2C0.6%2C1.3%29%2C%0D%0A+locate%28+-8%2C+6%2C+number_line%28+300%2C+-5%2C+5%2Cx+=+-1%2Cx+=+2+%29+%29%0D%0A+%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Finally, for the interval (2,&#8734;), lets use x = 3.<BR>\n" );
document.write( "(3 + 1)(3 - 2) = (4)(1) = 4 which is positive so (x+1)(x-2) > 0 on the interval (2,&#8734;).\r<BR>\n" );
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document.write( "This tells us that (x+1)(x-2) > 0 on the union, (-&#8734;,-1) U (2,&#8734;).\r<BR>\n" );
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document.write( "But remember, we also want to know where (x+1)(x-2) = 0 which we found to be at x = -1 and x = 2 so in our union we close the subintervals at -1 and 2 to obtain\r<BR>\n" );
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document.write( "(x+1)(x-2) &#8805; 0 when x &epsilon; (-&#8734;,-1] U [2,&#8734;). </SPAN>\n" );
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