document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1092731>Question 1092731</A>: <I><SPAN STYLE=\"word-break: break-all;\"> May I have some help with this question please?  I have to solve the equation and inequality and show all steps and work.  It's one homework problem from the summer and I am still not clear on it.  The \"0\" is confusing to me. \r<BR>\n" );
document.write( "\n" );
document.write( "(x-3)/(x+2) > 0    USE INTERVAL NOTATION\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Thanks. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #707323 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=707323\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> looking at this function on the graph will help you to see what's going on.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "here's the graph of y = (x-3) * (x+2)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2017/090701.jpg\" alt=\"$$$\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you can see that the graph crosses the x-axis at x = -2 and x = 3.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "that's where the value of y is equal to 0.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "when y = 0, the equation becomes 0 = (x-3) * (x+2)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "y is equal to 0 when x = 3 because y = (3-3) * (3+2) which is equal to 0 * 6 which is equal to 0.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "y is equal to 0 when x = -2 because y = (-2-3) * (-2+2) which is equal to 6 * 0 which is equal to 0.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "those are the roots of the equation and are found when you you set y = 0 in the equation of y = (x-3) * (x+2).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you are asked to find the value of x when (x-3) * (x+2) > 0.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if you set y = (x-3) * (x+2), then the question is to find the value of x when y > 0, because y = (x-3) * (x+2).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "y is used to represent the equation because we normally graph a 2 dimensional equation with x on the horizontal axis and y on the vertical axis.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "it doesn't have to be that way, but most of the graphing software is set up that way.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "some more sophisticated graphing software allow you to use any variable name you want, but those capabilities are not always available.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "for example, the ti-84 plus requires the vertical axis to be the value of y, while the horizontal axis requires the horizontal axis to be x.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if you're familiar with functional notation, then the equation would have been shown as f(x) = (x-3) * (x+2).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the use of f(x) and y are inter-changeable as they both represent the value of the equation after replacing the x variable with a value.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if you were using f(x) = (x-3) * (x+2), then the question would have been to find the value of x when f(x) > 0.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "so, .....\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you can find the answer to this problem by setting y = (x-3) * (x+2) and then graphing it and finding out where it crosses the x-axis.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "this will be at x = -2 and x = 3.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you would then test the intervals where x < -2 and -2 < x < 3 and x > 3 to see if the function is positive or negative in those intervals.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you can see that the function is positive when x < -2 and when x > 3, and that the function is negative when -2 < x < 3.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "that's your answer graphically.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if you were to do the same without the benefit of looking at a graph, you would solve in a similar way.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you would make your equation y = (x-3) * (x+2) and then set y = 0 to find the roots.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the equation would be 0 = (x-3) * (x+2)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "this is a quadratic equation because when you multiply the factors out, the leading term is x^2.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "in fact, the quadratic equation, before being factored, is y = x^2 - x - 6.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "factor that equation and you get y = (x-3) * (x+2)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "since it's already in factored form, it's easy to see that the roots of the equation are x = 3 and x = -2.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "that's when y = 0.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the rest is finding the interval when y is positive, which means when (x-3) * (x+3) is positive.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "since the function is continuous, if any point in the interval is positive, then all points in the interval will be positive until the function crosses the x-axis.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "likewise if any point in the interval is negative.<BR>\n" );
document.write( "it will remain negative until the function crosses the x-axis again.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "so just test a point in each interval and it will tell you whether that interval is positive or not.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the intervals to test are:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "x < -2<BR>\n" );
document.write( "-2 < x < 3<BR>\n" );
document.write( "x > 3\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "those are what are called the critical points in the graph which is the points where the function can change sign.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "there is a third way to look at it.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "you want to know when the function (x-3) * (x+2) is greater than 0.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if you let a = (x-3) and b = (x+2), then the function becomes a * b > 0\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "this occurs when a and b are both positive and when a and b are both negative.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "so, you are looking for when (x-3) is positive and when (x+2) is positive, and you are looking for when (x-3) is negative and when (x+2) is also negative.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "take a look at (x-3)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "it is positive when x > 3<BR>\n" );
document.write( "it is negative when x < 3\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "take a look at (x+2)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "it is positive when x > -2<BR>\n" );
document.write( "it is negative when x < -2\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "so, positive * positive is when x > 3 or x > -2.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "and, negative * negative is when x < 3 or x < -2\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "if you want to have the factors both positive at the same time, then it can only occur when x > 3.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "it is possible for x > -2 and not > 3, but it is not possible for x > 3 and not > -2.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "likewise, if you want to have the factors both negative at the same time, then it can only occur when x < -2.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "it is possible for x < 3 and not < -2, but it is not possible for x < -2 and not < 3.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "in all of this, you should wind up with the same conclusion.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "(x-3) * (x+2) > 0 when x < -2 and when x > 3.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "in interval notation this would be x = (-infinity, -2) union (3, infinity)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );