document.write( "Question 771004: How do I factor this out 6x^3>7x^2+3x (this is a polynomial and rational inequality) ???
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Algebra.Com's Answer #469907 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "6x³ > 7x² + 3x\r\n" );
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document.write( "There are three cases to consider, when\r\n" );
document.write( "x > 0, when x = 0, and when x < 0.\r\n" );
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document.write( "Case 1:  x > 0\r\n" );
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document.write( "6x³ > 7x² + 3x\r\n" );
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document.write( "Since x is positive, we may divide\r\n" );
document.write( "through by x without changing the\r\n" );
document.write( "direction of the inequality.\r\n" );
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document.write( "         6x² > 7x + 3\r\n" );
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document.write( "6x² - 7x - 3 > 0\r\n" );
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document.write( "(2x-3)(3x+1) > 0\r\n" );
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document.write( "Setting each of those two parentheses =\r\n" );
document.write( "0 gives critical numbers \"3%2F2\" and\r\n" );
document.write( "\"-1%2F3\".  We can ignore the negative\r\n" );
document.write( "since x is positive for this case, and\r\n" );
document.write( "so we take 0 as the lower critical\r\n" );
document.write( "number.  So the critical numbers are 0\r\n" );
document.write( "and \"3%2F2\".\r\n" );
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document.write( "If x is between the two critical\r\n" );
document.write( "valuee, say x=1, the inequality becomes\r\n" );
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document.write( "    [2(1)-3][3(1)+1] > 0\r\n" );
document.write( "             (-1)(4) > 0\r\n" );
document.write( "                  -4 > 0\r\n" );
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document.write( "That's false, so the inequality does\r\n" );
document.write( "not hold in the interval between the\r\n" );
document.write( "critical values 0 and \"3%2F2\".\r\n" );
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document.write( "If x > \"3%2F2\", say x=2, the\r\n" );
document.write( "inequality becomes\r\n" );
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document.write( "[2(2)-3][3(2)+1] > 0\r\n" );
document.write( "      (4-3)(6+1) > 0 \r\n" );
document.write( "          (1)(7) > 0\r\n" );
document.write( "               7 > 0\r\n" );
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document.write( "That's true so the inequality is\r\n" );
document.write( "true when x is in the interval (\"3%2F2\",∞)\r\n" );
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document.write( "So for case 1, the inequality has\r\n" );
document.write( "solution set\r\n" );
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document.write( "(\"3%2F2\",∞)\r\n" );
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document.write( "Case 2: x=0\r\n" );
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document.write( "This case is ruled out because\r\n" );
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document.write( "6x³ > 7x² + 3x becomes 0 > 0 which is\r\n" );
document.write( "false.\r\n" );
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document.write( "Case 3: x < 0\r\n" );
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document.write( "6x³ > 7x² + 3x\r\n" );
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document.write( "Since x is negative, if we divide\r\n" );
document.write( "through by x we must change the\r\n" );
document.write( "direction of the inequality.\r\n" );
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document.write( "         6x² < 7x + 3\r\n" );
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document.write( "6x² - 7x - 3 < 0\r\n" );
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document.write( "(2x-3)(3x+1) < 0\r\n" );
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document.write( "Setting each of those two parentheses =\r\n" );
document.write( "0 gives critical numbers \"3%2F2\" and\r\n" );
document.write( "\"-1%2F3\".  The inequality is not true \r\n" );
document.write( "at either of the critical numbers.  We\r\n" );
document.write( "can ignore the positive critical number\r\n" );
document.write( "since x is negative for this case, and\r\n" );
document.write( "so we take 0 as the upper critical\r\n" );
document.write( "number.  So the critical numbers are\r\n" );
document.write( "\"-1%2F3\" and 0.\r\n" );
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document.write( "If x < \"-1%2F3\", say x=-1, the\r\n" );
document.write( "inequality becomes\r\n" );
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document.write( "[2(-1)-3][3(-1)+1] < 0\r\n" );
document.write( "      (-2-3)(-3+1) < 0 \r\n" );
document.write( "          (-5)(-2) < 0\r\n" );
document.write( "                10 < 0\r\n" );
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document.write( "That's false so the inequality does not\r\n" );
document.write( "hold when x is less than \"-1%2F3\"\r\n" );
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document.write( "If x is between the two critical\r\n" );
document.write( "numbers, say x=-.01, the inequality\r\n" );
document.write( "becomes\r\n" );
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document.write( "    [2(-.01)-3][3(0)+1] < 0\r\n" );
document.write( "             (-3.02)(1) < 0\r\n" );
document.write( "                  -3.02 < 0\r\n" );
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document.write( "That's true, so Case 3 is true\r\n" );
document.write( "when (\"-1%2F3\",0).\r\n" );
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document.write( "So for case 3, the inequality has\r\n" );
document.write( "solution set\r\n" );
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document.write( "(\"-1%2F3\",0)\r\n" );
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document.write( "Answer:  The solution set for the given \r\n" );
document.write( "inequality is:\r\n" );
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document.write( "(\"-1%2F3\",0) U (\"3%2F2\",∞)\r\n" );
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document.write( "Edwin
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