document.write( "Question 792922: 2x^3 +x^2-12x+9 <0 \n" ); document.write( "
Algebra.Com's Answer #480193 by Edwin McCravy(20056) $\"\"$ $\"About$
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document.write( "2x³ + x² - 12x + 9 < 0\r\n" );
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document.write( "We begin by trying to factor.  A good way to start is\r\n" );
document.write( "by dividing it by x-1 or by x+1, because 1 is always \r\n" );
document.write( "a possible zero or root of a polynomial.  It may not work\r\n" );
document.write( "but it's always a good try.\r\n" );
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document.write( "1 | 2  1 -12  9\r\n" );
document.write( "  |    2   3 -9  \r\n" );
document.write( "    2  3  -9  0\r\n" );
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document.write( "Hot dog!  It worked!  So we have now factored\r\n" );
document.write( "it as\r\n" );
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document.write( "(x - 1)(2x² + x - 12) < 0\r\n" );
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document.write( "And now we can factor the quadratic:\r\n" );
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document.write( "(x - 1)(2x - 3)(x + 3) < 0\r\n" );
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document.write( "So there are three critical values, found by\r\n" );
document.write( "setting all factors = 0\r\n" );
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document.write( "x = 1; x = ); x = -3\r\n" );
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document.write( "Mark those on a number line:\r\n" );
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document.write( "-------------o-----------o-o------------------\r\n" );
document.write( "-6  -5  -4  -3  -1   0   1   2   3   4   5   6\r\n" );
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document.write( "They divide the number line into 4 intervals:\r\n" );
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document.write( "(-∞,-3), (-3,1), (1,), (,∞)\r\n" );
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document.write( "For any of these intervals between or beyond critical \r\n" );
document.write( "values, if (x - 1)(2x - 3)(x + 3) < 0 is true (respectively, \r\n" );
document.write( "false) for one value in that interval, then it is true \r\n" );
document.write( "(respectively, false) for all values in that interval, so we \r\n" );
document.write( "take a test value for x in each interval.\r\n" );
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document.write( "For the interval (-∞,-3), the easiest test point is -4:\r\n" );
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document.write( "(x - 1)(2x - 3)(x + 3) < 0\r\n" );
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document.write( "(-4 - 1)(2(-4) - 3)(-4 + 3) < 0\r\n" );
document.write( "             (-5)(-8-3)(-1) < 0\r\n" );
document.write( "              (-5)(-11)(-1) < 0\r\n" );
document.write( "                        -55 < 0\r\n" );
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document.write( "That is true, so (-∞,-3) is part of the solution:\r\n" );
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document.write( "<============o-----------o-o------------------\r\n" );
document.write( "-6  -5  -4  -3  -1   0   1   2   3   4   5   6\r\n" );
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document.write( "For the interval (-3,1), the easiest test point is 0:\r\n" );
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document.write( "(x - 1)(2x - 3)(x + 3) < 0\r\n" );
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document.write( "(0 - 1)(2(0) - 3)(0 + 3) < 0\r\n" );
document.write( "            (-1)(0-3)(3) < 0\r\n" );
document.write( "             (-1)(-3)(3) < 0\r\n" );
document.write( "                       3 < 0\r\n" );
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document.write( "That is false, so (-3,1) is not part of the solution:\r\n" );
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document.write( "For the interval (1,), , the easiest test point is 1.1:\r\n" );
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document.write( "(x - 1)(2x - 3)(x + 3) < 0\r\n" );
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document.write( "(1.1 - 1)(2(1.1) - 3)(-1.1 + 3) < 0\r\n" );
document.write( "               (0.1)(2.2-3)(-1) < 0\r\n" );
document.write( "               (0.1)(-0.8)(1.9) < 0\r\n" );
document.write( "                          -.152 < 0\r\n" );
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document.write( "That is true, so (1,)  is part of the solution:\r\n" );
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document.write( "<============o-----------o=o------------------\r\n" );
document.write( "-6  -5  -4  -3  -1   0   1   2   3   4   5   6\r\n" );
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document.write( "For the interval (, ∞), the easiest test point is 2:\r\n" );
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document.write( "  (x - 1)(2x - 3)(x + 3) < 0\r\n" );
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document.write( "(2 - 1)(2(2) - 3)(2 + 3) < 0\r\n" );
document.write( "             (1)(4-3)(5) < 0\r\n" );
document.write( "               (1)(1)(5) < 0\r\n" );
document.write( "                       5 < 0\r\n" );
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document.write( "That is false, so (, ∞) is not part of the solution:\r\n" );
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document.write( "<============o-----------o=o------------------\r\n" );
document.write( "-6  -5  -4  -3  -1   0   1   2   3   4   5   6\r\n" );
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document.write( "So the solution in interval notation is\r\n" );
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document.write( "(-∞,-3) U (1,)\r\n" );
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document.write( "What the problem is asking is: Where is the graph of \r\n" );
document.write( "y = 2x³ + x² - 12x + 9 below the x-axis?:\r\n" );
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document.write( "And we see that it is below the x-axis (less than 0)\r\n" );
document.write( "to the left of -3, and between 1 and . \r\n" );
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document.write( "Edwin
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