document.write( "Question 1155716: x^3-2x^2<15x
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Algebra.Com's Answer #778350 by ikleyn(52803) $\"\"$ $\"About$

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document.write( "    x^3 - 2x^2 < 15x\r\n" );
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document.write( "    x^3 - 2x^2 - 15x < 0\r\n" );
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document.write( "Factor x out\r\n" );
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document.write( "    x*(x^2-2x - 15) < 0\r\n" );
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document.write( "Factor further\r\n" );
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document.write( "    x*(x-5)*(x+3) < 0\r\n" );
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document.write( "1)  In the interval  (-oo,-3)  all three factors are negative;\r\n" );
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document.write( "    so, their product is negative.  \r\n" );
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document.write( "    Thus this interval  (-oo,-3)  is the solution (is the part of the solution set).\r\n" );
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document.write( "2)  Next interval is  (-3,0).  In this interval,\r\n" );
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document.write( "        the factor (x+3) is positive; two other factors are negative.\r\n" );
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document.write( "    So, the product of the three factors is positive,\r\n" );
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document.write( "    and hence this interval (-3,0) is not the solution to the inequality.\r\n" );
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document.write( "3)  Next interval is  (0,5).  In this interval,\r\n" );
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document.write( "        the factors (x+3) and x  are positive; the last factor is negative.\r\n" );
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document.write( "    So, the product of the three factors is negative,\r\n" );
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document.write( "    and hence this interval (0,5) is the solution to the inequality.\r\n" );
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document.write( "4)  Last interval is  (5,oo).  In this interval,\r\n" );
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document.write( "        all three factors (x+3) are positive; \r\n" );
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document.write( "    So, the product of the three factors is positive,\r\n" );
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document.write( "    and hence this interval (5,oo) is not the solution to the inequality.\r\n" );
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document.write( "ANSWER.  The solution set to the given inequality is the union of two intervals  (-oo,-3) U (0,5).\r\n" );
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\n" ); document.write( "\n" ); document.write( "To see many other similar solved problems, look into the lesson\r
\n" ); document.write( "\n" ); document.write( " - Solving inequalities for high degree polynomials factored into a product of linear binomials \r
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