document.write( "Question 188143: Solve: x^3+3x-2x^2-6=0\r
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Algebra.Com's Answer #141018 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
Any possible rational zero can be found through this formula\r
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\n" ); document.write( "\n" ); document.write( " where p and q are the factors of the last and first coefficients\r
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\n" ); document.write( "\n" ); document.write( "So let's list the factors of -6 (the last coefficient):\r
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\n" ); document.write( "\n" ); document.write( "Now let's list the factors of 1 (the first coefficient):\r
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\n" ); document.write( "\n" ); document.write( "Now let's divide each factor of the last coefficient by each factor of the first coefficient\r
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\n" ); document.write( "\n" ); document.write( "Now simplify\r
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\n" ); document.write( "\n" ); document.write( "These are all the distinct possible rational zeros of the function.\r
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\n" ); document.write( "\n" ); document.write( "Note: these are the possible zeros. The function may not even have rational zeros (they may be irrational or complex).\r
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\n" ); document.write( "\n" ); document.write( "Now simply use synthetic division to find the real rational zeros\r
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\n" ); document.write( "\n" ); document.write( "Let's see if the possible zero \"1\" is really a root for the function \"x%5E3-2x%5E2%2B3x-6\"\r
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\n" ); document.write( "\n" ); document.write( "So let's make the synthetic division table for the function \"x%5E3-2x%5E2%2B3x-6\" given the possible zero \"1\":\r
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1|1-23-6
| 1-12
1-12-4
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\n" ); document.write( "\n" ); document.write( "Since the remainder \"-4\" (the right most entry in the last row) is not equal to zero, this means that \"1\" is not a zero of \"x%5E3-2x%5E2%2B3x-6\"\r
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\n" ); document.write( "\n" ); document.write( "Let's see if the possible zero \"2\" is really a root for the function \"x%5E3-2x%5E2%2B3x-6\"\r
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\n" ); document.write( "\n" ); document.write( "So let's make the synthetic division table for the function \"x%5E3-2x%5E2%2B3x-6\" given the possible zero \"2\":\r
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2|1-23-6
| 206
1030
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\n" ); document.write( "\n" ); document.write( "Since the remainder \"0\" (the right most entry in the last row) is equal to zero, this means that \"2\" is a zero of \"x%5E3-2x%5E2%2B3x-6\"\r
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\n" ); document.write( "\n" ); document.write( "Because \"x=2\" is a zero, this means that \"x-2\" is a factor of \"x%5E3-2x%5E2%2B3x-6\"\r
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\n" ); document.write( "\n" ); document.write( "The first three numbers in the last row 1, 0, and 3 form the coefficients to the polynomial \"x%5E2%2B3\". So this consequently means that \"x%5E3-2x%5E2%2B3x-6=%28x-2%29%28x%5E2%2B3%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28x-2%29%28x%5E2%2B3%29=0\" Set the right side equal to zero\r
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\n" ); document.write( "\n" ); document.write( "\"x-2=0\" or \"x%5E2%2B3=0\" Set each factor equal to zero\r
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\n" ); document.write( "\n" ); document.write( "Since we know that \"x=2\" is already a zero, we can ignore the first equation. So simply solve the quadratic equation \"x%5E2%2B3=0\" to find the remaining solutions:\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2%2B3=0\" Start with the given equation\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2=-3\" Subtract 3 from both sides\r
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\n" ); document.write( "\n" ); document.write( "\"x=sqrt%28-3%29\" or \"x=-sqrt%28-3%29\" Take the square root of both sides (don't forget the \"plus/minus\")\r
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\n" ); document.write( "\n" ); document.write( "\"x=sqrt%283%29%2Ai\" or \"x=-sqrt%283%29%2Ai\" Simplify. Note: \"i=sqrt%28-1%29\"\r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So the solutions of \"x%5E3-2x%5E2%2B3x-6\" are \"x=2\", \"x=sqrt%283%29%2Ai\" or \"x=-sqrt%283%29%2Ai\"
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