document.write( "Question 112794: use the ac test to determine if trinomial is factored. find values of m and n.
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Algebra.Com's Answer #82222 by solver91311(24713)\"\" \"About 
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I think you meant to say \"use the ac test to determine if the trinomial is factorable,\" and I'm presuming the m and n you are referring to are the m and n in the following:\r
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\n" ); document.write( "\n" ); document.write( "\"%28x%2Bm%29%28x%2Bn%29=x%5E2-3x%2B7\"\r
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\n" ); document.write( "\n" ); document.write( "First the AC method. A and C refer to the coefficients on the equation in standard form, namely: \"ax%5E2%2Bbx%2Bc=0\"\r
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\n" ); document.write( "\n" ); document.write( "The first step of the AC method is to determine the product \"a%2Ac\". In this case it is \"1%2A7\".\r
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\n" ); document.write( "\n" ); document.write( "The second step of the AC method is to list the possible factor pairs of the product \"a%2Ac\". In this case, 7 is a prime number so the only factor pairs that exist are (1,7) and (-1,-7).\r
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\n" ); document.write( "\n" ); document.write( "The third step is to find a factor pair that adds up to b or -3 in this problem. But there is no such pair because 1 + 7 = 8 and (-1) + (-7) = -8. Therefore, the trinomial is not factorable.\r
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\n" ); document.write( "\n" ); document.write( "Since the trinomial is not factorable, we are left with completing the square to find values for m and n in \"%28x%2Bm%29%28x%2Bn%29=x%5E2-3x%2B7\".\r
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\n" ); document.write( "\n" ); document.write( "Completeing the square:
\n" ); document.write( "Step 1:\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2-3x%2B7=0\". Begin with the original trinomial set equal to zero.
\n" ); document.write( "\"x%5E2-3x=-7\". Add the negative of the constant term to both sides. (if the coefficient on the \"x%5E2\" term were other than 1, you would have the preliminary step of dividing both sides by that coefficient)
\n" ); document.write( "\"x%5E2-3x%2B%283%2F2%29%5E2=-7%2B%283%2F2%29%5E2\". Take the coefficient on the x term, divide by 2 and then square the result and add that result to both sides of the equation.
\n" ); document.write( "\"x%5E2-3x%2B%283%2F2%29%5E2=-19%2F4\". And simplify the right side.\r
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\n" ); document.write( "\n" ); document.write( "Now the left side of the equation is a perfect square that can be factored.\r
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\n" ); document.write( "\n" ); document.write( "\"%28x-3%2F2%29%5E2=-19%2F4\"
\n" ); document.write( "\"x=3%2F2%2B-sqrt%28-19%2F4%29\"\r
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\n" ); document.write( "\n" ); document.write( "But we have a problem because \"sqrt%28-1%29\" doesn't exist. So we define the imaginary number \"i\" such that \"i%5E2=-1\". Now we can write:\r
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\n" ); document.write( "\n" ); document.write( "\"x=3%2F2%2B-i%2Asqrt%2819%29%2F2\", So,\r
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\n" ); document.write( "\n" ); document.write( "\"x=3%2F2%2Bi%2Asqrt%2819%29%2F2\" or \"x=3%2F2-i%2Asqrt%2819%29%2F2\", which is to say:\r
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\n" ); document.write( "\n" ); document.write( "\"x-%283%2F2%2Bi%2Asqrt%2819%29%2F2%29=0\" or \"x-%283%2F2-i%2Asqrt%2819%29%2F2%29=0\". And now we can say that:\r
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\n" ); document.write( "\n" ); document.write( " and finally,\r
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\n" ); document.write( "\n" ); document.write( "\"m=-%283%2F2%2Bi%2Asqrt%2819%29%2F2%29\" and \"n=-%283%2F2-i%2Asqrt%2819%29%2F2%29\"\r
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\n" ); document.write( "\n" ); document.write( "Now we still have to check the answer:\r
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\n" ); document.write( "\n" ); document.write( "Is this statement true? \r
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\n" ); document.write( "\n" ); document.write( "Apply the FOIL method to the left side:\r
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\n" ); document.write( "\n" ); document.write( " Remembering that \"i%5E2=-1\"\r
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\n" ); document.write( "\n" ); document.write( "Now collect terms and watch all of those pesky radicals go away:\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2-3x%2F2-3x%2F2%2B9%2F4%2B19%2F4\"
\n" ); document.write( "\"x%5E2-3x%2B28%2F4\"
\n" ); document.write( "\"x%5E2-3x%2B7\"\r
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\n" ); document.write( "\n" ); document.write( "And the answer checks.\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps.\r
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\n" ); document.write( "\n" ); document.write( "John
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