document.write( "Question 76274: Use the ac test to determine which of the following trinomials can be factored. Find the values of m and n for each trinomial that can be factored.\r
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Algebra.Com's Answer #55126 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
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Solved by pluggable solver: Factoring Quadratics with a leading coefficient of 1 (a=1)
In order to factor \"1%2Ax%5E2%2B2%2Ax%2B-15\", first we need to ask ourselves: What two numbers multiply to -15 and add to 2? Lets find out by listing all of the possible factors of -15
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\n" ); document.write( " Factors:
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\n" ); document.write( " 1,3,5,15,
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\n" ); document.write( " -1,-3,-5,-15,List the negative factors as well. This will allow us to find all possible combinations
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\n" ); document.write( " These factors pair up to multiply to -15.
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\n" ); document.write( " (-1)*(15)=-15
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\n" ); document.write( " (-3)*(5)=-15
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\n" ); document.write( " Now which of these pairs add to 2? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 2
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First Number|Second Number|Sum
1|-15|1+(-15)=-14
3|-5|3+(-5)=-2
-1|15|(-1)+15=14
-3|5|(-3)+5=2
We can see from the table that -3 and 5 add to 2.So the two numbers that multiply to -15 and add to 2 are: -3 and 5\r\n" ); document.write( " \r\n" ); document.write( " Now we substitute these numbers into a and b of the general equation of a product of linear factors which is:\r\n" ); document.write( " \r\n" ); document.write( " \"%28x%2Ba%29%28x%2Bb%29\"substitute a=-3 and b=5\r\n" ); document.write( " \r\n" ); document.write( " So the equation becomes:\r\n" ); document.write( " \r\n" ); document.write( " (x-3)(x+5)\r\n" ); document.write( " \r\n" ); document.write( " Notice that if we foil (x-3)(x+5) we get the quadratic \"1%2Ax%5E2%2B2%2Ax%2B-15\" again\n" ); document.write( "

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\n" ); document.write( "\n" ); document.write( "So our polynomial factors to\r
\n" ); document.write( "\n" ); document.write( "\"%28x-3%29%28x%2B5%29\"\r
\n" ); document.write( "\n" ); document.write( "Notice \"%28x-3%29%28x%2B5%29\" foils back to \"x%5E2%2B2x-15\" \n" ); document.write( "
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