document.write( "Question 299476: trial factoring or reverse foil\r
\n" ); document.write( "\n" ); document.write( "x^2-7x+10 \n" ); document.write( "

Algebra.Com's Answer #215165 by jim_thompson5910(35256) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "Looking at the expression $\"x%5E2-7x%2B10\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"-7\"$ , and the last term is $\"10\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient $\"1\"$ by the last term $\"10\"$ to get $\"%281%29%2810%29=10\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now the question is: what two whole numbers multiply to $\"10\"$ (the previous product) and add to the second coefficient $\"-7\"$ ?\r
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\n" ); document.write( "\n" ); document.write( "To find these two numbers, we need to list all of the factors of $\"10\"$ (the previous product).\r
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\n" ); document.write( "\n" ); document.write( "Factors of $\"10\"$ :\r
\n" ); document.write( "\n" ); document.write( "1,2,5,10\r
\n" ); document.write( "\n" ); document.write( "-1,-2,-5,-10\r
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\n" ); document.write( "\n" ); document.write( "Note: list the negative of each factor. This will allow us to find all possible combinations.\r
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\n" ); document.write( "\n" ); document.write( "These factors pair up and multiply to $\"10\"$ .\r
\n" ); document.write( "\n" ); document.write( "1*10 = 10
\n" ); document.write( "2*5 = 10
\n" ); document.write( "(-1)*(-10) = 10
\n" ); document.write( "(-2)*(-5) = 10\r
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\n" ); document.write( "\n" ); document.write( "Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"-7\"$ :\r
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First Number	Second Number	Sum
1	10	1+10=11
2	5	2+5=7
-1	-10	-1+(-10)=-11
-2	-5	-2+(-5)=-7

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\n" ); document.write( "\n" ); document.write( "From the table, we can see that the two numbers $\"-2\"$ and $\"-5\"$ add to $\"-7\"$ (the middle coefficient).\r
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\n" ); document.write( "\n" ); document.write( "So the two numbers $\"-2\"$ and $\"-5\"$ both multiply to $\"10\"$ and add to $\"-7\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now replace the middle term $\"-7x\"$ with $\"-2x-5x\"$ . Remember, $\"-2\"$ and $\"-5\"$ add to $\"-7\"$ . So this shows us that $\"-2x-5x=-7x\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"x%5E2%2Bhighlight%28-2x-5x%29%2B10\"$ Replace the second term $\"-7x\"$ with $\"-2x-5x\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"%28x%5E2-2x%29%2B%28-5x%2B10%29\"$ Group the terms into two pairs.\r
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\n" ); document.write( "\n" ); document.write( " $\"x%28x-2%29%2B%28-5x%2B10%29\"$ Factor out the GCF $\"x\"$ from the first group.\r
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\n" ); document.write( "\n" ); document.write( " $\"x%28x-2%29-5%28x-2%29\"$ Factor out $\"5\"$ from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.\r
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\n" ); document.write( "\n" ); document.write( " $\"%28x-5%29%28x-2%29\"$ Combine like terms. Or factor out the common term $\"x-2\"$ \r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So $\"x%5E2-7x%2B10\"$ factors to $\"%28x-5%29%28x-2%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "In other words, $\"x%5E2-7x%2B10=%28x-5%29%28x-2%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Note: you can check the answer by expanding $\"%28x-5%29%28x-2%29\"$ to get $\"x%5E2-7x%2B10\"$ or by graphing the original expression and the answer (the two graphs should be identical).
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