document.write( "Question 195729: factor the trinomial\r
\n" ); document.write( "\n" ); document.write( "a^2+10a+21 \n" ); document.write( "

Algebra.Com's Answer #146772 by jim_thompson5910(35256) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "Looking at the expression $\"a%5E2%2B10a%2B21\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"10\"$ , and the last term is $\"21\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient $\"1\"$ by the last term $\"21\"$ to get $\"%281%29%2821%29=21\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now the question is: what two whole numbers multiply to $\"21\"$ (the previous product) and add to the second coefficient $\"10\"$ ?\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 $\"21\"$ (the previous product).\r
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\n" ); document.write( "\n" ); document.write( "Factors of $\"21\"$ :\r
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\n" ); document.write( "\n" ); document.write( "-1,-3,-7,-21\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 $\"21\"$ .\r
\n" ); document.write( "\n" ); document.write( "1*21
\n" ); document.write( "3*7
\n" ); document.write( "(-1)*(-21)
\n" ); document.write( "(-3)*(-7)\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 $\"10\"$ :\r
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First Number	Second Number	Sum
1	21	1+21=22
3	7	3+7=10
-1	-21	-1+(-21)=-22
-3	-7	-3+(-7)=-10

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\n" ); document.write( "\n" ); document.write( "From the table, we can see that the two numbers $\"3\"$ and $\"7\"$ add to $\"10\"$ (the middle coefficient).\r
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\n" ); document.write( "\n" ); document.write( "So the two numbers $\"3\"$ and $\"7\"$ both multiply to $\"21\"$ and add to $\"10\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now replace the middle term $\"10a\"$ with $\"3a%2B7a\"$ . Remember, $\"3\"$ and $\"7\"$ add to $\"10\"$ . So this shows us that $\"3a%2B7a=10a\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"a%5E2%2Bhighlight%283a%2B7a%29%2B21\"$ Replace the second term $\"10a\"$ with $\"3a%2B7a\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"%28a%5E2%2B3a%29%2B%287a%2B21%29\"$ Group the terms into two pairs.\r
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\n" ); document.write( "\n" ); document.write( " $\"a%28a%2B3%29%2B%287a%2B21%29\"$ Factor out the GCF $\"a\"$ from the first group.\r
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\n" ); document.write( "\n" ); document.write( " $\"a%28a%2B3%29%2B7%28a%2B3%29\"$ Factor out $\"7\"$ 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( " $\"%28a%2B7%29%28a%2B3%29\"$ Combine like terms. Or factor out the common term $\"a%2B3\"$ \r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So $\"a%5E2%2B10a%2B21\"$ factors to $\"%28a%2B7%29%28a%2B3%29\"$ .\r
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