document.write( "Question 151747: Factor 7x^2+49x+42 \n" ); document.write( "

Algebra.Com's Answer #111553 by jim_thompson5910(35256) $\"\"$ $\"About$

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\n" ); document.write( " $\"7x%5E2%2B49x%2B42\"$ Start with the given expression\r
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\n" ); document.write( "\n" ); document.write( " $\"7%28x%5E2%2B7x%2B6%29\"$ Factor out the GCF $\"7\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now let's focus on the inner expression $\"x%5E2%2B7x%2B6\"$ \r
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\n" ); document.write( "\n" ); document.write( "Looking at the expression $\"x%5E2%2B7x%2B6\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"7\"$ , and the last term is $\"6\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient $\"1\"$ by the last term $\"6\"$ to get $\"%281%29%286%29=6\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now the question is: what two whole numbers multiply to $\"6\"$ (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 $\"6\"$ (the previous product).\r
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\n" ); document.write( "\n" ); document.write( "Factors of $\"6\"$ :\r
\n" ); document.write( "\n" ); document.write( "1,2,3,6\r
\n" ); document.write( "\n" ); document.write( "-1,-2,-3,-6\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 $\"6\"$ .\r
\n" ); document.write( "\n" ); document.write( "1*6
\n" ); document.write( "2*3
\n" ); document.write( "(-1)*(-6)
\n" ); document.write( "(-2)*(-3)\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	6	1+6=7
2	3	2+3=5
-1	-6	-1+(-6)=-7
-2	-3	-2+(-3)=-5

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\n" ); document.write( "\n" ); document.write( "From the table, we can see that the two numbers $\"1\"$ and $\"6\"$ add to $\"7\"$ (the middle coefficient).\r
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\n" ); document.write( "\n" ); document.write( "So the two numbers $\"1\"$ and $\"6\"$ both multiply to $\"6\"$ and add to $\"7\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now replace the middle term $\"7x\"$ with $\"x%2B6x\"$ . Remember, $\"1\"$ and $\"6\"$ add to $\"7\"$ . So this shows us that $\"x%2B6x=7x\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"x%5E2%2Bhighlight%28x%2B6x%29%2B6\"$ Replace the second term $\"7x\"$ with $\"x%2B6x\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"%28x%5E2%2Bx%29%2B%286x%2B6%29\"$ Group the terms into two pairs.\r
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\n" ); document.write( "\n" ); document.write( " $\"x%28x%2B1%29%2B%286x%2B6%29\"$ Factor out the GCF $\"x\"$ from the first group.\r
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\n" ); document.write( "\n" ); document.write( " $\"x%28x%2B1%29%2B6%28x%2B1%29\"$ Factor out $\"6\"$ 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( " $\"%28x%2B6%29%28x%2B1%29\"$ Combine like terms. Or factor out the common term $\"x%2B1\"$ \r
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\n" ); document.write( "\n" ); document.write( "So our expression goes from $\"7%28x%5E2%2B7x%2B6%29\"$ and factors further to $\"7%28x%2B6%29%28x%2B1%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "So $\"7x%5E2%2B49x%2B42\"$ factors to $\"7%28x%2B6%29%28x%2B1%29\"$ \r
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