document.write( "Question 716968: x^2+2x+1 \n" ); document.write( "

Algebra.Com's Answer #440035 by jim_thompson5910(35256) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "Looking at the expression $\"x%5E2%2B2x%2B1\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"2\"$ , and the last term is $\"1\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient $\"1\"$ by the last term $\"1\"$ to get $\"%281%29%281%29=1\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now the question is: what two whole numbers multiply to $\"1\"$ (the previous product) and add to the second coefficient $\"2\"$ ?\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 $\"1\"$ (the previous product).\r
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\n" ); document.write( "\n" ); document.write( "Factors of $\"1\"$ :\r
\n" ); document.write( "\n" ); document.write( "1\r
\n" ); document.write( "\n" ); document.write( "-1\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 $\"1\"$ .\r
\n" ); document.write( "\n" ); document.write( "1*1 = 1
\n" ); document.write( "(-1)*(-1) = 1\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 $\"2\"$ :\r
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First Number	Second Number	Sum
1	1	1+1=2
-1	-1	-1+(-1)=-2

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\n" ); document.write( "\n" ); document.write( "From the table, we can see that the two numbers $\"1\"$ and $\"1\"$ add to $\"2\"$ (the middle coefficient).\r
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\n" ); document.write( "\n" ); document.write( "So the two numbers $\"1\"$ and $\"1\"$ both multiply to $\"1\"$ and add to $\"2\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now replace the middle term $\"2x\"$ with $\"x%2Bx\"$ . Remember, $\"1\"$ and $\"1\"$ add to $\"2\"$ . So this shows us that $\"x%2Bx=2x\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"x%5E2%2Bhighlight%28x%2Bx%29%2B1\"$ Replace the second term $\"2x\"$ with $\"x%2Bx\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"%28x%5E2%2Bx%29%2B%28x%2B1%29\"$ Group the terms into two pairs.\r
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\n" ); document.write( "\n" ); document.write( " $\"x%28x%2B1%29%2B%28x%2B1%29\"$ Factor out the GCF $\"x\"$ from the first group.\r
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\n" ); document.write( "\n" ); document.write( " $\"x%28x%2B1%29%2B1%28x%2B1%29\"$ Factor out $\"1\"$ 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( " $\"%28x%2B1%29%28x%2B1%29\"$ Combine like terms. Or factor out the common term $\"x%2B1\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"%28x%2B1%29%5E2\"$ Condense the terms.\r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So $\"x%5E2%2B2x%2B1\"$ factors to $\"%28x%2B1%29%5E2\"$ .\r
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\n" ); document.write( "\n" ); document.write( "In other words, $\"x%5E2%2B2x%2B1=%28x%2B1%29%5E2\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Note: you can check the answer by expanding $\"%28x%2B1%29%5E2\"$ to get $\"x%5E2%2B2x%2B1\"$ or by graphing the original expression and the answer (the two graphs should be identical).
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