document.write( "Question 191474: Factor the trinomial completely\r
\n" ); document.write( "\n" ); document.write( "x^2+18xy+81y^2 \n" ); document.write( "

Algebra.Com's Answer #143680 by jim_thompson5910(35256) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "Looking at $\"x%5E2%2B18xy%2B81y%5E2\"$ we can see that the first term is $\"x%5E2\"$ and the last term is $\"81y%5E2\"$ where the coefficients are 1 and 81 respectively.\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient 1 and the last coefficient 81 to get 81. Now what two numbers multiply to 81 and add to the middle coefficient 18? Let's list all of the factors of 81:\r
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\n" ); document.write( "\n" ); document.write( "Factors of 81:\r
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\n" ); document.write( "\n" ); document.write( "-1,-3,-9,-27 ...List the negative factors as well. 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 81\r
\n" ); document.write( "\n" ); document.write( "1*81\r
\n" ); document.write( "\n" ); document.write( "3*27\r
\n" ); document.write( "\n" ); document.write( "9*9\r
\n" ); document.write( "\n" ); document.write( "(-1)*(-81)\r
\n" ); document.write( "\n" ); document.write( "(-3)*(-27)\r
\n" ); document.write( "\n" ); document.write( "(-9)*(-9)\r
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\n" ); document.write( "\n" ); document.write( "note: remember two negative numbers multiplied together make a positive number\r
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\n" ); document.write( "\n" ); document.write( "Now which of these pairs add to 18? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 18\r
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First Number	Second Number	Sum
1	81	1+81=82
3	27	3+27=30
9	9	9+9=18
-1	-81	-1+(-81)=-82
-3	-27	-3+(-27)=-30
-9	-9	-9+(-9)=-18

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\n" ); document.write( "\n" ); document.write( "From this list we can see that 9 and 9 add up to 18 and multiply to 81\r
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\n" ); document.write( "\n" ); document.write( "Now looking at the expression $\"x%5E2%2B18xy%2B81y%5E2\"$ , replace $\"18xy\"$ with $\"9xy%2B9xy\"$ (notice $\"9xy%2B9xy\"$ adds up to $\"18xy\"$ . So it is equivalent to $\"18xy\"$ )\r
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\n" ); document.write( "\n" ); document.write( " $\"x%5E2%2Bhighlight%289xy%2B9xy%29%2B81y%5E2\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now let's factor $\"x%5E2%2B9xy%2B9xy%2B81y%5E2\"$ by grouping:\r
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\n" ); document.write( "\n" ); document.write( " $\"%28x%5E2%2B9xy%29%2B%289xy%2B81y%5E2%29\"$ Group like terms\r
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\n" ); document.write( "\n" ); document.write( " $\"x%28x%2B9y%29%2B9y%28x%2B9y%29\"$ Factor out the GCF of $\"x\"$ out of the first group. Factor out the GCF of $\"9y\"$ out of the second group\r
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\n" ); document.write( "\n" ); document.write( " $\"%28x%2B9y%29%28x%2B9y%29\"$ Since we have a common term of $\"x%2B9y\"$ , we can combine like terms\r
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\n" ); document.write( "\n" ); document.write( "So $\"x%5E2%2B9xy%2B9xy%2B81y%5E2\"$ factors to $\"%28x%2B9y%29%28x%2B9y%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "So this also means that $\"x%5E2%2B18xy%2B81y%5E2\"$ factors to $\"%28x%2B9y%29%28x%2B9y%29\"$ (since $\"x%5E2%2B18xy%2B81y%5E2\"$ is equivalent to $\"x%5E2%2B9xy%2B9xy%2B81y%5E2\"$ )\r
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\n" ); document.write( "\n" ); document.write( "note: $\"%28x%2B9y%29%28x%2B9y%29\"$ is equivalent to $\"%28x%2B9y%29%5E2\"$ since the term $\"x%2B9y\"$ occurs twice. So $\"x%5E2%2B18xy%2B81y%5E2\"$ also factors to $\"%28x%2B9y%29%5E2\"$ \r
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\n" ); document.write( "\n" ); document.write( " Answer:\r
\n" ); document.write( "\n" ); document.write( "So $\"x%5E2%2B18xy%2B81y%5E2\"$ factors to $\"%28x%2B9y%29%5E2\"$
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