document.write( "Question 200432: X^3 y+2x^2 y^2+xy^3= \n" ); document.write( "
Algebra.Com's Answer #150699 by jim_thompson5910(35256)\"\" \"About 
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\n" ); document.write( "\"x%5E3y%2B2x%5E2y%5E2%2Bxy%5E3\" Start with the given expression\r
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\n" ); document.write( "\n" ); document.write( "\"xy%28x%5E2%2B2xy%2By%5E2%29\" Factor out the GCF \"xy\"\r
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\n" ); document.write( "\n" ); document.write( "Now let's focus on the inner expression \"x%5E2%2B2xy%2By%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "Looking at \"x%5E2%2B2xy%2By%5E2\" we can see that the first term is \"x%5E2\" and the last term is \"y%5E2\" where the coefficients are 1 and 1 respectively.\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient 1 and the last coefficient 1 to get 1. Now what two numbers multiply to 1 and add to the middle coefficient 2? Let's list all of the factors of 1:\r
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\n" ); document.write( "\n" ); document.write( "Factors of 1:\r
\n" ); document.write( "\n" ); document.write( "1\r
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\n" ); document.write( "\n" ); document.write( "-1 ...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 1\r
\n" ); document.write( "\n" ); document.write( "1*1\r
\n" ); document.write( "\n" ); document.write( "(-1)*(-1)\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 2? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 2\r
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First NumberSecond NumberSum
111+1=2
-1-1-1+(-1)=-2
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\n" ); document.write( "\n" ); document.write( "From this list we can see that 1 and 1 add up to 2 and multiply to 1\r
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\n" ); document.write( "\n" ); document.write( "Now looking at the expression \"x%5E2%2B2xy%2By%5E2\", replace \"2xy\" with \"xy%2Bxy\" (notice \"xy%2Bxy\" adds up to \"2xy\". So it is equivalent to \"2xy\")\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2%2Bhighlight%28xy%2Bxy%29%2By%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "Now let's factor \"x%5E2%2Bxy%2Bxy%2By%5E2\" by grouping:\r
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\n" ); document.write( "\n" ); document.write( "\"%28x%5E2%2Bxy%29%2B%28xy%2By%5E2%29\" Group like terms\r
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\n" ); document.write( "\n" ); document.write( "\"x%28x%2By%29%2By%28x%2By%29\" Factor out the GCF of \"x\" out of the first group. Factor out the GCF of \"y\" out of the second group\r
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\n" ); document.write( "\n" ); document.write( "\"%28x%2By%29%28x%2By%29\" Since we have a common term of \"x%2By\", we can combine like terms\r
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\n" ); document.write( "\n" ); document.write( "So \"x%5E2%2Bxy%2Bxy%2By%5E2\" factors to \"%28x%2By%29%28x%2By%29\"\r
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\n" ); document.write( "\n" ); document.write( "So this also means that \"x%5E2%2B2xy%2By%5E2\" factors to \"%28x%2By%29%28x%2By%29\" (since \"x%5E2%2B2xy%2By%5E2\" is equivalent to \"x%5E2%2Bxy%2Bxy%2By%5E2\")\r
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\n" ); document.write( "\n" ); document.write( "note: \"%28x%2By%29%28x%2By%29\" is equivalent to \"%28x%2By%29%5E2\" since the term \"x%2By\" occurs twice. So \"x%5E2%2B2xy%2By%5E2\" also factors to \"%28x%2By%29%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "So our expression goes from \"xy%28x%5E2%2B2xy%2By%5E2%29\" and factors further to \"xy%28x%2By%29%5E2\"\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%5E3y%2B2x%5E2y%5E2%2Bxy%5E3\" completely factors to \"xy%28x%2By%29%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "In other words, \"x%5E3y%2B2x%5E2y%5E2%2Bxy%5E3=xy%28x%2By%29%5E2\"\r
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