document.write( "Question 191472: Factor the trinomial completely\r
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Algebra.Com's Answer #143681 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "Looking at \"7a%5E2%2B50ab%2B7b%5E2\" we can see that the first term is \"7a%5E2\" and the last term is \"7b%5E2\" where the coefficients are 7 and 7 respectively.\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient 7 and the last coefficient 7 to get 49. Now what two numbers multiply to 49 and add to the middle coefficient 50? Let's list all of the factors of 49:\r
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\n" ); document.write( "\n" ); document.write( "Factors of 49:\r
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\n" ); document.write( "\n" ); document.write( "-1,-7 ...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 49\r
\n" ); document.write( "\n" ); document.write( "1*49\r
\n" ); document.write( "\n" ); document.write( "7*7\r
\n" ); document.write( "\n" ); document.write( "(-1)*(-49)\r
\n" ); document.write( "\n" ); document.write( "(-7)*(-7)\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 50? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 50\r
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First NumberSecond NumberSum
1491+49=50
777+7=14
-1-49-1+(-49)=-50
-7-7-7+(-7)=-14
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\n" ); document.write( "\n" ); document.write( "From this list we can see that 1 and 49 add up to 50 and multiply to 49\r
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\n" ); document.write( "\n" ); document.write( "Now looking at the expression \"7a%5E2%2B50ab%2B7b%5E2\", replace \"50ab\" with \"ab%2B49ab\" (notice \"ab%2B49ab\" adds up to \"50ab\". So it is equivalent to \"50ab\")\r
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\n" ); document.write( "\n" ); document.write( "\"7a%5E2%2Bhighlight%28ab%2B49ab%29%2B7b%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "Now let's factor \"7a%5E2%2Bab%2B49ab%2B7b%5E2\" by grouping:\r
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\n" ); document.write( "\n" ); document.write( "\"%287a%5E2%2Bab%29%2B%2849ab%2B7b%5E2%29\" Group like terms\r
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\n" ); document.write( "\n" ); document.write( "\"a%287a%2Bb%29%2B7b%287a%2Bb%29\" Factor out the GCF of \"a\" out of the first group. Factor out the GCF of \"7b\" out of the second group\r
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\n" ); document.write( "\n" ); document.write( "\"%28a%2B7b%29%287a%2Bb%29\" Since we have a common term of \"7a%2Bb\", we can combine like terms\r
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\n" ); document.write( "\n" ); document.write( "So \"7a%5E2%2B1ab%2B49ab%2B7b%5E2\" factors to \"%28a%2B7b%29%287a%2Bb%29\"\r
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\n" ); document.write( "\n" ); document.write( "So this also means that \"7a%5E2%2B50ab%2B7b%5E2\" factors to \"%28a%2B7b%29%287a%2Bb%29\" (since \"7a%5E2%2B50ab%2B7b%5E2\" is equivalent to \"7a%5E2%2B1ab%2B49ab%2B7b%5E2\")\r
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\n" ); document.write( "\n" ); document.write( " Answer:\r
\n" ); document.write( "\n" ); document.write( "So \"7a%5E2%2B50ab%2B7b%5E2\" factors to \"%28a%2B7b%29%287a%2Bb%29\"
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