document.write( "Question 597922: 3x^2-36x+60. Can you help me solve this as a grouping polynomial.
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Algebra.Com's Answer #378399 by jim_thompson5910(35256) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( " $\"3x%5E2-36x%2B60\"$ Start with the given expression.\r
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\n" ); document.write( "\n" ); document.write( " $\"3%28x%5E2-12x%2B20%29\"$ Factor out the GCF $\"3\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now let's try to factor the inner expression $\"x%5E2-12x%2B20\"$ \r
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\n" ); document.write( "\n" ); document.write( "Looking at the expression $\"x%5E2-12x%2B20\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"-12\"$ , and the last term is $\"20\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient $\"1\"$ by the last term $\"20\"$ to get $\"%281%29%2820%29=20\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now the question is: what two whole numbers multiply to $\"20\"$ (the previous product) and add to the second coefficient $\"-12\"$ ?\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 $\"20\"$ (the previous product).\r
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\n" ); document.write( "\n" ); document.write( "Factors of $\"20\"$ :\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 $\"20\"$ .\r
\n" ); document.write( "\n" ); document.write( "1*20 = 20
\n" ); document.write( "2*10 = 20
\n" ); document.write( "4*5 = 20
\n" ); document.write( "(-1)*(-20) = 20
\n" ); document.write( "(-2)*(-10) = 20
\n" ); document.write( "(-4)*(-5) = 20\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 $\"-12\"$ :\r
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First Number	Second Number	Sum
1	20	1+20=21
2	10	2+10=12
4	5	4+5=9
-1	-20	-1+(-20)=-21
-2	-10	-2+(-10)=-12
-4	-5	-4+(-5)=-9

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\n" ); document.write( "\n" ); document.write( "From the table, we can see that the two numbers $\"-2\"$ and $\"-10\"$ add to $\"-12\"$ (the middle coefficient).\r
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\n" ); document.write( "\n" ); document.write( "So the two numbers $\"-2\"$ and $\"-10\"$ both multiply to $\"20\"$ and add to $\"-12\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now replace the middle term $\"-12x\"$ with $\"-2x-10x\"$ . Remember, $\"-2\"$ and $\"-10\"$ add to $\"-12\"$ . So this shows us that $\"-2x-10x=-12x\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"x%5E2%2Bhighlight%28-2x-10x%29%2B20\"$ Replace the second term $\"-12x\"$ with $\"-2x-10x\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"%28x%5E2-2x%29%2B%28-10x%2B20%29\"$ Group the terms into two pairs.\r
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\n" ); document.write( "\n" ); document.write( " $\"x%28x-2%29%2B%28-10x%2B20%29\"$ Factor out the GCF $\"x\"$ from the first group.\r
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\n" ); document.write( "\n" ); document.write( " $\"x%28x-2%29-10%28x-2%29\"$ Factor out $\"10\"$ 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( " $\"%28x-10%29%28x-2%29\"$ Combine like terms. Or factor out the common term $\"x-2\"$ \r
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\n" ); document.write( "\n" ); document.write( "So $\"3%28x%5E2-12x%2B20%29\"$ then factors further to $\"3%28x-10%29%28x-2%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So $\"3x%5E2-36x%2B60\"$ completely factors to $\"3%28x-10%29%28x-2%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "In other words, $\"3x%5E2-36x%2B60=3%28x-10%29%28x-2%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Note: you can check the answer by expanding $\"3%28x-10%29%28x-2%29\"$ to get $\"3x%5E2-36x%2B60\"$ or by graphing the original expression and the answer (the two graphs should be identical).
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