document.write( "Question 790717: I need to factor the trinomial completely.
\n" ); document.write( "4a^3+12a^2+5a \n" ); document.write( "

Algebra.Com's Answer #479278 by jim_thompson5910(35256) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( " $\"4a%5E3%2B12a%5E2%2B5a\"$ Start with the given expression.\r
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\n" ); document.write( "\n" ); document.write( " $\"a%284a%5E2%2B12a%2B5%29\"$ Factor out the GCF $\"a\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now let's try to factor the inner expression $\"4a%5E2%2B12a%2B5\"$ \r
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\n" ); document.write( "\n" ); document.write( "Looking at the expression $\"4a%5E2%2B12a%2B5\"$ , we can see that the first coefficient is $\"4\"$ , the second coefficient is $\"12\"$ , and the last term is $\"5\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient $\"4\"$ by the last term $\"5\"$ to get $\"%284%29%285%29=20\"$ .\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 $\"12a\"$ with $\"2a%2B10a\"$ . Remember, $\"2\"$ and $\"10\"$ add to $\"12\"$ . So this shows us that $\"2a%2B10a=12a\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"4a%5E2%2Bhighlight%282a%2B10a%29%2B5\"$ Replace the second term $\"12a\"$ with $\"2a%2B10a\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"%284a%5E2%2B2a%29%2B%2810a%2B5%29\"$ Group the terms into two pairs.\r
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\n" ); document.write( "\n" ); document.write( " $\"2a%282a%2B1%29%2B%2810a%2B5%29\"$ Factor out the GCF $\"2a\"$ from the first group.\r
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\n" ); document.write( "\n" ); document.write( " $\"2a%282a%2B1%29%2B5%282a%2B1%29\"$ Factor out $\"5\"$ 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( " $\"%282a%2B5%29%282a%2B1%29\"$ Combine like terms. Or factor out the common term $\"2a%2B1\"$ \r
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\n" ); document.write( "\n" ); document.write( "So $\"a%284a%5E2%2B12a%2B5%29\"$ then factors further to $\"a%282a%2B5%29%282a%2B1%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So $\"4a%5E3%2B12a%5E2%2B5a\"$ completely factors to $\"a%282a%2B5%29%282a%2B1%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "In other words, $\"4a%5E3%2B12a%5E2%2B5a=a%282a%2B5%29%282a%2B1%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Note: you can check the answer by expanding $\"a%282a%2B5%29%282a%2B1%29\"$ to get $\"4a%5E3%2B12a%5E2%2B5a\"$ or by graphing the original expression and the answer (the two graphs should be identical).
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