document.write( "Question 276797: Factor.
\n" ); document.write( " 2y^2+17y+30 \n" ); document.write( "

Algebra.Com's Answer #201671 by jim_thompson5910(35256) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "Looking at the expression $\"2y%5E2%2B17y%2B30\"$ , we can see that the first coefficient is $\"2\"$ , the second coefficient is $\"17\"$ , and the last term is $\"30\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient $\"2\"$ by the last term $\"30\"$ to get $\"%282%29%2830%29=60\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now the question is: what two whole numbers multiply to $\"60\"$ (the previous product) and add to the second coefficient $\"17\"$ ?\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 $\"60\"$ (the previous product).\r
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\n" ); document.write( "\n" ); document.write( "Factors of $\"60\"$ :\r
\n" ); document.write( "\n" ); document.write( "1,2,3,4,5,6,10,12,15,20,30,60\r
\n" ); document.write( "\n" ); document.write( "-1,-2,-3,-4,-5,-6,-10,-12,-15,-20,-30,-60\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 $\"60\"$ .\r
\n" ); document.write( "\n" ); document.write( "1*60 = 60
\n" ); document.write( "2*30 = 60
\n" ); document.write( "3*20 = 60
\n" ); document.write( "4*15 = 60
\n" ); document.write( "5*12 = 60
\n" ); document.write( "6*10 = 60
\n" ); document.write( "(-1)*(-60) = 60
\n" ); document.write( "(-2)*(-30) = 60
\n" ); document.write( "(-3)*(-20) = 60
\n" ); document.write( "(-4)*(-15) = 60
\n" ); document.write( "(-5)*(-12) = 60
\n" ); document.write( "(-6)*(-10) = 60\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 $\"17\"$ :\r
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First Number	Second Number	Sum
1	60	1+60=61
2	30	2+30=32
3	20	3+20=23
4	15	4+15=19
5	12	5+12=17
6	10	6+10=16
-1	-60	-1+(-60)=-61
-2	-30	-2+(-30)=-32
-3	-20	-3+(-20)=-23
-4	-15	-4+(-15)=-19
-5	-12	-5+(-12)=-17
-6	-10	-6+(-10)=-16

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\n" ); document.write( "\n" ); document.write( "From the table, we can see that the two numbers $\"5\"$ and $\"12\"$ add to $\"17\"$ (the middle coefficient).\r
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\n" ); document.write( "\n" ); document.write( "So the two numbers $\"5\"$ and $\"12\"$ both multiply to $\"60\"$ and add to $\"17\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now replace the middle term $\"17y\"$ with $\"5y%2B12y\"$ . Remember, $\"5\"$ and $\"12\"$ add to $\"17\"$ . So this shows us that $\"5y%2B12y=17y\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"2y%5E2%2Bhighlight%285y%2B12y%29%2B30\"$ Replace the second term $\"17y\"$ with $\"5y%2B12y\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"%282y%5E2%2B5y%29%2B%2812y%2B30%29\"$ Group the terms into two pairs.\r
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\n" ); document.write( "\n" ); document.write( " $\"y%282y%2B5%29%2B%2812y%2B30%29\"$ Factor out the GCF $\"y\"$ from the first group.\r
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\n" ); document.write( "\n" ); document.write( " $\"y%282y%2B5%29%2B6%282y%2B5%29\"$ Factor out $\"6\"$ 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( " $\"%28y%2B6%29%282y%2B5%29\"$ Combine like terms. Or factor out the common term $\"2y%2B5\"$ \r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So $\"2y%5E2%2B17y%2B30\"$ factors to $\"%28y%2B6%29%282y%2B5%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "In other words, $\"2y%5E2%2B17y%2B30=%28y%2B6%29%282y%2B5%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Note: you can check the answer by expanding $\"%28y%2B6%29%282y%2B5%29\"$ to get $\"2y%5E2%2B17y%2B30\"$ or by graphing the original expression and the answer (the two graphs should be identical).
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