document.write( "Question 618477: factorise it 3y^2+13y+14
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Algebra.Com's Answer #388930 by jim_thompson5910(35256) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "Looking at the expression $\"3y%5E2%2B13y%2B14\"$ , we can see that the first coefficient is $\"3\"$ , the second coefficient is $\"13\"$ , and the last term is $\"14\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient $\"3\"$ by the last term $\"14\"$ to get $\"%283%29%2814%29=42\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now the question is: what two whole numbers multiply to $\"42\"$ (the previous product) and add to the second coefficient $\"13\"$ ?\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 $\"42\"$ (the previous product).\r
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\n" ); document.write( "\n" ); document.write( "Factors of $\"42\"$ :\r
\n" ); document.write( "\n" ); document.write( "1,2,3,6,7,14,21,42\r
\n" ); document.write( "\n" ); document.write( "-1,-2,-3,-6,-7,-14,-21,-42\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 $\"42\"$ .\r
\n" ); document.write( "\n" ); document.write( "1*42 = 42
\n" ); document.write( "2*21 = 42
\n" ); document.write( "3*14 = 42
\n" ); document.write( "6*7 = 42
\n" ); document.write( "(-1)*(-42) = 42
\n" ); document.write( "(-2)*(-21) = 42
\n" ); document.write( "(-3)*(-14) = 42
\n" ); document.write( "(-6)*(-7) = 42\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 $\"13\"$ :\r
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First Number	Second Number	Sum
1	42	1+42=43
2	21	2+21=23
3	14	3+14=17
6	7	6+7=13
-1	-42	-1+(-42)=-43
-2	-21	-2+(-21)=-23
-3	-14	-3+(-14)=-17
-6	-7	-6+(-7)=-13

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\n" ); document.write( "\n" ); document.write( "From the table, we can see that the two numbers $\"6\"$ and $\"7\"$ add to $\"13\"$ (the middle coefficient).\r
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\n" ); document.write( "\n" ); document.write( "So the two numbers $\"6\"$ and $\"7\"$ both multiply to $\"42\"$ and add to $\"13\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now replace the middle term $\"13y\"$ with $\"6y%2B7y\"$ . Remember, $\"6\"$ and $\"7\"$ add to $\"13\"$ . So this shows us that $\"6y%2B7y=13y\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"3y%5E2%2Bhighlight%286y%2B7y%29%2B14\"$ Replace the second term $\"13y\"$ with $\"6y%2B7y\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"%283y%5E2%2B6y%29%2B%287y%2B14%29\"$ Group the terms into two pairs.\r
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\n" ); document.write( "\n" ); document.write( " $\"3y%28y%2B2%29%2B%287y%2B14%29\"$ Factor out the GCF $\"3y\"$ from the first group.\r
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\n" ); document.write( "\n" ); document.write( " $\"3y%28y%2B2%29%2B7%28y%2B2%29\"$ Factor out $\"7\"$ 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( " $\"%283y%2B7%29%28y%2B2%29\"$ Combine like terms. Or factor out the common term $\"y%2B2\"$ \r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So $\"3y%5E2%2B13y%2B14\"$ factors to $\"%283y%2B7%29%28y%2B2%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "In other words, $\"3y%5E2%2B13y%2B14=%283y%2B7%29%28y%2B2%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Note: you can check the answer by expanding $\"%283y%2B7%29%28y%2B2%29\"$ to get $\"3y%5E2%2B13y%2B14\"$ or by graphing the original expression and the answer (the two graphs should be identical).
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