document.write( "Question 278244: Solve by Factoring:
\n" ); document.write( "2r^2(Squared)+6r-80=0\r
\n" ); document.write( "\n" ); document.write( "I have 2(r^2+3r-40)=0....so far, is that right.?
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Algebra.Com's Answer #202485 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
Good job. Now let's factor $\"r%5E2%2B3r-40\"$ \r
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\n" ); document.write( "\n" ); document.write( "Looking at the expression $\"r%5E2%2B3r-40\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"3\"$ , and the last term is $\"-40\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient $\"1\"$ by the last term $\"-40\"$ to get $\"%281%29%28-40%29=-40\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now the question is: what two whole numbers multiply to $\"-40\"$ (the previous product) and add to the second coefficient $\"3\"$ ?\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 $\"-40\"$ (the previous product).\r
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\n" ); document.write( "\n" ); document.write( "Factors of $\"-40\"$ :\r
\n" ); document.write( "\n" ); document.write( "1,2,4,5,8,10,20,40\r
\n" ); document.write( "\n" ); document.write( "-1,-2,-4,-5,-8,-10,-20,-40\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 $\"-40\"$ .\r
\n" ); document.write( "\n" ); document.write( "1*(-40) = -40
\n" ); document.write( "2*(-20) = -40
\n" ); document.write( "4*(-10) = -40
\n" ); document.write( "5*(-8) = -40
\n" ); document.write( "(-1)*(40) = -40
\n" ); document.write( "(-2)*(20) = -40
\n" ); document.write( "(-4)*(10) = -40
\n" ); document.write( "(-5)*(8) = -40\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 $\"3\"$ :\r
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First Number	Second Number	Sum
1	-40	1+(-40)=-39
2	-20	2+(-20)=-18
4	-10	4+(-10)=-6
5	-8	5+(-8)=-3
-1	40	-1+40=39
-2	20	-2+20=18
-4	10	-4+10=6
-5	8	-5+8=3

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\n" ); document.write( "\n" ); document.write( "From the table, we can see that the two numbers $\"-5\"$ and $\"8\"$ add to $\"3\"$ (the middle coefficient).\r
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\n" ); document.write( "\n" ); document.write( "So the two numbers $\"-5\"$ and $\"8\"$ both multiply to $\"-40\"$ and add to $\"3\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now replace the middle term $\"3r\"$ with $\"-5r%2B8r\"$ . Remember, $\"-5\"$ and $\"8\"$ add to $\"3\"$ . So this shows us that $\"-5r%2B8r=3r\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"r%5E2%2Bhighlight%28-5r%2B8r%29-40\"$ Replace the second term $\"3r\"$ with $\"-5r%2B8r\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"%28r%5E2-5r%29%2B%288r-40%29\"$ Group the terms into two pairs.\r
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\n" ); document.write( "\n" ); document.write( " $\"r%28r-5%29%2B%288r-40%29\"$ Factor out the GCF $\"r\"$ from the first group.\r
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\n" ); document.write( "\n" ); document.write( " $\"r%28r-5%29%2B8%28r-5%29\"$ Factor out $\"8\"$ 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( " $\"%28r%2B8%29%28r-5%29\"$ Combine like terms. Or factor out the common term $\"r-5\"$ \r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So $\"2r%5E2%2B6r-80\"$ completely factors to $\"2%28r%2B8%29%28r-5%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "In other words, $\"2r%5E2%2B6r-80=2%28r%2B8%29%28r-5%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "I'll let you continue on to solve for 'r'. \n" ); document.write( "

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