document.write( "Question 214983: I need help with sloving quadractic equations by factoring:
\n" ); document.write( " $\"2x%5E2-3x-5=0\"$ \n" ); document.write( "

Algebra.Com's Answer #162447 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
First, let's factor $\"2x%5E2-3x-5\"$ \r
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\n" ); document.write( "\n" ); document.write( "Looking at the expression $\"2x%5E2-3x-5\"$ , we can see that the first coefficient is $\"2\"$ , the second coefficient is $\"-3\"$ , and the last term is $\"-5\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient $\"2\"$ by the last term $\"-5\"$ to get $\"%282%29%28-5%29=-10\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now the question is: what two whole numbers multiply to $\"-10\"$ (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 $\"-10\"$ (the previous product).\r
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\n" ); document.write( "\n" ); document.write( "Factors of $\"-10\"$ :\r
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\n" ); document.write( "\n" ); document.write( "-1,-2,-5,-10\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 $\"-10\"$ .\r
\n" ); document.write( "\n" ); document.write( "1*(-10) = -10
\n" ); document.write( "2*(-5) = -10
\n" ); document.write( "(-1)*(10) = -10
\n" ); document.write( "(-2)*(5) = -10\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	-10	1+(-10)=-9
2	-5	2+(-5)=-3
-1	10	-1+10=9
-2	5	-2+5=3

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\n" ); document.write( "\n" ); document.write( "From the table, we can see that the two numbers $\"2\"$ and $\"-5\"$ add to $\"-3\"$ (the middle coefficient).\r
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\n" ); document.write( "\n" ); document.write( "So the two numbers $\"2\"$ and $\"-5\"$ both multiply to $\"-10\"$ and add to $\"-3\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now replace the middle term $\"-3x\"$ with $\"2x-5x\"$ . Remember, $\"2\"$ and $\"-5\"$ add to $\"-3\"$ . So this shows us that $\"2x-5x=-3x\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"2x%5E2%2Bhighlight%282x-5x%29-5\"$ Replace the second term $\"-3x\"$ with $\"2x-5x\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"%282x%5E2%2B2x%29%2B%28-5x-5%29\"$ Group the terms into two pairs.\r
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\n" ); document.write( "\n" ); document.write( " $\"2x%28x%2B1%29%2B%28-5x-5%29\"$ Factor out the GCF $\"2x\"$ from the first group.\r
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\n" ); document.write( "\n" ); document.write( " $\"2x%28x%2B1%29-5%28x%2B1%29\"$ 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( " $\"%282x-5%29%28x%2B1%29\"$ Combine like terms. Or factor out the common term $\"x%2B1\"$ \r
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\n" ); document.write( "\n" ); document.write( "So $\"2x%5E2-3x-5\"$ factors to $\"%282x-5%29%28x%2B1%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "In other words, $\"2x%5E2-3x-5=%282x-5%29%28x%2B1%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "-------------------------------------------------------\r
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\n" ); document.write( "\n" ); document.write( "Now let's use this factorization to solve $\"2x%5E2-3x-5=0\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"2x%5E2-3x-5=0\"$ Start with the given equation\r
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\n" ); document.write( "\n" ); document.write( " $\"%282x-5%29%28x%2B1%29=0\"$ Factor the left side (using the factorization from above)\r
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\n" ); document.write( "\n" ); document.write( "Now set each factor equal to zero:\r
\n" ); document.write( "\n" ); document.write( " $\"2x-5=0\"$ or $\"x%2B1=0\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"x=5%2F2\"$ or $\"x=-1\"$ Now solve for x in each case\r
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\n" ); document.write( "\n" ); document.write( "So the solutions are $\"x=5%2F2\"$ or $\"x=-1\"$ \n" ); document.write( "

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