document.write( "Question 196046: Factor completely:
\n" ); document.write( "x^2 - 5x - 7x + 35 = x^2 - 12x + 35
\n" ); document.write( "2x^2 + 11x + 5
\n" ); document.write( "x^4y - 16y = x^4 - 15y
\n" ); document.write( "Thank you so much! \n" ); document.write( "

Algebra.Com's Answer #146971 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
I'm assuming that you want to factor $\"x%5E2+-+12x+%2B+35\"$ \r
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\n" ); document.write( "\n" ); document.write( "Looking at the expression $\"x%5E2-12x%2B35\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"-12\"$ , and the last term is $\"35\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient $\"1\"$ by the last term $\"35\"$ to get $\"%281%29%2835%29=35\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now the question is: what two whole numbers multiply to $\"35\"$ (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 $\"35\"$ (the previous product).\r
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\n" ); document.write( "\n" ); document.write( "Factors of $\"35\"$ :\r
\n" ); document.write( "\n" ); document.write( "1,5,7,35\r
\n" ); document.write( "\n" ); document.write( "-1,-5,-7,-35\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 $\"35\"$ .\r
\n" ); document.write( "\n" ); document.write( "1*35
\n" ); document.write( "5*7
\n" ); document.write( "(-1)*(-35)
\n" ); document.write( "(-5)*(-7)\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	35	1+35=36
5	7	5+7=12
-1	-35	-1+(-35)=-36
-5	-7	-5+(-7)=-12

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\n" ); document.write( "\n" ); document.write( "From the table, we can see that the two numbers $\"-5\"$ and $\"-7\"$ add to $\"-12\"$ (the middle coefficient).\r
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\n" ); document.write( "\n" ); document.write( "So the two numbers $\"-5\"$ and $\"-7\"$ both multiply to $\"35\"$ and add to $\"-12\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now replace the middle term $\"-12x\"$ with $\"-5x-7x\"$ . Remember, $\"-5\"$ and $\"-7\"$ add to $\"-12\"$ . So this shows us that $\"-5x-7x=-12x\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"x%5E2%2Bhighlight%28-5x-7x%29%2B35\"$ Replace the second term $\"-12x\"$ with $\"-5x-7x\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"%28x%5E2-5x%29%2B%28-7x%2B35%29\"$ Group the terms into two pairs.\r
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\n" ); document.write( "\n" ); document.write( " $\"x%28x-5%29%2B%28-7x%2B35%29\"$ Factor out the GCF $\"x\"$ from the first group.\r
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\n" ); document.write( "\n" ); document.write( " $\"x%28x-5%29-7%28x-5%29\"$ 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( " $\"%28x-7%29%28x-5%29\"$ Combine like terms. Or factor out the common term $\"x-5\"$ \r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So $\"x%5E2-12x%2B35\"$ factors to $\"%28x-7%29%28x-5%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Note: you can check the answer by FOILing $\"%28x-7%29%28x-5%29\"$ to get $\"x%5E2-12x%2B35\"$ or by graphing the original expression and the answer (the two graphs should be identical).\r
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\n" ); document.write( "\n" ); document.write( "# 2\r
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\n" ); document.write( "\n" ); document.write( "Looking at the expression $\"2x%5E2%2B11x%2B5\"$ , we can see that the first coefficient is $\"2\"$ , the second coefficient is $\"11\"$ , 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 $\"%282%29%285%29=10\"$ .\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 $\"11\"$ ?\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
\n" ); document.write( "\n" ); document.write( "1,2,5,10\r
\n" ); document.write( "\n" ); document.write( "-1,-2,-5,-10\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Note: list the negative of each factor. This will allow us to find all possible combinations.\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "These factors pair up and multiply to $\"10\"$ .\r
\n" ); document.write( "\n" ); document.write( "1*10
\n" ); document.write( "2*5
\n" ); document.write( "(-1)*(-10)
\n" ); document.write( "(-2)*(-5)\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 $\"11\"$ :\r
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First Number	Second Number	Sum
1	10	1+10=11
2	5	2+5=7
-1	-10	-1+(-10)=-11
-2	-5	-2+(-5)=-7

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\n" ); document.write( "\n" ); document.write( "From the table, we can see that the two numbers $\"1\"$ and $\"10\"$ add to $\"11\"$ (the middle coefficient).\r
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\n" ); document.write( "\n" ); document.write( "So the two numbers $\"1\"$ and $\"10\"$ both multiply to $\"10\"$ and add to $\"11\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now replace the middle term $\"11x\"$ with $\"x%2B10x\"$ . Remember, $\"1\"$ and $\"10\"$ add to $\"11\"$ . So this shows us that $\"x%2B10x=11x\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"2x%5E2%2Bhighlight%28x%2B10x%29%2B5\"$ Replace the second term $\"11x\"$ with $\"x%2B10x\"$ .\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " $\"%282x%5E2%2Bx%29%2B%2810x%2B5%29\"$ Group the terms into two pairs.\r
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\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " $\"x%282x%2B1%29%2B%2810x%2B5%29\"$ Factor out the GCF $\"x\"$ from the first group.\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " $\"x%282x%2B1%29%2B5%282x%2B1%29\"$ 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( " $\"%28x%2B5%29%282x%2B1%29\"$ Combine like terms. Or factor out the common term $\"2x%2B1\"$ \r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So $\"2x%5E2%2B11x%2B5\"$ factors to $\"%28x%2B5%29%282x%2B1%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Note: you can check the answer by FOILing $\"%28x%2B5%29%282x%2B1%29\"$ to get $\"2x%5E2%2B11x%2B5\"$ or by graphing the original expression and the answer (the two graphs should be identical).\r
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\n" ); document.write( "\n" ); document.write( "I'm assuming you want to factor $\"x%5E4y+-+16y\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"x%5E4y+-+16y\"$ Start with the given expression\r
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\n" ); document.write( "\n" ); document.write( " $\"y%28x%5E4+-+16%29\"$ Factor out the GCF\r
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\n" ); document.write( "\n" ); document.write( " $\"y%28x%5E2+-+4%29%28x%5E2+%2B4%29\"$ Factor $\"x%5E4+-+16\"$ using the difference of squares formula\r
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\n" ); document.write( "\n" ); document.write( " $\"y%28x-2%29%28x%2B2%29%28x%5E2+%2B4%29\"$ Factor $\"x%5E2+-+4\"$ using the difference of squares formula\r
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\n" ); document.write( "\n" ); document.write( "So $\"x%5E4y+-+16y\"$ completely factors to $\"y%28x-2%29%28x%2B2%29%28x%5E2+%2B4%29\"$
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