document.write( "Question 880974: \"Completely factor each polynomial. If the polynomial cannot be factored, state so.\"\r
\n" ); document.write( "\n" ); document.write( "25y^3-10y^2+y\r
\n" ); document.write( "\n" ); document.write( "y(25y^2-10y+1)\r
\n" ); document.write( "\n" ); document.write( "I know this is a perfect square polynomial, but I'm not sure what to do after this. Thanks!!! \n" ); document.write( "

Algebra.Com's Answer #531828 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let's factor $\"25y%5E2-10y%2B1\"$ \r
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\n" ); document.write( "\n" ); document.write( "Looking at the expression $\"25y%5E2-10y%2B1\"$ , we can see that the first coefficient is $\"25\"$ , the second coefficient is $\"-10\"$ , and the last term is $\"1\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient $\"25\"$ by the last term $\"1\"$ to get $\"%2825%29%281%29=25\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Now the question is: what two whole numbers multiply to $\"25\"$ (the previous product) and add to the second coefficient $\"-10\"$ ?\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 $\"25\"$ (the previous product).\r
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\n" ); document.write( "\n" ); document.write( "Factors of $\"25\"$ :\r
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\n" ); document.write( "\n" ); document.write( "-1,-5,-25\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 $\"25\"$ .\r
\n" ); document.write( "\n" ); document.write( "1*25 = 25
\n" ); document.write( "5*5 = 25
\n" ); document.write( "(-1)*(-25) = 25
\n" ); document.write( "(-5)*(-5) = 25\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 $\"-10\"$ :\r
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First Number	Second Number	Sum
1	25	1+25=26
5	5	5+5=10
-1	-25	-1+(-25)=-26
-5	-5	-5+(-5)=-10

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\n" ); document.write( "\n" ); document.write( "From the table, we can see that the two numbers $\"-5\"$ and $\"-5\"$ add to $\"-10\"$ (the middle coefficient).\r
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\n" ); document.write( "\n" ); document.write( "So the two numbers $\"-5\"$ and $\"-5\"$ both multiply to $\"25\"$ and add to $\"-10\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now replace the middle term $\"-10y\"$ with $\"-5y-5y\"$ . Remember, $\"-5\"$ and $\"-5\"$ add to $\"-10\"$ . So this shows us that $\"-5y-5y=-10y\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"25y%5E2%2Bhighlight%28-5y-5y%29%2B1\"$ Replace the second term $\"-10y\"$ with $\"-5y-5y\"$ .\r
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\n" ); document.write( "\n" ); document.write( " $\"%2825y%5E2-5y%29%2B%28-5y%2B1%29\"$ Group the terms into two pairs.\r
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\n" ); document.write( "\n" ); document.write( " $\"5y%285y-1%29%2B%28-5y%2B1%29\"$ Factor out the GCF $\"5y\"$ from the first group.\r
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\n" ); document.write( "\n" ); document.write( " $\"5y%285y-1%29-1%285y-1%29\"$ Factor out $\"1\"$ 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( " $\"%285y-1%29%285y-1%29\"$ Combine like terms. Or factor out the common term $\"5y-1\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"%285y-1%29%5E2\"$ Condense the terms.\r
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\n" ); document.write( "\n" ); document.write( "So $\"25y%5E2-10y%2B1\"$ factors to $\"%285y-1%29%5E2\"$ .\r
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\n" ); document.write( "\n" ); document.write( "In other words, $\"25y%5E2-10y%2B1=%285y-1%29%5E2\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Note: you can check the answer by expanding $\"%285y-1%29%5E2\"$ to get $\"25y%5E2-10y%2B1\"$ 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( "So $\"y%2825y%5E2-10y%2B1%29+\"$ turns into $\"y%285y-1%29%5E2+\"$ \r
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\n" ); document.write( "\n" ); document.write( "Final Answer: $\"25y%5E3-10y%5E2%2By+\"$ completely factors to $\"y%285y-1%29%5E2\"$ \n" ); document.write( "

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