document.write( "Question 337623: 16y^2-8y+1 \n" ); document.write( "

Algebra.Com's Answer #242134 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
I'm assuming you want to factor this\r
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\n" ); document.write( "\n" ); document.write( "Looking at $\"16y%5E2-8y%2B1\"$ we can see that the first term is $\"16y%5E2\"$ and the last term is $\"1\"$ where the coefficients are 16 and 1 respectively.\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient 16 and the last coefficient 1 to get 16. Now what two numbers multiply to 16 and add to the middle coefficient -8? Let's list all of the factors of 16:\r
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\n" ); document.write( "\n" ); document.write( "Factors of 16:\r
\n" ); document.write( "\n" ); document.write( "1,2,4,8\r
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\n" ); document.write( "\n" ); document.write( "-1,-2,-4,-8 ...List the negative factors as well. 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 16\r
\n" ); document.write( "\n" ); document.write( "1*16\r
\n" ); document.write( "\n" ); document.write( "2*8\r
\n" ); document.write( "\n" ); document.write( "4*4\r
\n" ); document.write( "\n" ); document.write( "(-1)*(-16)\r
\n" ); document.write( "\n" ); document.write( "(-2)*(-8)\r
\n" ); document.write( "\n" ); document.write( "(-4)*(-4)\r
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\n" ); document.write( "\n" ); document.write( "note: remember two negative numbers multiplied together make a positive number\r
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\n" ); document.write( "\n" ); document.write( "Now which of these pairs add to -8? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -8\r
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First Number	Second Number	Sum
1	16	1+16=17
2	8	2+8=10
4	4	4+4=8
-1	-16	-1+(-16)=-17
-2	-8	-2+(-8)=-10
-4	-4	-4+(-4)=-8

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\n" ); document.write( "\n" ); document.write( "From this list we can see that -4 and -4 add up to -8 and multiply to 16\r
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\n" ); document.write( "\n" ); document.write( "Now looking at the expression $\"16y%5E2-8y%2B1\"$ , replace $\"-8y\"$ with $\"-4y%2B-4y\"$ (notice $\"-4y%2B-4y\"$ adds up to $\"-8y\"$ . So it is equivalent to $\"-8y\"$ )\r
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\n" ); document.write( "\n" ); document.write( " $\"16y%5E2%2Bhighlight%28-4y%2B-4y%29%2B1\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now let's factor $\"16y%5E2-4y-4y%2B1\"$ by grouping:\r
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\n" ); document.write( "\n" ); document.write( " $\"%2816y%5E2-4y%29%2B%28-4y%2B1%29\"$ Group like terms\r
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\n" ); document.write( "\n" ); document.write( " $\"4y%284y-1%29-1%284y-1%29\"$ Factor out the GCF of $\"4y\"$ out of the first group. Factor out the GCF of $\"-1\"$ out of the second group\r
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\n" ); document.write( "\n" ); document.write( " $\"%284y-1%29%284y-1%29\"$ Since we have a common term of $\"4y-1\"$ , we can combine like terms\r
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\n" ); document.write( "\n" ); document.write( "So $\"16y%5E2-4y-4y%2B1\"$ factors to $\"%284y-1%29%284y-1%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "So this also means that $\"16y%5E2-8y%2B1\"$ factors to $\"%284y-1%29%284y-1%29\"$ (since $\"16y%5E2-8y%2B1\"$ is equivalent to $\"16y%5E2-4y-4y%2B1\"$ )\r
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\n" ); document.write( "\n" ); document.write( "note: $\"%284y-1%29%284y-1%29\"$ is equivalent to $\"%284y-1%29%5E2\"$ since the term $\"4y-1\"$ occurs twice. So $\"16y%5E2-8y%2B1\"$ also factors to $\"%284y-1%29%5E2\"$ \r
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\n" ); document.write( "\n" ); document.write( " Answer:\r
\n" ); document.write( "\n" ); document.write( "So $\"16y%5E2-8y%2B1\"$ factors to $\"%284y-1%29%5E2\"$ \r
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\n" ); document.write( "\n" ); document.write( "If you need more help, email me at jim_thompson5910@hotmail.com\r
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\n" ); document.write( "\n" ); document.write( "Also, feel free to check out my tutoring website. \r
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\n" ); document.write( "\n" ); document.write( "Jim
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