Question 337623
I'm assuming you want to factor this



Looking at {{{16y^2-8y+1}}} we can see that the first term is {{{16y^2}}} and the last term is {{{1}}} where the coefficients are 16 and 1 respectively.


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:




Factors of 16:

1,2,4,8


-1,-2,-4,-8 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to 16

1*16

2*8

4*4

(-1)*(-16)

(-2)*(-8)

(-4)*(-4)


note: remember two negative numbers multiplied together make a positive number



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


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">16</td><td>1+16=17</td></tr><tr><td align="center">2</td><td align="center">8</td><td>2+8=10</td></tr><tr><td align="center">4</td><td align="center">4</td><td>4+4=8</td></tr><tr><td align="center">-1</td><td align="center">-16</td><td>-1+(-16)=-17</td></tr><tr><td align="center">-2</td><td align="center">-8</td><td>-2+(-8)=-10</td></tr><tr><td align="center">-4</td><td align="center">-4</td><td>-4+(-4)=-8</td></tr></table>



From this list we can see that -4 and -4 add up to -8 and multiply to 16



Now looking at the expression {{{16y^2-8y+1}}}, replace {{{-8y}}} with {{{-4y+-4y}}} (notice {{{-4y+-4y}}} adds up to {{{-8y}}}. So it is equivalent to {{{-8y}}})


{{{16y^2+highlight(-4y+-4y)+1}}}



Now let's factor {{{16y^2-4y-4y+1}}} by grouping:



{{{(16y^2-4y)+(-4y+1)}}} Group like terms



{{{4y(4y-1)-1(4y-1)}}} Factor out the GCF of {{{4y}}} out of the first group. Factor out the GCF of {{{-1}}} out of the second group



{{{(4y-1)(4y-1)}}} Since we have a common term of {{{4y-1}}}, we can combine like terms


So {{{16y^2-4y-4y+1}}} factors to {{{(4y-1)(4y-1)}}}



So this also means that {{{16y^2-8y+1}}} factors to {{{(4y-1)(4y-1)}}} (since {{{16y^2-8y+1}}} is equivalent to {{{16y^2-4y-4y+1}}})



note:  {{{(4y-1)(4y-1)}}} is equivalent to  {{{(4y-1)^2}}} since the term {{{4y-1}}} occurs twice. So {{{16y^2-8y+1}}} also factors to {{{(4y-1)^2}}}




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     Answer:

So {{{16y^2-8y+1}}} factors to {{{(4y-1)^2}}}



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Jim