Question 131627
Do you want to factor?





Looking at {{{4m^2-4m+1}}} we can see that the first term is {{{4m^2}}} and the last term is {{{1}}} where the coefficients are 4 and 1 respectively.


Now multiply the first coefficient 4 and the last coefficient 1 to get 4. Now what two numbers multiply to 4 and add to the  middle coefficient -4? Let's list all of the factors of 4:




Factors of 4:

1,2


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


These factors pair up and multiply to 4

1*4

2*2

(-1)*(-4)

(-2)*(-2)


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



Now which of these pairs add to -4? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -4


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">4</td><td>1+4=5</td></tr><tr><td align="center">2</td><td align="center">2</td><td>2+2=4</td></tr><tr><td align="center">-1</td><td align="center">-4</td><td>-1+(-4)=-5</td></tr><tr><td align="center">-2</td><td align="center">-2</td><td>-2+(-2)=-4</td></tr></table>



From this list we can see that -2 and -2 add up to -4 and multiply to 4



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


{{{4m^2+highlight(-2m+-2m)+1}}}



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



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



{{{2m(2m-1)-1(2m-1)}}} Factor out the GCF of {{{2m}}} out of the first group. Factor out the GCF of {{{-1}}} out of the second group



{{{(2m-1)(2m-1)}}} Since we have a common term of {{{2m-1}}}, we can combine like terms


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



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



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




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     Answer:

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