Question 168972


{{{4k^2-16kv+16v^2}}} Start with the given expression



{{{4(k^2-4kv+4v^2)}}} Factor out the GCF {{{4}}}



Now let's focus on the inner expression {{{k^2-4kv+4v^2}}}





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


Now multiply the first coefficient 1 and the last coefficient 4 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 {{{1k^2-4kv+4v^2}}}, replace {{{-4kv}}} with {{{-2kv+-2kv}}} (notice {{{-2kv+-2kv}}} adds up to {{{-4kv}}}. So it is equivalent to {{{-4kv}}})


{{{1k^2+highlight(-2kv+-2kv)+4v^2}}}



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



{{{(1k^2-2kv)+(-2kv+4v^2)}}} Group like terms



{{{k(k-2v)-2v(k-2v)}}} Factor out the GCF of {{{k}}} out of the first group. Factor out the GCF of {{{-2v}}} out of the second group



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


So {{{1k^2-2kv-2kv+4v^2}}} factors to {{{(k-2v)(k-2v)}}}



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



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




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So our expression goes from {{{4(k^2-4kv+4v^2)}}} and factors further to {{{4(k-2v)^2}}}



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Answer:


So {{{4k^2-16kv+16v^2}}} factors to {{{4(k-2v)^2}}}