Question 405542

{{{-4n^4+40n^3-100n^2}}} Start with the given expression



{{{-4n^2(n^2-10n+25)}}} Factor out the GCF {{{-4n^2}}}



Now let's focus on the inner expression {{{n^2-10n+25}}}





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Looking at {{{n^2-10n+25}}} we can see that the first term is {{{n^2}}} and the last term is {{{25}}} where the coefficients are 1 and 25 respectively.


Now multiply the first coefficient 1 and the last coefficient 25 to get 25. Now what two numbers multiply to 25 and add to the  middle coefficient -10? Let's list all of the factors of 25:




Factors of 25:

1,5


-1,-5 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to 25

1*25

5*5

(-1)*(-25)

(-5)*(-5)


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



Now which of these pairs add to -10? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -10


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">25</td><td>1+25=26</td></tr><tr><td align="center">5</td><td align="center">5</td><td>5+5=10</td></tr><tr><td align="center">-1</td><td align="center">-25</td><td>-1+(-25)=-26</td></tr><tr><td align="center">-5</td><td align="center">-5</td><td>-5+(-5)=-10</td></tr></table>



From this list we can see that -5 and -5 add up to -10 and multiply to 25



Now looking at the expression {{{n^2-10n+25}}}, replace {{{-10n}}} with {{{-5n-5n}}} (notice {{{-5n-5n}}} combines back to {{{-10n}}}. So it is equivalent to {{{-10n}}})


{{{n^2+highlight(-5n-5n)+25}}}



Now let's factor {{{n^2-5n-5n+25}}} by grouping:



{{{(n^2-5n)+(-5n+25)}}} Group like terms



{{{n(n-5)-5(n-5)}}} Factor out the GCF of {{{n}}} out of the first group. Factor out the GCF of {{{-5}}} out of the second group



{{{(n-5)(n-5)}}} Since we have a common term of {{{n-5}}}, we can combine like terms


So {{{n^2-5n-5n+25}}} factors to {{{(n-5)(n-5)}}}



So this also means that {{{n^2-10n+25}}} factors to {{{(n-5)(n-5)}}} (since {{{n^2-10n+25}}} is equivalent to {{{n^2-5n-5n+25}}})



note:  {{{(n-5)(n-5)}}} is equivalent to  {{{(n-5)^2}}} since the term {{{n-5}}} occurs twice. So {{{n^2-10n+25}}} also factors to {{{(n-5)^2}}}




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So our expression goes from {{{-4n^2(n^2-10n+25)}}} and factors further to {{{-4n^2(n-5)^2}}}



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Answer:


So {{{-4n^4+40n^3-100n^2}}} factors to {{{-4n^2(n-5)^2}}}

    

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