Question 339939


{{{c^3+20c^2+100c}}} Start with the given expression



{{{c(c^2+20c+100)}}} Factor out the GCF {{{c}}}



Now let's focus on the inner expression {{{c^2+20c+100}}}





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Looking at {{{1c^2+20c+100}}} we can see that the first term is {{{1c^2}}} and the last term is {{{100}}} where the coefficients are 1 and 100 respectively.


Now multiply the first coefficient 1 and the last coefficient 100 to get 100. Now what two numbers multiply to 100 and add to the  middle coefficient 20? Let's list all of the factors of 100:




Factors of 100:

1,2,4,5,10,20,25,50


-1,-2,-4,-5,-10,-20,-25,-50 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to 100

1*100

2*50

4*25

5*20

10*10

(-1)*(-100)

(-2)*(-50)

(-4)*(-25)

(-5)*(-20)

(-10)*(-10)


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



Now which of these pairs add to 20? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 20


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">100</td><td>1+100=101</td></tr><tr><td align="center">2</td><td align="center">50</td><td>2+50=52</td></tr><tr><td align="center">4</td><td align="center">25</td><td>4+25=29</td></tr><tr><td align="center">5</td><td align="center">20</td><td>5+20=25</td></tr><tr><td align="center">10</td><td align="center">10</td><td>10+10=20</td></tr><tr><td align="center">-1</td><td align="center">-100</td><td>-1+(-100)=-101</td></tr><tr><td align="center">-2</td><td align="center">-50</td><td>-2+(-50)=-52</td></tr><tr><td align="center">-4</td><td align="center">-25</td><td>-4+(-25)=-29</td></tr><tr><td align="center">-5</td><td align="center">-20</td><td>-5+(-20)=-25</td></tr><tr><td align="center">-10</td><td align="center">-10</td><td>-10+(-10)=-20</td></tr></table>



From this list we can see that 10 and 10 add up to 20 and multiply to 100



Now looking at the expression {{{1c^2+20c+100}}}, replace {{{20c}}} with {{{10c+10c}}} (notice {{{10c+10c}}} adds up to {{{20c}}}. So it is equivalent to {{{20c}}})


{{{1c^2+highlight(10c+10c)+100}}}



Now let's factor {{{1c^2+10c+10c+100}}} by grouping:



{{{(1c^2+10c)+(10c+100)}}} Group like terms



{{{c(c+10)+10(c+10)}}} Factor out the GCF of {{{c}}} out of the first group. Factor out the GCF of {{{10}}} out of the second group



{{{(c+10)(c+10)}}} Since we have a common term of {{{c+10}}}, we can combine like terms


So {{{1c^2+10c+10c+100}}} factors to {{{(c+10)(c+10)}}}



So this also means that {{{1c^2+20c+100}}} factors to {{{(c+10)(c+10)}}} (since {{{1c^2+20c+100}}} is equivalent to {{{1c^2+10c+10c+100}}})



note:  {{{(c+10)(c+10)}}} is equivalent to  {{{(c+10)^2}}} since the term {{{c+10}}} occurs twice. So {{{1c^2+20c+100}}} also factors to {{{(c+10)^2}}}




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So our expression goes from {{{c(c^2+20c+100)}}} and factors further to {{{c(c+10)^2}}}



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Answer:


So {{{c^3+20c^2+100c}}} factors to {{{c(c+10)^2}}}

    

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