SOLUTION: Factor Completely
c^3 + 20c^2 + 100c
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Question 339939: Factor Completely
c^3 + 20c^2 + 100c
Found 2 solutions by jim_thompson5910, Fombitz:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Start with the given expression
Factor out the GCF
Now let's focus on the inner expression
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Looking at we can see that the first term is and the last term is 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
| First Number | Second Number | Sum | | 1 | 100 | 1+100=101 |
| 2 | 50 | 2+50=52 |
| 4 | 25 | 4+25=29 |
| 5 | 20 | 5+20=25 |
| 10 | 10 | 10+10=20 |
| -1 | -100 | -1+(-100)=-101 |
| -2 | -50 | -2+(-50)=-52 |
| -4 | -25 | -4+(-25)=-29 |
| -5 | -20 | -5+(-20)=-25 |
| -10 | -10 | -10+(-10)=-20 |
From this list we can see that 10 and 10 add up to 20 and multiply to 100
Now looking at the expression , replace with (notice adds up to . So it is equivalent to )
Now let's factor by grouping:
Group like terms
Factor out the GCF of out of the first group. Factor out the GCF of out of the second group
Since we have a common term of , we can combine like terms
So factors to
So this also means that factors to (since is equivalent to )
note: is equivalent to since the term occurs twice. So also factors to
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So our expression goes from and factors further to
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Answer:
So factors to
If you need more help, email me at jim_thompson5910@hotmail.com
Also, feel free to check out my tutoring website
Jim
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
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