SOLUTION: Factor: 20r^2+100rg+125g^2 So far I have 5r(4r+10g)+5g(10r+25g) now I am completely stuck as to where to go from here. Thanks!

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Factor: 20r^2+100rg+125g^2 So far I have 5r(4r+10g)+5g(10r+25g) now I am completely stuck as to where to go from here. Thanks!      Log On


   

Question 303540: Factor:
20r^2+100rg+125g^2
So far I have 5r(4r+10g)+5g(10r+25g)
now I am completely stuck as to where to go from here.
Thanks!
Found 2 solutions by jim_thompson5910, mananth:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

20r%5E2%2B100rg%2B125g%5E2 Start with the given expression


5%284r%5E2%2B20rg%2B25g%5E2%29 Factor out the GCF 5


Now let's focus on the inner expression 4r%5E2%2B20rg%2B25g%5E2




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Looking at 4r%5E2%2B20rg%2B25g%5E2 we can see that the first term is 4r%5E2 and the last term is 25g%5E2 where the coefficients are 4 and 25 respectively.

Now multiply the first coefficient 4 and the last coefficient 25 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 NumberSecond NumberSum
11001+100=101
2502+50=52
4254+25=29
5205+20=25
101010+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 4r%5E2%2B20rg%2B25g%5E2, replace 20rg with 10rg%2B10rg (notice 10rg%2B10rg adds up to 20rg. So it is equivalent to 20rg)

4r%5E2%2Bhighlight%2810rg%2B10rg%29%2B25g%5E2


Now let's factor 4r%5E2%2B10rg%2B10rg%2B25g%5E2 by grouping:


%284r%5E2%2B10rg%29%2B%2810rg%2B25g%5E2%29 Group like terms


2r%282r%2B5g%29%2B5g%282r%2B5g%29 Factor out the GCF of 2r out of the first group. Factor out the GCF of 5g out of the second group


%282r%2B5g%29%282r%2B5g%29 Since we have a common term of 2r%2B5g, we can combine like terms

So 4r%5E2%2B10rg%2B10rg%2B25g%5E2 factors to %282r%2B5g%29%282r%2B5g%29


So this also means that 4r%5E2%2B20rg%2B25g%5E2 factors to %282r%2B5g%29%282r%2B5g%29 (since 4r%5E2%2B20rg%2B25g%5E2 is equivalent to 4r%5E2%2B10rg%2B10rg%2B25g%5E2)


note: %282r%2B5g%29%282r%2B5g%29 is equivalent to %282r%2B5g%29%5E2 since the term 2r%2B5g occurs twice. So 4r%5E2%2B20rg%2B25g%5E2 also factors to %282r%2B5g%29%5E2



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So our expression goes from 5%284r%5E2%2B20rg%2B25g%5E2%29 and factors further to 5%282r%2B5g%29%5E2


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Answer:

So 20r%5E2%2B100rg%2B125g%5E2 factors to 5%282r%2B5g%29%5E2

In other words, 20r%5E2%2B100rg%2B125g%5E2=5%282r%2B5g%29%5E2

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
20r^2+100rg+125g^2
5(4r^2+20rg+25g^2)
5{4r^2+10rg+10rg=25g^2}
5{2r(2r+5g)+5g(2r+5g)}
5{2r+5g)^2
yOU SHOULD HAVE SEPARATED THE COMMON CO-EFFICIENTS
ANANTH