SOLUTION: Factor: 5x^3+20x^2+20x

Algebra ->  Systems-of-equations -> SOLUTION: Factor: 5x^3+20x^2+20x      Log On


   



Question 144025: Factor:
5x^3+20x^2+20x

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

5x%5E3%2B20x%5E2%2B20x Start with the given expression


5x%28x%5E2%2B4x%2B4%29 Factor out the GCF 5x


Now let's focus on the inner expression x%5E2%2B4x%2B4




------------------------------------------------------------



Looking at 1x%5E2%2B4x%2B4 we can see that the first term is 1x%5E2 and the last term is 4 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

First NumberSecond NumberSum
141+4=5
222+2=4
-1-4-1+(-4)=-5
-2-2-2+(-2)=-4



From this list we can see that 2 and 2 add up to 4 and multiply to 4


Now looking at the expression 1x%5E2%2B4x%2B4, replace 4x with 2x%2B2x (notice 2x%2B2x adds up to 4x. So it is equivalent to 4x)

1x%5E2%2Bhighlight%282x%2B2x%29%2B4


Now let's factor 1x%5E2%2B2x%2B2x%2B4 by grouping:


%281x%5E2%2B2x%29%2B%282x%2B4%29 Group like terms


x%28x%2B2%29%2B2%28x%2B2%29 Factor out the GCF of x out of the first group. Factor out the GCF of 2 out of the second group


%28x%2B2%29%28x%2B2%29 Since we have a common term of x%2B2, we can combine like terms

So 1x%5E2%2B2x%2B2x%2B4 factors to %28x%2B2%29%28x%2B2%29


So this also means that 1x%5E2%2B4x%2B4 factors to %28x%2B2%29%28x%2B2%29 (since 1x%5E2%2B4x%2B4 is equivalent to 1x%5E2%2B2x%2B2x%2B4)


note: %28x%2B2%29%28x%2B2%29 is equivalent to %28x%2B2%29%5E2 since the term x%2B2 occurs twice. So 1x%5E2%2B4x%2B4 also factors to %28x%2B2%29%5E2



------------------------------------------------------------




So our expression goes from 5x%28x%5E2%2B4x%2B4%29 and factors further to 5x%28x%2B2%29%5E2


------------------
Answer:

So 5x%5E3%2B20x%5E2%2B20x factors to 5x%28x%2B2%29%5E2