SOLUTION: Factor the expression completely: x^3 + 2x^2 + x I would like help solving this. I factored out x so I arrived at : x (x^2 + 2x) ... now I can't figure it out... lost...

Algebra ->  Distributive-associative-commutative-properties -> SOLUTION: Factor the expression completely: x^3 + 2x^2 + x I would like help solving this. I factored out x so I arrived at : x (x^2 + 2x) ... now I can't figure it out... lost...      Log On


   

Question 634992: Factor the expression completely:
x^3 + 2x^2 + x
I would like help solving this. I factored out x so I arrived at :
x (x^2 + 2x) ... now I can't figure it out... lost...
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

x%5E3%2B2x%5E2%2Bx Start with the given expression.


x%28x%5E2%2B2x%2B1%29 Factor out the GCF x.


Now let's try to factor the inner expression x%5E2%2B2x%2B1


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


Looking at the expression x%5E2%2B2x%2B1, we can see that the first coefficient is 1, the second coefficient is 2, and the last term is 1.


Now multiply the first coefficient 1 by the last term 1 to get %281%29%281%29=1.


Now the question is: what two whole numbers multiply to 1 (the previous product) and add to the second coefficient 2?


To find these two numbers, we need to list all of the factors of 1 (the previous product).


Factors of 1:
1
-1


Note: list the negative of each factor. This will allow us to find all possible combinations.


These factors pair up and multiply to 1.
1*1 = 1
(-1)*(-1) = 1

Now let's add up each pair of factors to see if one pair adds to the middle coefficient 2:


First NumberSecond NumberSum
111+1=2
-1-1-1+(-1)=-2



From the table, we can see that the two numbers 1 and 1 add to 2 (the middle coefficient).


So the two numbers 1 and 1 both multiply to 1 and add to 2


Now replace the middle term 2x with x%2Bx. Remember, 1 and 1 add to 2. So this shows us that x%2Bx=2x.


x%5E2%2Bhighlight%28x%2Bx%29%2B1 Replace the second term 2x with x%2Bx.


%28x%5E2%2Bx%29%2B%28x%2B1%29 Group the terms into two pairs.


x%28x%2B1%29%2B%28x%2B1%29 Factor out the GCF x from the first group.


x%28x%2B1%29%2B1%28x%2B1%29 Factor out 1 from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.


%28x%2B1%29%28x%2B1%29 Combine like terms. Or factor out the common term x%2B1


%28x%2B1%29%5E2 Condense the terms.


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


So x%28x%5E2%2B2x%2B1%29 then factors further to x%28x%2B1%29%5E2


===============================================================


Answer:


So x%5E3%2B2x%5E2%2Bx completely factors to x%28x%2B1%29%5E2.


In other words, x%5E3%2B2x%5E2%2Bx=x%28x%2B1%29%5E2.


Note: you can check the answer by expanding x%28x%2B1%29%5E2 to get x%5E3%2B2x%5E2%2Bx or by graphing the original expression and the answer (the two graphs should be identical).

--------------------------------------------------------------------------------------------------------------
If you need more help, email me at jim_thompson5910@hotmail.com

Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you

Jim
--------------------------------------------------------------------------------------------------------------