Answer:
![x(x+1)^2](https://tex.z-dn.net/?f=x%28x%2B1%29%5E2)
Step-by-step explanation:
Given:
The expression to factor is given as:
![x^3+2x^2+x](https://tex.z-dn.net/?f=x%5E3%2B2x%5E2%2Bx)
In order to factor it, we write the factors of each of the terms of the given polynomial. So,
The factors of the three terms are:
![x^3=x\times x\times x\\\\2x^2=2\times x\times x\\\\x=x](https://tex.z-dn.net/?f=x%5E3%3Dx%5Ctimes%20x%5Ctimes%20x%5C%5C%5C%5C2x%5E2%3D2%5Ctimes%20x%5Ctimes%20x%5C%5C%5C%5Cx%3Dx)
Now, 'x' is a common factor for all the three terms. So, we factor it out. This gives,
![x(\frac{x^3}{x}+2\frac{x^2}{x}+\frac{x}{x})\\\\x(x^2+2x+1)](https://tex.z-dn.net/?f=x%28%5Cfrac%7Bx%5E3%7D%7Bx%7D%2B2%5Cfrac%7Bx%5E2%7D%7Bx%7D%2B%5Cfrac%7Bx%7D%7Bx%7D%29%5C%5C%5C%5Cx%28x%5E2%2B2x%2B1%29)
Now, we know a identity which is given as:
![(a+b)^2=a^2+2ab+b^2](https://tex.z-dn.net/?f=%28a%2Bb%29%5E2%3Da%5E2%2B2ab%2Bb%5E2)
Here,
can be rewritten as ![x^2+2(1)(x)+1^2](https://tex.z-dn.net/?f=x%5E2%2B2%281%29%28x%29%2B1%5E2)
So, ![a=x\ and\ b=1](https://tex.z-dn.net/?f=a%3Dx%5C%20and%5C%20b%3D1)
Thus, ![x^2+2(1)(x)+1^2= (x+1)^2](https://tex.z-dn.net/?f=x%5E2%2B2%281%29%28x%29%2B1%5E2%3D%20%28x%2B1%29%5E2)
Therefore, the complete factorization of the given expression is:
![x^3+2x^2+x=x(x+1)^2](https://tex.z-dn.net/?f=x%5E3%2B2x%5E2%2Bx%3Dx%28x%2B1%29%5E2)