Answer:
![x(x+1)^2](https://img.qammunity.org/2020/formulas/mathematics/college/a6mg6uoxkxdqh9cdlw51e12hv2ewjqpiph.png)
Explanation:
Given:
The expression to factor is given as:
![x^3+2x^2+x](https://img.qammunity.org/2020/formulas/mathematics/college/tra6syhtiad8pg2rfmualdwywnwu5fyh3p.png)
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* x* x\\\\2x^2=2* x* x\\\\x=x](https://img.qammunity.org/2020/formulas/mathematics/college/y07vr5mdp6sq0i0z3w7lc4kzhbiilwsz07.png)
Now, 'x' is a common factor for all the three terms. So, we factor it out. This gives,
![x((x^3)/(x)+2(x^2)/(x)+(x)/(x))\\\\x(x^2+2x+1)](https://img.qammunity.org/2020/formulas/mathematics/college/7b14hxj0x0xapk1iq1otrb3uiqvsagiu2r.png)
Now, we know a identity which is given as:
![(a+b)^2=a^2+2ab+b^2](https://img.qammunity.org/2020/formulas/mathematics/college/kjx1dnhjbtrcty2dpa3khsnj2at6j6c6g3.png)
Here,
can be rewritten as
![x^2+2(1)(x)+1^2](https://img.qammunity.org/2020/formulas/mathematics/college/xy5utk5445wyvdbbn6mx6lsdmfb2z14ql0.png)
So,
![a=x\ and\ b=1](https://img.qammunity.org/2020/formulas/mathematics/college/9cqlen5b6y1gcnkn6clktc9bwvdlhysf54.png)
Thus,
![x^2+2(1)(x)+1^2= (x+1)^2](https://img.qammunity.org/2020/formulas/mathematics/college/nt531x4ft9tltv3oaduen50nummi5xz15a.png)
Therefore, the complete factorization of the given expression is:
![x^3+2x^2+x=x(x+1)^2](https://img.qammunity.org/2020/formulas/mathematics/college/3v2p8vmudlqmatq8qxkp8oi9wj2fvj0tms.png)