Option B
ANSWER:
The factors of
is
![$x(x+1)^(2)$](https://img.qammunity.org/2020/formulas/mathematics/middle-school/f7qfxqhpj3jnvkqdidafhgcmuuk0srdkpg.png)
SOLUTION:
Given, cubic expression is
![$x^(3)+2 x^(2)+x$](https://img.qammunity.org/2020/formulas/mathematics/middle-school/zdh2vjg5axkgcr6g9lsl9tzzmawf23r1wu.png)
Now, we have to find the factors of above equation.
To factorize the given equation, follow the below steps:
![$\mathrm{x}^(3)+2 \mathrm{x}^(2)+\mathrm{x}$](https://img.qammunity.org/2020/formulas/mathematics/middle-school/l0djiybyyuic7rmy3yrcmdxl6p3kt5q8ct.png)
Since x is common in every term of expression, we can take it as common
![$x\left(x^(2)+2 x+1\right)$](https://img.qammunity.org/2020/formulas/mathematics/middle-school/3i63vpxo2hlmwmihnglukkdw615vcuwwgw.png)
“2x” can be rewritten as “x + x”, the above equation becomes,
![$x\left(x^(2)+x+x+1\right)$](https://img.qammunity.org/2020/formulas/mathematics/middle-school/zn5put2iy2of22j629p7tnlzte0uomj6vt.png)
Taking the common terms out of bracket. we get
x(x (x + 1) + 1 (x + 1))
Taking (x + 1) as common., we get
x ((x + 1)(x + 1))
![$x(x+1)^(2)$](https://img.qammunity.org/2020/formulas/mathematics/middle-school/f7qfxqhpj3jnvkqdidafhgcmuuk0srdkpg.png)
Hence, the second option b is correct.