# How do I factor this expression completely?

## $4 {x}^{2} \left({x}^{2} + 1\right) {\left(x + 2\right)}^{3} + 36 {\left({x}^{2} + 1\right)}^{2} {\left(x + 2\right)}^{4}$

Jun 30, 2016

#### Answer:

$4 {x}^{2} \left({x}^{2} + 1\right) {\left(x + 2\right)}^{2} + 36 {\left({x}^{2} + 1\right)}^{2} {\left(x + 2\right)}^{4} = 4 \left({x}^{2} + 1\right) {\left(x + 2\right)}^{2} \left(9 {x}^{4} + 36 {x}^{3} + 46 {x}^{2} + 36 x + 36\right)$

#### Explanation:

It is observed that $4 {x}^{2} \left({x}^{2} + 1\right) {\left(x + 2\right)}^{2}$ and $36 {\left({x}^{2} + 1\right)}^{2} {\left(x + 2\right)}^{4}$ have common factors

$4$, $\left({x}^{2} + 1\right)$ and ${\left(x + 2\right)}^{2}$.

Note that for $\left({x}^{2} + 1\right)$ and $\left(x + 2\right)$, we have selected the minimal power of these among two expressions. Similarly between $4$ and $36$, $4$ is common factor.

Hence taking out these common factors, we get

$4 {x}^{2} \left({x}^{2} + 1\right) {\left(x + 2\right)}^{2} + 36 {\left({x}^{2} + 1\right)}^{2} {\left(x + 2\right)}^{4}$

= $4 \left({x}^{2} + 1\right) {\left(x + 2\right)}^{2} \left[{x}^{2} + 9 \left({x}^{2} + 1\right) {\left(x + 2\right)}^{2}\right]$

= $4 \left({x}^{2} + 1\right) {\left(x + 2\right)}^{2} \left[{x}^{2} + 9 \left({x}^{2} + 1\right) \left({x}^{2} + 4 x + 4\right)\right]$

= $4 \left({x}^{2} + 1\right) {\left(x + 2\right)}^{2} \left[{x}^{2} + 9 \left({x}^{4} + 4 {x}^{3} + 4 {x}^{2} + {x}^{2} + 4 x + 4\right)\right]$

= $4 \left({x}^{2} + 1\right) {\left(x + 2\right)}^{2} \left(9 {x}^{4} + 36 {x}^{3} + 46 {x}^{2} + 36 x + 36\right)$