# How do you factor 12j^2k - 36j^6k^6 + 12j^2?

May 21, 2016

$12 {j}^{2} k - 36 {j}^{6} {k}^{6} + 12 {j}^{2} = 12 {j}^{2} \left(k - 3 {j}^{4} {k}^{6} + 1\right)$

#### Explanation:

We can write $12 {j}^{2} k = {2}^{2} \cdot 3 \cdot {j}^{2} \cdot k$

$36 \cdot {j}^{6} \cdot {k}^{6} = {2}^{2} \cdot {3}^{2} \cdot {j}^{6} \cdot {k}^{6}$ and

$12 {j}^{2} = {2}^{2} \cdot 3 \cdot {j}^{2}$

Hence $12 {j}^{2} k - 36 \cdot {j}^{6} \cdot {k}^{6} + 12 {j}^{2}$

= ${2}^{2} \cdot 3 \cdot {j}^{2} \cdot k - {2}^{2} \cdot {3}^{2} \cdot {j}^{6} \cdot {k}^{6} + {2}^{2} \cdot 3 \cdot {j}^{2}$

Now minimum power for $2$ is $2$; for $3$ is $1$; for $j$ is $2$ and for $k$ is not there in last moomial. Taking this as common, we get

$12 {j}^{2} k - 36 {j}^{6} {k}^{6} + 12 {j}^{2}$

= ${2}^{2} \cdot 3 \cdot {j}^{2} \left(k - 3 {j}^{4} \cdot {k}^{6} + 1\right)$

= $12 {j}^{2} \left(k - 3 {j}^{4} {k}^{6} + 1\right)$