# How do you simplify 7m ^ { 2} n + 4m ^ { 2} n ^ { 2} - 4m ^ { 2} n?

May 25, 2018

${m}^{2} n \left(4 n + 3\right)$

#### Explanation:

In this expression, there are no common integers, but there are variables that are common to each element that can be grouped together to avoid writing them multiple times. That is what is meant by simplifying the expression.

It is easy to see that ${m}^{2}$ appears in every element, so it can be pulled out to stand alone. Not so easy to see is an $n$ that is also in every element, but disguised as an ${n}^{2}$ at one location. That can also be pulled out.

From: $7 {m}^{2} n + 4 {m}^{2} {n}^{2} - 4 {m}^{2} n$

We now have: ${m}^{2} n \left(7 + 4 n - 4\right)$

Resulting in: ${m}^{2} n \left(4 n + 3\right)$

If you want to check the answer, choose some numbers for m&n.

If $m = 1 \mathmr{and} n = 2$ in our expressions then:

$7 {m}^{2} n + 4 {m}^{2} {n}^{2} - 4 {m}^{2} n = {m}^{2} n \left(4 n + 3\right)$

$7 {\left(1\right)}^{2} \left(2\right) + 4 {\left(1\right)}^{2} {\left(2\right)}^{2} - 4 {\left(1\right)}^{2} \left(2\right) = {\left(1\right)}^{2} \left(2\right) \left(4 \left(2\right) + 3\right)$

$14 + 16 - 8 = 2 \left(8 + 3\right)$

$22 = 22$