# How do you multiply (4m^2-2n)^2?

Apr 24, 2017

See the solution process below:

#### Explanation:

To multiply this expression we can use this rule:

${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$

If we let: $a = 4 {m}^{2}$ and $b = 2 n$ and substitute we get:

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

$16 {m}^{4} - 16 {m}^{2} n + 4 {n}^{2}$

Another way to multiply this is to first rewrite the expression as:

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

Then, to multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

$\left(\textcolor{red}{4 {m}^{2}} - \textcolor{red}{2 n}\right) \left(\textcolor{b l u e}{4 {m}^{2}} - \textcolor{b l u e}{2 n}\right)$ becomes:

$\left(\textcolor{red}{4 {m}^{2}} \times \textcolor{b l u e}{4 {m}^{2}}\right) - \left(\textcolor{red}{4 {m}^{2}} \times \textcolor{b l u e}{2 n}\right) - \left(\textcolor{red}{2 n} \times \textcolor{b l u e}{4 {m}^{2}}\right) + \left(\textcolor{red}{2 n} \times \textcolor{b l u e}{2 n}\right)$

$16 {m}^{4} - 8 {m}^{2} n - 8 {m}^{2} n + 4 {n}^{2}$

We can now combine like terms:

$16 {m}^{4} + \left(- 8 - 8\right) {m}^{2} n + 4 {n}^{2}$

$16 {m}^{4} + \left(- 16\right) {m}^{2} n + 4 {n}^{2}$

$16 {m}^{4} - 16 {m}^{2} n + 4 {n}^{2}$