# How do you simplify  (2+5i)^2?

Jan 27, 2016

$- 21 + 20 i$

#### Explanation:

First, distribute this using FOIL as you normally would a binomial.

${\left(2 + 5 i\right)}^{2} = \left(2 + 5 i\right) \left(2 + 5 i\right) =$

overbrace(2xx2)^("First")+overbrace(2xx5i)^"Outside"+overbrace(5ixx2)^"Inside"+overbrace(5ixx5i)^"Last"

This gives

$4 + 10 i + 10 i + 25 {i}^{2}$

or

$4 + 20 i + 25 {i}^{2}$

While this may look simplified, we can go one step further.

Since $i = \sqrt{- 1}$, we know that ${i}^{2} = - 1$, so we can replace ${i}^{2}$ with $- 1$, giving:

$4 + 20 i + 25 \left(- 1\right)$

Now, subtract the $25$ from $4$, giving our answer in the $a + b i$ form of a complex number

$- 21 + 20 i$