# How do you simplify  (sqrt(-4)+3)(2sqrt(-9)-1)?

Mar 1, 2016

$- 15 + 16 i$

#### Explanation:

Recall the definition of $i$:

$i = \sqrt{- 1}$

This allows us to deal with square roots with negative numbers inside of them. The $2$ in this problem are:

$\sqrt{- 4} = \sqrt{{2}^{2} \times - 1} = 2 i$

$\sqrt{- 9} = \sqrt{{3}^{2} \times - 1} = 3 i$

Substituting this into the original expression, we get:

$= \left(2 i + 3\right) \left(2 \left(3 i\right) - 1\right)$

$= \left(2 i + 3\right) \left(6 i - 1\right)$

Now, distribute using the FOIL method.

$= {\underbrace{2 i \cdot 6 i}}_{\text{First"+underbrace(2i * -1)_ "Outside"+underbrace(3 * 6i)_ "Inside"+underbrace(3 * -1)_ "Last}}$

$= 12 {i}^{2} - 2 i + 18 i - 3$

$= 12 {i}^{2} + 16 i - 3$

However, we are not done, since ${i}^{2}$ can be simplified. Recall that since $i = \sqrt{- 1}$, we know that ${i}^{2} = - 1$.

$= 12 \left(- 1\right) + 16 i - 3$

$= - 15 + 16 i$