Question

apply the distributive property to show $3i(2i) + (-3i - 5)$ in its simplest form. (1 point)
$\square - \square i$

Explanation:

Step1: Apply Distributive Property

The distributive property states that \( a(b + c + d)=ab+ac + ad \). So, for \( 3i[(2i)+(-3i - 5)] \), we distribute \( 3i \) to each term inside the brackets:
\( 3i\times(2i)+3i\times(-3i)+3i\times(-5) \)

Step2: Simplify each term

• For \( 3i\times(2i) \): Using the rule \( i^2=- 1 \) and \( a\times b\times c\times d=(a\times b)\times(c\times d) \), we have \( 3\times2\times i\times i = 6i^{2}=6\times(-1)=-6 \)
• For \( 3i\times(-3i) \): \( 3\times(-3)\times i\times i=-9i^{2}=-9\times(-1) = 9 \)
• For \( 3i\times(-5) \): \( 3\times(-5)\times i=-15i \)

Now, combine the terms: \( - 6+9-15i \)
Simplify the real - part: \( (-6 + 9)-15i=3-15i \)

Answer:

\( 3-15i \), so the first box is \( 3 \) and the second box is \( 15 \)