Question

multiply.
$-4a^{2}(3a^{2}-4a+3)$

$-4a^{2}(3a^{2}-4a+3)=\square$ (simplify your answer.)

Explanation:

Step1: Distribute $-4a^2$ to each term

$-4a^2 \cdot 3a^2 + (-4a^2) \cdot (-4a) + (-4a^2) \cdot 3$

Step2: Calculate each product

$$\begin{align} -4a^2 \cdot 3a^2 &= -12a^{2+2} = -12a^4 \\ (-4a^2) \cdot (-4a) &= 16a^{2+1} = 16a^3 \\ (-4a^2) \cdot 3 &= -12a^2 \end{align}$$

Step3: Combine the terms

$-12a^4 + 16a^3 - 12a^2$

Answer:

$-12a^4 + 16a^3 - 12a^2$