Question

simplify the following expression.
$6a^2(4a^3 - 8)$
$?a^{\square} + \square a^2$

Explanation:

Step1: Apply the distributive property

We use the distributive property (also known as the distributive law of multiplication over subtraction) which states that \( c(a + b)=ca+cb \). Here, \( c = 6a^{2}\), \( a=4a^{3}\) and \( b=- 8\). So we have:
\(6a^{2}(4a^{3}-8)=6a^{2}\times4a^{3}-6a^{2}\times8\)

Step2: Multiply the coefficients and add the exponents of like bases

For the first term \(6a^{2}\times4a^{3}\), we multiply the coefficients \(6\times4 = 24\) and for the variables with base \(a\), we use the rule \(a^{m}\times a^{n}=a^{m + n}\), so \(a^{2}\times a^{3}=a^{2 + 3}=a^{5}\). So the first term becomes \(24a^{5}\).
For the second term \(6a^{2}\times8\), we multiply the coefficients \(6\times8 = 48\), and since we are multiplying by \(a^{2}\) and the other factor is a constant, the second term is \(- 48a^{2}\) (because \(6a^{2}\times(- 8)=-48a^{2}\))

Answer:

The first box (coefficient of \(a^{5}\)) is \(24\), the exponent of \(a\) in the first term is \(5\), and the coefficient of \(a^{2}\) in the second term is \(- 48\). So filling in the boxes: \(\boldsymbol{24}a^{\boldsymbol{5}}+\boldsymbol{-48}a^{2}\)