Question

1. what is the correct factored form of $54r^3 + 16$?

Explanation:

Step1: Find the GCF of 54 and 16

First, factorize 54 and 16:

• Prime factors of \( 54 = 2\times3^3 \)
• Prime factors of \( 16 = 2^4 \)

The greatest common factor (GCF) of 54 and 16 is \( 2 \).

Step2: Factor out the GCF from the expression

The given expression is \( 54r^3 + 16 \). Factor out \( 2 \):
\( 54r^3 + 16 = 2(27r^3 + 8) \)

Step3: Recognize the sum of cubes

Notice that \( 27r^3 = (3r)^3 \) and \( 8 = 2^3 \). The sum of cubes formula is \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). Here, \( a = 3r \) and \( b = 2 \).

Step4: Apply the sum of cubes formula

Factor \( 27r^3 + 8 \) using the sum of cubes formula:
\( (3r)^3 + 2^3 = (3r + 2)((3r)^2 - (3r)(2) + 2^2) \)
Simplify the second factor:
\( (3r)^2 - 6r + 4 = 9r^2 - 6r + 4 \)

Step5: Combine the factors

Putting it all together, the factored form of \( 54r^3 + 16 \) is:
\( 2(3r + 2)(9r^2 - 6r + 4) \)

Answer:

\( 2(3r + 2)(9r^2 - 6r + 4) \)