Question

david found and factored out the gcf of the polynomial \\(80b^4 - 32b^2c^3 + 48b^4c\\). his work is below.

1. gcf of 80, 32, and 48: 16
2. gcf of \\(b^4\\), \\(b^2\\), and \\(b^4\\): \\(b^2\\)
3. gcf of \\(c^3\\) and \\(c\\): \\(c\\)
4. gcf of the polynomial: \\(16b^2c\\)
5. rewrite as a product of the gcf:

\\(16b^2c(5b^2) - 16b^2c(2c^2) + 16b^2c(3b^2)\\)

1. factor out gcf: \\(16b^2c(5b^2 - 2c^2 + 3b^2)\\)

which statements are true about davids work? check all that apply.

• the gcf of the coefficients is correct.
• the gcf of the variable \\(b\\) should be \\(b^4\\) instead of \\(b^2\\).
• the variable \\(c\\) is not common to all terms, so a power of \\(c\\) should not have been factored out.
• the expression in step 5 is equivalent to the given polynomial.
• in step 6, david applied the distributive property.

Explanation:

Analyze the GCF of the coefficients

Using the Greatest Common Factor knowledge point

$$ \text{GCF}(80, 32, 48) = 16 $$

This matches David's step 1. Thus, the first statement is true.

Analyze the GCF of the variables

Using the Greatest Common Factor of Monomials knowledge point

$$
LATEXBLOCK0
$$

Since \(c\) is missing from the first term \(80b^4\), it cannot be part of the GCF of the polynomial. Thus, David's step 3 and step 4 are incorrect, and the third statement is true.

Verify equivalence in step 5

Using the Factoring Polynomials knowledge point

$$
LATEXBLOCK1
$$

The expression in step 5 is not equivalent to the original polynomial because of the extra \(c\) in the first term. Thus, the fourth statement is false.

Evaluate the distributive property in step 6

Using the Factoring Polynomials knowledge point

$$ 16b^2c(5b^2) - 16b^2c(2c^2) + 16b^2c(3b^2) = 16b^2c(5b^2 - 2c^2 + 3b^2) $$

David factored out the common monomial \(16b^2c\) from each term, which is the reverse application of the distributive property. Thus, the fifth statement is true.

Answer:

• The GCF of the coefficients is correct. (Correct answer)
• The GCF of the variable \(b\) should be \(b^4\) instead of \(b^2\).
• The variable \(c\) is not common to all terms, so a power of \(c\) should not have been factored out. (Correct answer)
• The expression in step 5 is equivalent to the given polynomial.
• In step 6, David applied the distributive property. (Correct answer)