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:
$16b^2c(5b^2) - 16b^2c(2c^2) + 16b^2c(3b^2)$
6. 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.

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:
$16b^2c(5b^2) - 16b^2c(2c^2) + 16b^2c(3b^2)$
6. 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.

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Greatest Common Factor of Monomials", "Greatest Common Factor", "Factoring Polynomials" ], "new_concepts": [], "current_concepts": [ "Greatest Common Factor of Monomials", "Greatest Common Factor", "Factoring Polynomials" ] } ## 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 $$ \begin{aligned} &\text{Terms: } 80b^4,\quad -32b^2c^3,\quad 48b^4c\ &\text{Variable } b \text{ powers: } b^4, b^2, b^4 \implies \text{GCF}(b^4, b^2, b^4) = b^2\ &\text{Variable } c \text{ powers: } c^0, c^3, c^1 \implies c \text{ is not common to all terms} \end{aligned} $$ 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 $$ \begin{aligned} &16b^2c(5b^2) - 16b^2c(2c^2) + 16b^2c(3b^2)\ &= 80b^4c - 32b^2c^3 + 48b^4c\ & e 80b^4 - 32b^2c^3 + 48b^4c \end{aligned} $$ 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.

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.

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Factoring Polynomials" ] }

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QUESTION IMAGE

Question

Explanation:

Analyze the GCF of the coefficients

Analyze the GCF of the variables

Verify equivalence in step 5

Evaluate the distributive property in step 6

Answer: