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David found and factored out the GCF of the polynomial 80b4 – 32b2c3 + 48b4c. His work is below. GFC of 80, 32, and 48: 16 GCF of b4, b2, and b4: b2 GCF of c3 and c: c GCF of the polynomial: 16b2c Rewrite as a product of the GCF: 16b2c(5b2) – 16b2c(2c2) + 16b2c(3b2) Factor out GCF: 16b2c(5b2 – 2c2 + 3b2) Which statements are true about David’s work? Check all that apply. The GCF of the coefficients is correct. The GCF of the variable b should be b4 instead of b2. 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.

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Answer: The correct statements are

The GCF of the coefficients is correct.

The variable c is not common to all terms, so a power of c should not have been factored out.

David applied the distributive property.

Step-by-step explanation:

GCF = Greatest common factor

1) GCF of coefficients : (80,32,48)

80 = 2 × 2 × 2 × 2 × 5

32 = 2 × 2 × 2 × 2 × 2

48 = 2 × 2 × 2 × 2 × 3

GCF of coefficients : (80,32,48) is 16.

2) GCF of variables :([tex]b^4,b^2,b^4[/tex])

[tex]b^4[/tex]= b × b × b × b

[tex]b^2[/tex] = b × b

[tex]b^4[/tex] =b × b × b × b

GCF of variables :([tex]b^4,b^2,b^4[/tex]) is [tex]b^2[/tex]

3) GCF of [tex]c^3[/tex] and c: c is not the GCF of the polynomial. The variable c is not common to all terms, so a power of c should not have been factored out.

4) [tex]80b^4-32b^2c^3+48b^4c[/tex]

[tex]=16b^2(5b^2-2c^3+3b^2c)[/tex]

David applied the distributive property.

citlalilovessss

Answer:

The GCF of the coefficients is correct.

The variable c is not common to all terms, so a power of c should not have been factored out.

In step 6, David applied the distributive property.

Step-by-step explanation:

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