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.
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 :(\(b^4,b^2,b^4\))
\(b^4\)= b × b × b × b
\(b^2\) = b × b
\(b^4\) =b × b × b × b
GCF of variables :(\(b^4,b^2,b^4\)) is \(b^2\)
3) GCF of \(c^3\) 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) \(80b^4-32b^2c^3+48b^4c\)
\(=16b^2(5b^2-2c^3+3b^2c)\)
David applied the distributive property.
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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