Question

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 1

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 2

Ok, so notice there is no c in 80b^4

answer is gcf of coefients are corrrect
variable c is not common to all terms so it shouldn't be factored out
step 6 davide applied reverse distributive

Answer 3

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.