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fomenos
3 years ago
6

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 o

f 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.
Mathematics
3 answers:
BlackZzzverrR [31]3 years ago
7 0

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.

Vsevolod [243]3 years ago
7 0
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
aleyah 3 years ago
1 0

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.

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