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)

given polynomial 80b^4 – 32b^2c^3 + 48b^4c.

to find: factor put gcf.

solution: in order to factor a gcf of ploynomial, we always look over coefficent of each term and factor out gcf of those coefficents .

then we choose first variable terms in all terms and find gcf of first variable terms.

then choose another variable and factor out gcf of those terms.

david choose 80, 32 and 48 and factored out gcf of those numbers as 16.

so first step is taken by david is correct.

then he choose b variable factors   b^4,b^2, b^4 and found gcf b^2 which is correct.

then in third step he choose c^3, and c and found gcf c.

here devid mistaken. there are three terms and we need to take gcf of variable(s) that would be all terms.

variable c is not there in first term 80b^4.

therefore, final gcf would be just product of gcf found in first two steps.

that is 16 times b^2 = 16b^2.

therefore, following statements would be correct statements :

a)the gcf of the coefficients is correct.

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

c) in step 6, david applied the distributive property.

it is b)

hope this : )

step-by-step explanation:

hi how are you doing today? if you ever need just tell me

step-by-step explanation: