Question 1: Which equation shows p (x) = x^6-1 factored completely over the integers? (Hint: You will need to use more than one method to complete this problem.)

a. p (x) = (x^3+1) (x^3-1)

b. p (x) = (x^2-1) (x^4+x^2+1)

c. p (x) = (x-1) (x^2+x+1) (x+1) (x^2-x+1)

d. p (x) = (x-1) (x+1) (x4+x^2+1)

Question 2: Which expression is the expanded form of p (x) = 4 (x-7) (2x^2+3) ?

a. 32x^3-224x^2+48x-336

b. - 48x^2+12x-84

c. 8x^3-56x^2+12x-84

d. 8x^3+56x^2-12x-84

Question #1: Option C, (x - 1) (x^2 + x + 1) (x + 1) (x^2 - x + 1)

Question #2: Option C, 8x^3-56x^2+12x-84

Step-by-step explanation:

Question #1

Step 1: Factor

p (x) = x^6 - 1

p (x) = (x + 1) (x - 1) (x^2 + x + 1) (x^2 - x + 1)

Answer: Option C, (x - 1) (x^2 + x + 1) (x + 1) (x^2 - x + 1)

Question #2

Step 1: Expand

p (x) = 4 (x - 7) (2x^2 + 3)

p (x) = (4x - 28) (2x^2 + 3)

p (x) = 8x^3 + 12x - 56x^2 - 84

p (x) = 8x^3 - 56x^2 + 12x - 84

Answer: Option C, 8x^3-56x^2+12x-84