Question

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

Answer 1

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

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