## 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 compl

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

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19 hours 2021-09-10T14:48:46+00:00 1 Answer 0

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