Ask Question
10 April, 16:54

Spencer is asked to factor the polynomial 256x^4y^2-y^2 completely over the integers. His work is shown below.

256x^4 y^2-y^2=y^2 (256x^4-1)

y2 (256x^4-1) = y^2 (16x^2-1) (16x^2+1)

Did Spencer factor the polynomial completely over the integers? Why or why not?

a. Spencer did factor the polynomial completely; he identified the GCF and applied the difference of squares method.

b. Spencer did not factor the polynomial completely; 16x^2-1 can be factored over the integers.

c. Spencer did not factor the polynomial completely; 16x^2+1 can be factored over the integers.

d. Spencer did factor the polynomial completely; he identified the GCF and applied the difference of cubes method.

+4
Answers (2)
  1. 10 April, 17:02
    0
    Option B, Spencer did not factor the polynomial completely; 16x^2-1 can be factored over the integers.

    Step-by-step explanation:

    Step 1: Factor

    256x^4y^2-y^2

    y^2 (256x^4 - 1)

    y^2 (16x^2 - 1) (16x^2 + 1)

    y^2 (4x + 1) (4x - 1) (16x^2 + 1)

    Answer: Option B, Spencer did not factor the polynomial completely; 16x^2-1 can be factored over the integers.
  2. 10 April, 17:23
    0
    answer B
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Spencer is asked to factor the polynomial 256x^4y^2-y^2 completely over the integers. His work is shown below. 256x^4 y^2-y^2=y^2 ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers