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.

Question

Quinn Wagner

Mathematics

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)

Know the Answer?

August Cain · Answer 1

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.