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^

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(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.

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Liliana 2 weeks 2021-09-10T17:23:01+00:00 2 Answers 0

Answers ( )

    0
    2021-09-10T17:24:27+00:00

    Answer:

    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.

    0
    2021-09-10T17:24:42+00:00

    Answer:

    answer B

    Step-by-step explanation:

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45:7+7-4:2-5:5*4+35:2 =? ( )