Answered

Which expression correctly shows 2x^5y−2x^2y factored completely over the integers?

A.−2x^2y(x+1)(x^2−x+1)

B.2xy(x−1)(x^2+x+1)

C.2x^2(x−y)(x+y)

D.2x^2y(x−1)(x^2+x+1)

E.2x^2(x−y)(x^2+xy+y^2)

Answer :

igoroleshko156

Answer:

Option D, 2x^2y(x - 1)(x^2 + x + 1)

Step-by-step explanation:

Step 1:  Factor

2x^5y - 2x^2y

2x^2y(x^3 - 1)

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

Answer:  Option D, 2x^2y(x - 1)(x^2 + x + 1)

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