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Factor 16x^2 + 16x + 4 completely a. (16x + 4)(x + 1) b. (4x + 2)(4x - 2) c. 4(2x + 1)^2

Factor 16x^2 + 16x + 4 completely a. (16x + 4)(x + 1) b. (4x + 2)(4x - 2) c. 4(2x + 1)^2
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5 votes
Factor 16x^2 + 16x + 4 completely

a. (16x + 4)(x + 1)
b. (4x + 2)(4x - 2)
c. 4(2x + 1)^2

asked
User Artem Luzhanovskyi
by
8.6k points

1 Answer

5 votes
16x^2 + 16x + 4...common factor is 4
4(4x^2 + 4x + 1) =
4(2x + 1)(2x + 1) =
4(2x + 1)^2 <===
answered
User Natan Rubinstein
by
8.2k points
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