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Mariulka [41]
3 years ago
15

Factor -2bk^2 + 6bk - 2b. -2b(k^2 - 3k - 1) -2b(k^2 + 3k + 1) -2b(k^2 - 3k + 1)

Mathematics
2 answers:
Aleks [24]3 years ago
7 0

Answer:

-2b(k²-3k+1)

Step-by-step explanation:

<u>The question is on factorization</u>

-2bk²+6bk-2b

<u>Factor out 2b</u>

-2b(k²-3k+1)

<u>When you open brackets to check the answer</u>

-2b×k²-2b×-3k-2b×1= -2bk²+6bk+2b

Nikitich [7]3 years ago
5 0

For this case we must factor the following expression:

-2bk ^ 2 + 6bk-2b

It is observed that the three terms have in common a ce divisible between 2 and the variable "b".

Then, we can extract common factor "2b":

2b (-k ^ 2 + 3k-1)

If we take out common factor "-" we have:

-2b (k ^ 2-3k + 1)

Answer:

Option C

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