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3 years ago

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:

The question is on factorization

-2bk²+6bk-2b

Factor out 2b

-2b(k²-3k+1)

When you open brackets to check the answer

-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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