Factorizing 2ab - 4ac + bd - 2de | Help Appreciated

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The expression 2ab - 4ac + bd - 2de can be factorized as 2a(b - 2c) + d(b - 2e). There is a suggestion that if there is a typo in the expression, changing 'e' to 'c' would allow for a different factorization of (2a + d)(b - 2c). The discussion emphasizes the importance of correctly identifying common factors for successful factorization. The primary method discussed is extraction or grouping of common factors. The final consensus is that the best factorization remains 2a(b - 2c) + d(b - 2e).
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Factorize the following

2ab - 4ac + bd - 2de

cant get this at all

any help is appreciated
 
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Dave06 said:
Factorize the following

2ab - 4ac + bd - 2de

cant get this at all

any help is appreciated
Immediately you can get 2a(b-2c)+d(b-2e).

If you have a typo (e should be c?), then you have (2a+d)(b-2c).
 
im afraid not

the question asks

Factorise the following expressions by extraction or grouping of the common factors

2ab - 4ac + bd - 2de
 
Then the best you can do is, as mathman said, 2a(b- 2c)+ d(b- 2e).
 

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