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

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Discussion Overview

The discussion revolves around the factorization of the expression 2ab - 4ac + bd - 2de. Participants are seeking assistance with the factorization process, specifically through extraction or grouping of common factors.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant requests help with factorizing the expression and expresses difficulty in doing so.
  • Another participant suggests that the expression can be factored as 2a(b-2c) + d(b-2e) and questions if there is a typo in the expression.
  • A different participant insists that the question specifically asks for factorization by extraction or grouping of common factors.
  • Another reply reiterates the previous suggestion of 2a(b-2c) + d(b-2e) as the best approach.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the factorization method, with some suggesting different approaches and one participant emphasizing the requirement for extraction or grouping.

Contextual Notes

There is uncertainty regarding whether the expression contains a typo, as one participant suggests that 'e' may need to be 'c'. The discussion does not resolve this ambiguity.

Dave06
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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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