Factoring trick by determining a zero value

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SUMMARY

The discussion revolves around the factorization of algebraic expressions, specifically a³-b³ and a²-4b². The zero value method is effective for a³-b³, yielding the factor (a-b)(a²+ab+b²). However, this method fails for a²-4b², where the zero value occurs at a=4b, but (a-4b) is not a valid factor. Instead, the correct factorization is (a+2b)(a-2b). The confusion arises from misinterpreting the zero value condition.

PREREQUISITES
  • Understanding of polynomial factorization
  • Familiarity with algebraic identities, particularly a²-b²
  • Knowledge of the zero product property
  • Basic skills in manipulating algebraic expressions
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  • Study polynomial identities, focusing on a²-b² and a³-b³
  • Learn about the zero product property in depth
  • Explore advanced factorization techniques for quadratic expressions
  • Practice problems from "Algebra" by I.M. Gelfand and A. Shen
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Students studying algebra, educators teaching factorization methods, and anyone looking to deepen their understanding of polynomial expressions.

CynicusRex
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I'm currently working through the book Algebra by I.M. Gelfand and A. Shen on my own. (As advised here: https://www.physicsforums.com/insights/self-study-basic-high-school-mathematics/#toggle-id-1)
This isn't really homework so I wasn't quite sure where to post this. Anyway, I've got a question regarding the following:

"Problem 113. Factor a³-b³.
Solution: The expression a³-b³ has a zero value when a = b. So it is reasonable to expect a factor a - b..."
Now, it is in fact a factor: (a-b)(a²+ab+b²) = a³-b³. So determining the zero value is useful in at least some cases. But when using this method in problem 117; Factor a² - 4b², it doesn't work. We get a zero value when a = 4b, but (a-4b) is not a factor.
I noticed it was similar to a²-b² and figured (a+2b)(a-2b) was likely the answer, and it is. However I was curious if the former method would work.

Why doesn't the zero value 'trick' work here?
 
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TheBlackAdder said:
I'm currently working through the book Algebra by I.M. Gelfand and A. Shen on my own. (As advised here: https://www.physicsforums.com/insights/self-study-basic-high-school-mathematics/#toggle-id-1)
This isn't really homework so I wasn't quite sure where to post this. Anyway, I've got a question regarding the following:

"Problem 113. Factor a³-b³.
Solution: The expression a³-b³ has a zero value when a = b. So it is reasonable to expect a factor a - b..."
Now, it is in fact a factor: (a-b)(a²+ab+b²) = a³-b³. So determining the zero value is useful in at least some cases. But when using this method in problem 117; Factor a² - 4b², it doesn't work. We get a zero value when a = 4b, but (a-4b) is not a factor.
I noticed it was similar to a²-b² and figured (a+2b)(a-2b) was likely the answer, and it is. However I was curious if the former method would work.

Why doesn't the zero value 'trick' work here?

You mean ##a=2b##.
 
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Ooooh I see my mistake. 4b² ≠ (4b)²
If I substitute a with 4b, I get (4b)² which is 16b².

Thanks.

PS I guess this should've been posted in the homework section then.
 

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