B Factoring trick by determining a zero value

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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