Factoring trick by determining a zero value

In summary, the conversation discusses the use of the zero value method in factoring algebraic expressions. While it works for some cases, it is not always applicable as seen in Problem 117 where the expression a² - 4b² does not have a factor of (a-4b). The mistake of substituting a with 4b instead of 2b is also mentioned. The conversation concludes with the realization that this discussion should have been posted in the homework section.
  • #1
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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  • #2
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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  • #3
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.
 

FAQ: Factoring trick by determining a zero value

What is the "factoring trick" and how does it work?

The factoring trick is a method used to factor quadratic equations by determining a zero value. It involves setting the equation equal to 0 and finding the values of x that make the equation true. These values can then be used as factors to factor the original quadratic equation.

Why is the factoring trick useful?

The factoring trick is useful because it provides a quick and efficient way to factor quadratic equations without the need for complex formulas or methods. It can also be used to solve quadratic equations that cannot be easily solved by other methods.

What types of equations can be factored using the factoring trick?

The factoring trick can be used to factor any quadratic equation, which is an equation in the form of ax² + bx + c = 0, where a, b, and c are constants and x is the variable. It can also be used to factor higher degree polynomials if the equation can be rewritten as a quadratic equation.

Can the factoring trick be used for all zero values?

No, the factoring trick can only be used for zero values that are factors of the constant term (c) in the quadratic equation. If the zero value is not a factor of c, then the factoring trick cannot be used to factor the equation.

Are there any drawbacks to using the factoring trick?

One drawback of using the factoring trick is that it may not always work for more complex quadratic equations. In these cases, other methods such as the quadratic formula may be more effective. Additionally, the factoring trick may not be as useful for equations with non-integer coefficients.

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