How to factor when no common factors?

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Homework Help Overview

The discussion revolves around the factorization of the expression 25x² - 49y², specifically addressing the challenge of factoring when there are no common factors between the coefficients.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the concept of factoring a difference of squares and suggest that the expression can be rewritten using the identity a² - b² = (a + b)(a - b). There are questions about the implications of having no common factors and the methods to approach the factorization.

Discussion Status

Participants have offered various insights into the factorization process, with some suggesting the use of the difference of squares method. There is ongoing dialogue about confirming the correctness of the factorization through expansion, indicating a productive exploration of the topic.

Contextual Notes

Some participants note that both 25 and 49, as well as x² and y², share no common factors, which influences the approach to the problem. There is an emphasis on developing confidence in one's mathematical abilities and the importance of verifying work independently.

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How can I foctorise / factor this given that 25 and 49 share no common factors?

25x2 - 49y2
 
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Note that 25=52 and [tex]a^nb^n=(ab)^n[/tex]
 


Gringo123 said:
How can I foctorise / factor this given that 25 and 49 share no common factors?

25x2 - 49y2
This is a difference of squares.
 


(a+b)(a-b) = a2-b2

Make sure to make a and b not squared..So instead of 25, 5.
 


Gringo123 said:
given that 25 and 49 share no common factors?
Neither do x2 and y2 share any common factors, so this should imply you're looking for another factoring method other than finding a common factor. In this case, difference of two squares.
 


Thanks everyone. So am I right in saying that the answer would be:
(5x - 7y) (5x + 7y)
?
 


Rather than asking for confirmation from us, why don't you expand your answer to check it?
 


Gringo123 said:
Thanks everyone. So am I right in saying that the answer would be:
(5x - 7y) (5x + 7y)
?

That would be correct.
 


Thanks Raid!
 
  • #10


If you prefer the easier way out (and the lazy approach at that), that's up to you, but you should reconsider becoming more trusting in yourself and not always of those around you.

In this problem, if you expanded your answer it will confirm whether you factorized correctly or not. In other topics you have been studying, for example the quadratic factoring, if asked to solve

[tex]x^2-x-6=0[/tex] and you factorize it as so [tex](x-3)(x+2)[/tex] you can check by expanding, or even plug your values x=3 and x=-2 into the quadratic to see if it's correct.

I'm just saying that there are ways to check your work so you become more independent with your mathematics. Having low confidence in your answers would be bad in tests, considering you're all alone in those situations.
 

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