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Have I factorised this completely?

Factorised 15ax^2 - 9a^2x + 6a^2x^2

to become

3ax (5x - 3a + 2ax)

But I'm not sure if this is the most efficient/smallest factorisation? Is there a better way?
Yes, 3ax is the highest common factor

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Original post by chubchorub
Factorised 15ax^2 - 9a^2x + 6a^2x^2

to become

3ax (5x - 3a + 2ax) (...)


If you multiply the term with the common factor out, you get 15ax² - 9a²x + 6a²x². Is that the term above? your writing confuses me. If yes, so you are right.
Reply 3
Original post by Kallisto
If you multiply the term with the common factor out, you get 15ax² - 9a²x + 6a²x². Is that the term above? your writing confuses me. If yes, so you are right.


Yes, but that would work with any factor, not just the largest one :wink:
Original post by Sataris
Yes, but that would work with any factor, not just the largest one :wink:


That's true but as I see it the thread starter has asked after the largest one, id est the common factor, and 3ax is it.
Reply 5
Original post by Kallisto
That's true but as I see it the thread starter has asked after the largest one, id est the common factor, and 3ax is it.


That's true, but your method of checking only confirms half of the story. I also think I should drop this now.
Original post by Sataris
That's true, but your method of checking only confirms half of the story. I also think I should drop this now.


My method was intended for checking the result. Its a well-tried way to prevent a mistake. :wink:
Reply 7
Original post by Kallisto
My method was intended for checking the result. Its a well-tried way to prevent a mistake. :wink:


Well of course! It checks that the factor works, but not that it is the largest you could have, which is what I was getting at :tongue:

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