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I've just started studying for an A-Level in Mathematics. This is probably a simple question but when I factorized the quadratic equation

$15x^2+42x-9$

I took out the common factor $3$ to get

$3(5x^2 + 14x - 3)$

Then factorized as follows:

$3(5x^2 - 1x + 15x - 1) = \\ 3(x(5x -1) + 3(5x - 1)) = \\ 3(x+3)(5x-1).$

When I checked the answer it was

$3(5x-1)(x+3)$

My question is this answer the same as the one I arrived to but with a different arrangement, or was my answer simply wrong?

This is my first question on math.stackexchange.com so I apologize in advance if I'm no adhering to the site's rules, e.g. what not to ask on the site.

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    $\begingroup$ Your answer is fine! Remember that the unknowns in the parentheses represent numbers, and we know that the order of multiplication does not matter. $2\times 3=3\times 2$ and so forth. $\endgroup$ – Andrew Nov 12 '12 at 4:32
  • $\begingroup$ 3(x+3)(5x−1) = 3(5x−1)(x+3) $\endgroup$ – Eric Angle Nov 12 '12 at 4:32
  • $\begingroup$ Wow, you got a 100 points without even asking a question. I asked 28 and only got 100. But to answer your question the arrangements doesn't matter. What you did was right! $\endgroup$ – Q.matin Nov 12 '12 at 4:33
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    $\begingroup$ @Q.matin Association Bonus because I have an account on stackoverflow.com where I have 2200+ points. $\endgroup$ – Taurayi Nov 12 '12 at 4:38
  • $\begingroup$ @Andrew since you answered first, do you think you could answer the question officially so I can mark the question as answered? $\endgroup$ – Taurayi Nov 12 '12 at 4:40
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Yes the answer is the same if you work in a classical set of numbers ($\mathbb R$ for example). (I suppose that since this concerne the A-level.) In that sets wa have : $a \times b = b \times a$ for all $a$ and $b$ in it. We say that: ' Multiplication is commutative in these sets'.

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    $\begingroup$ I was going to wait for Andrew, but I guess you answered the question first, so I'll mark this as the answer to the question, thank you. $\endgroup$ – Taurayi Nov 12 '12 at 4:47
  • $\begingroup$ Thank you and good luck in your career in mathematics! $\endgroup$ – Mohamed Nov 12 '12 at 4:53

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