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DougieFresh
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Whats the trick on finding factors larger than 12? 10 Sep 2016, 17:00
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1) \(x^2 + 13-40 = 8\) -> \(x^2 + 13x - 48 = 0\)
2) \(3x^2 - 9x - 70 = 14\) -> \(3x^2 - 9x - 84 = 0\)

Whats the trick on finding factor pairs larger than 12? Is there one?

For instance, 1) factors into (x-3) and (x+16) .. should I just memorize my multiplication tables to 16? That seems a little strange to me because thats a lot of work, not that im agianst that, and where do I stop? Is there a better way?

Also, 2) factors into \(3(x^2 - 3x -28)\). Which I would just divide 84 by 3 to get 28....via long division if i dont immediately recognize it......is that the proper/fastest technique?


So I guess what Im asking here is whats the thought process I should have when approaching factor pairs larger than 12.

Sorry, I dont know what level this is at. First post, evaah!
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Whats the trick on finding factors larger than 12? 10 Sep 2016, 21:23
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You don't need to remember any of the tables, but if you do you could speed yourself up to calculations wherever they may be required. For the problems above you could break down the multiplication into prime factors and then combine to arrive at the exact pair. e.g 48 can be written as 2 * 2 * 2 * 2 * 3. now combine pairs to get the total summation to whatever is required in this case 13. therefore the combination would be 2*2*2*2 and 3 i.e 16 and 3

Same logic should work for the second problem as well.
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Whats the trick on finding factors larger than 12? 10 Sep 2016, 22:57
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I would suggest go for Vedic Maths, if you can.

You will learn the BEST techniques there to solve the operations in a fraction of a second. :)
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Whats the trick on finding factors larger than 12? 11 Sep 2016, 01:45
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With abhimahna

Vedic maths techniques can work wonders in any Quantz Based exams , providing 2 ebooks for help -

Attachment:
vedic-maths.pdf [1.36 MiB]
Downloaded 82 times

Attachment:
Complete_Vedic_Maths.pdf [647.29 KiB]
Downloaded 75 times

Hope these help you..
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Whats the trick on finding factors larger than 12? 11 Sep 2016, 14:19
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Thanks, all.

Im gonna give this Vedic stuff a try...cant hurt.


Edit: WHAT?! Why didnt anyone tell me about this sooner.....Im in a moral dilemma now. Part of me wants to share this with my friends but the other part of me wants to use this as a parlor trick for free beer.
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Whats the trick on finding factors larger than 12? 11 Sep 2016, 17:56
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Abhishek009,

Whats with Chapter 2 in the second document? I cant find it.
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Whats the trick on finding factors larger than 12? 12 Sep 2016, 10:31
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DougieFresh
1) \(x^2 + 13-40 = 8\) -> \(x^2 + 13x - 48 = 0\)
2) \(3x^2 - 9x - 70 = 14\) -> \(3x^2 - 9x - 84 = 0\)

Whats the trick on finding factor pairs larger than 12? Is there one?

For instance, 1) factors into (x-3) and (x+16) .. should I just memorize my multiplication tables to 16? That seems a little strange to me because thats a lot of work, not that im agianst that, and where do I stop? Is there a better way?

Also, 2) factors into \(3(x^2 - 3x -28)\). Which I would just divide 84 by 3 to get 28....via long division if i dont immediately recognize it......is that the proper/fastest technique?


So I guess what Im asking here is whats the thought process I should have when approaching factor pairs larger than 12.

Sorry, I dont know what level this is at. First post, evaah!

Here's how I'd recommend approaching the first one: you're not just looking for any pair of numbers that multiply to make -48, you also need them to sum to 13. One of the two numbers will definitely have to be negative, or else their product couldn't be -48. So, the other number will have to be bigger than 13. Start checking methodically, either in your head or on paper:

14+(-1) = 13, but 14*-1 definitely doesn't equal -48
15 + (-2) = 13, but 15*-2 = -30... getting closer.
16 + (-3) = 13, and 16*-3 = -48. One clue that this is going to be the right pair is that the units digit will be the same as the units digit of 6*3, which is 8.

That is - you don't need to look at every pair of factors of 48 - you only need to find the pair of factors that also sums to 13. So in a lot of cases, it's actually easier to start with the 13.
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