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# Problem factoring difficult quadratic equation

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Manager
Joined: 12 Sep 2010
Posts: 232
Concentration: Healthcare, General Management

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30 Aug 2013, 10:18
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I had problem factoring x^2 + 12x - 540 when I first saw this problem. I now know what factors will work after 5 minutes. I broke 540 into primes: 2^2, 3^4, and 5, but it was still not obivous what number would work. Then I tried to listed all the factor pairs, but that method also took a long time. Can someone please suggest any method that can be used to factor a complicated quadratic equation? Also, what is a efficient method to list all the factor pairs. Thanks.

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Magoosh GMAT Instructor
Joined: 28 Dec 2011
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30 Aug 2013, 14:46
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Samwong wrote:
I had problem factoring x^2 + 12x - 540 when I first saw this problem. I now know what factors will work after 5 minutes. I broke 540 into primes: 2^2, 3^4, and 5, but it was still not obivous what number would work. Then I tried to listed all the factor pairs, but that method also took a long time. Can someone please suggest any method that can be used to factor a complicated quadratic equation? Also, what is a efficient method to list all the factor pairs. Thanks.

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Dear Samwong,
I'm happy to help.

The GMAT is not going to ask you to do factor of that level. That, I believe, is beyond the arithmetic they would expect, except on perhaps a very hard 700+ level question.

Yes, that's a hard problem. Here's what I would do if I wanted to factor this quickly. Product = -540 = (2^2)(3^3)(5) = 2*2*3*3*3*5, and sum = +12. Finding the prime factorization of 540 is extremely helpful!
Now, I know if the factors were far apart (e.g. 1 and 540), then there's no way the factors would have a difference of 12. Given the size of the product, 540, the difference is awfully small. If we start checking factors at the smallest factors {1, 2, 3, 4, ...} then we will be taking the long way. Let's start where the factors are closest together.
The factors of a number are closest together around the number's square-root (if the number is a perfect square, then the closest of all factors are the two factors of its square root --- e.g. 8 and 8 are factors of 64). We don't need to know the exact square root of 540 (which would be some ugly decimal) --- we'll just estimate in broad strokes 20^2 = 400 and 30^2 = 900, so the square root of 540 must be somewhere between 20 and 30. That's where we'll start checking.
27 = 3*3*3 could be a factor ---- 540 = 2*2*3*3*3*5, so 27*20 --- too close --- we have to move out from there --- from 20, go down to 18, the next possible factor ---18 = 2*3*3, so 540 = 2*2*3*3*3*5, so 18*30 --- BINGO! That's it. That's a relatively fast way to find the right pair.

Does all this make sense?
Mike
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Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Manager
Joined: 12 Sep 2010
Posts: 232
Concentration: Healthcare, General Management

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30 Aug 2013, 15:07
Thanks Mike for the awesome post!

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Math Expert
Joined: 02 Sep 2009
Posts: 52296

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31 Aug 2013, 05:10
Samwong wrote:
I had problem factoring x^2 + 12x - 540 when I first saw this problem. I now know what factors will work after 5 minutes. I broke 540 into primes: 2^2, 3^4, and 5, but it was still not obivous what number would work. Then I tried to listed all the factor pairs, but that method also took a long time. Can someone please suggest any method that can be used to factor a complicated quadratic equation? Also, what is a efficient method to list all the factor pairs. Thanks.

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Hope it helps.
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27 Mar 2017, 11:48
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Re: Problem factoring difficult quadratic equation &nbs [#permalink] 27 Mar 2017, 11:48
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