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Advanced Factorization

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Advanced Factorization [#permalink]

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New post 25 Jul 2011, 12:16
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I've been having some trouble solving equations like 0=5x^2-34x+24
How does factorization work in when there is a 5 in front of the x^2? Usually I look at the last number, see which factors are in 24 and then I compare it the the second number in the equation. This doesn't work when there is a number other than 1 in front of x^2. Is there a better method except for the quadratic formula? The OG factors this equation without further explanation to 0=(5x-4)(x-6)

Thanks for help

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Re: Advanced Factorization [#permalink]

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New post 25 Jul 2011, 13:04
Hello heyholetsgo,

I like to create a list of the factors of both sets of numbers. For example,

for the equation, 5x^2-34x+24=0,
For 5 the factors are:
1 x 5
For 24 the factors are:
1 x 24
2 x 12
3 x 8
4 x 6

so you have (5x +/- y)(x +/- z)=0. Since 24 is positive, y and z must be same sign, either both + or both -. The negative sign of the second term, -34, tells you y and z are negative, so our solution is now (5x -y)(x-z)=0

Now plug in your numbers:
(5x - 1)(x-24) = 5x^2 - 121x + 24 = 0 NO
(5x - 2)(x-12) = 5x^2 - 62x + 24 = 0 NO
(5x -12)(x - 2) = 5x^2 - 22x + 24 = 0 NO
until we arrive at (5x-4)(x-6) = 5x^2 - 34x + 24 = 0 YES

If you have an equation like 6x^2 - 7x -24 = 0, then the first term has more factor combinations, i.e. 1 x 6 and 2 x 3, but the process is the same.
(Solution is (2x + 3)(3x - 8) = 0 or 6x^2 - 16x + 9x -24 = 0 or 6x^2 - 7x -24 = 0).

As you practice, this will become almost intuitive.. .you'll be solving without writing them in no time. Hope this explanation was helpful.

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Re: Advanced Factorization [#permalink]

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New post 25 Jul 2011, 21:10
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heyholetsgo wrote:
I've been having some trouble solving equations like 0=5x^2-34x+24
How does factorization work in when there is a 5 in front of the x^2? Usually I look at the last number, see which factors are in 24 and then I compare it the the second number in the equation. This doesn't work when there is a number other than 1 in front of x^2. Is there a better method except for the quadratic formula? The OG factors this equation without further explanation to 0=(5x-4)(x-6)

Thanks for help


You will almost never need to use the formula on GMAT. Whenever you have a quadratic that you need to factor, you will almost always get two integral roots (in GMAT). If I am looking at a GMAT question and am unable to find the factors, I will go back and check my quadratic to see if it is correct rather than try to use this formula... I would be that sure of not needing to use it!

When you have something other than 1 as the co-efficient of x^2, you compare the product of the last term and the co-efficient of x^2 with the middle term.
i.e. if the given equation is px^2 + qx + r = 0, you need to find the numbers that multiply to give p*r and add up to give q (when p = 1, you find numbers that multiply to give r (which is 1*r so essentially it is p*r) and add up to give q)

e.g. the question above: 5x^2-34x+24
If you need to factorize it, you need to find two numbers a and b such that:

a + b = -34

a*b = 5*24

Step 1: Prime factorize the product.
a*b = 2*2*2*3*5

Step 2: Check the signs and decide what you need. Here sum is -ve while product is positive. This means a and b both are negative. Both will add to give a negative number and multiply to give a positive number. It also means that a and b both are smaller than 34 (since they will add up to give 34. If one of them were negative and other positive, one number would have been greater than 34. In that case, the product would have been negative.)

Step 3: Try to split the prime numbers into two groups such that their sum is 34. Try the most obvious group first i.e.
2*2*2 and 3*5 ---- 8 and 15 add up to give 23.
But you need 34, i.e. a number greater than 23.

Before we discuss the next step, let me explain one thing:
Let's say we have 2*2*5*5.. I split it into 2 groups 2*5 and 2*5 (10 and 10). Their sum is 20.
I split in in another way 2*2 and 5*5 (4 and 25). Their sum is 29. The sum increased.
I split in in another way 2 and 2*5*5 (2 and 50). Their sum is 52. The sum increased again.

Notice that farther apart the numbers are, the greater is their sum. We get the least sum (20) when the numbers are equal.

Going back to the original question, 8+15 gave us a sum of 23. We need 34 so we need to get the numbers farther from each other but not too far either. Let's say, I pick a 2 from 8 and give it to 15. I get two numbers 4 and 30. They are farther apart and their sum is 34. So the numbers we are looking for are -4 and -30 (to get -34 as sum)

Taking another example: 8x^2 - 47x - 63

a + b = -47
a*b = 8*(-63) = - 2*2*2*3*3*7

One of a and b is negative since the product is negative. So one of a and b is greater than 47.
I split the primes into two groups: 2*2*2 and 3*3*7 to get 8 and 63 but 63 - 8 = 55. We need to go lower than 55 so we need to get the numbers closer together. 63 is way greater than 8.
Take off a 3 from 63 and give it to 8 to get 21 and 24. Too close.
Rather take off 7 from 63 and give it to 8 to get 9 and 56. Now, 56 - 9 = 47 so you have your numbers as 56 and 9. The greater one has to be negative since the sum is negative so you split the middle term as: -56 and 9.

With a little bit of practice, the hardest questions can be easily solved...
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Re: Advanced Factorization [#permalink]

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New post 26 Jul 2011, 11:52
Just break "-34x" to" -30x-4x" and then factorize as the below:
5x^2-34x+24 = 5x^2-30x-4x+24=5x(x-6)-4(x-6)=(5x-4)(x-6)

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New post 26 Jul 2011, 13:58
Thank you guys. Awesome!!!

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New post 27 Jul 2011, 03:41
Karishma, thanks for a clear and detailed explanation!

But I am unable to understand one thing: for a quadratic eq: Ax^2+Bx+C = 0,
there are standard 2 solutions: x= (-B +/- sqrt B^2-4AC)/2A, from these solutions what I get is that:

a+b = -B/A and a*b = C/A

whereas you state that a+b = B and a*b=A*C I am unable to reconcile this difference. can you please elaborate on this?
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Re: Advanced Factorization [#permalink]

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New post 27 Jul 2011, 04:33
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vinayrsm wrote:
Karishma, thanks for a clear and detailed explanation!

But I am unable to understand one thing: for a quadratic eq: Ax^2+Bx+C = 0,
there are standard 2 solutions: x= (-B +/- sqrt B^2-4AC)/2A, from these solutions what I get is that:

a+b = -B/A and a*b = C/A

whereas you state that a+b = B and a*b=A*C I am unable to reconcile this difference. can you please elaborate on this?


Good Question.
The two pairs of a and b are different.
When we use the formula and say the roots are a and b
a+b = -B/A and ab = C/A

When we say find a and b such that a+b = B and ab = AC, we are trying to split up the middle term of the equation. The a and b we get are not the roots.
e.g. The roots of 5x^2-34x+24 are not -4 and -30. We have split the middle term into -4 and -30.
5x^2-34x+24 = 0
5x^2-30x - 4x+24 = 0
5x(x - 6) -4(x - 6) = 0
(x - 6)(5x - 4) = 0
Roots are 6 and 4/5.
You see that 6+4/5 = 34/5 (-B/A) and 6*4/5 = 24/5 (C/A)

or you can say that once you figure out this pair of a and b, the roots will be -a/A and -b/A where A is the co-efficient of x^2. Put in the values to confirm.

Sum of Roots:
(-a/A) + (-b/A) = -(a+b)/A = -B/A

Product of Roots:
(-a/A)(-b/A) = ab/A^2 = CA/A^2 = C/A
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New post 27 Jul 2011, 06:17
Karishma - Great, thanks for clarifying! very helpful.
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New post 07 Aug 2011, 04:24
i agree. good explannation

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New post 19 Sep 2017, 06:36
Hello from the GMAT Club BumpBot!

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Re: Advanced Factorization   [#permalink] 19 Sep 2017, 06:36
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