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Re: Hard Factoring question [#permalink]
10 Feb 2011, 00:32

1

This post was BOOKMARKED

tinki wrote:

GUYS, Any idea how to factor ? (i came across this in OG though have no idea how they factored it )

5x^2-34x+24

You can try this small trick: Add and substract the same numbers. This opperation does not change the value of the expression, but allows you to factor it:

Hard Factoring question [#permalink]
10 Feb 2011, 01:11

craky wrote:

tinki wrote:

GUYS, Any idea how to factor ? (i came across this in OG though have no idea how they factored it )

5x^2-34x+24

You can try this small trick: Add and substract the same numbers. This opperation does not change the value of the expression, but allows you to factor it:

Re: Hard Factoring question [#permalink]
10 Feb 2011, 01:42

the equation in the question that you have posted is a quadratic equation of the form ax^2 + bx + c where a is coefficient of x^2 term, b is coefficient of X term and C is the constant.

now ax^2+ bx + c can be factored into (x-r1)(x-r2)

where (r1,r2) = ((-b+sqrt(b^2-4ac)/2a),(-b-sqrt(b^2-4ac)/2a))

Trust it is a simple and important concept, please follow the link I pasted below. get back if you want further details!

GUYS, Any idea how to factor ? (i came across this in OG though have no idea how they factored it )

5x^2-34x+24

You can try this small trick: Add and substract the same numbers. This opperation does not change the value of the expression, but allows you to factor it:

Re: Hard Factoring question [#permalink]
10 Feb 2011, 04:12

Yes... The method quoted by Fluke may also be used

This is the general formula for roots of a quadratic equation. It will be helpful to remember that. Aslo, 1. if the term under the root is less than zero, the equation has no real solution 2. if the term is 0, the equation has only one solution 3. if the term is greater than zero, the equation has two real and distinct solution. _________________

"Wherever you go, go with all your heart" - Confucius

I was going to post this as an alternative, thank you for beating me to it. Although this method would take me about a minute and a half, it would be shorter than using the splitting method (I'm just not good at it). _________________

"Only by going too far, can one find out how far one can go."

Re: Hard Factoring question [#permalink]
11 Feb 2011, 13:48

4

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

TwoThrones wrote:

I was going to post this as an alternative, thank you for beating me to it. Although this method would take me about a minute and a half, it would be shorter than using the splitting method (I'm just not good at it).

Actually 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!

And as for '(I'm just not good at it)', it is just about a little bit of practice. I assume you know that you need to find the factors that add up to give the middle term and multiply to give the product of the constant term and co-efficient of x^2. What you may have problems with is splitting the middle term.

e.g. the question above: \(5x^2-34x+24\) If you need to factorize it, you need 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... _________________

Re: Hard Factoring question [#permalink]
26 Feb 2011, 01:36

Hi Pardon me. I don't know if you mean "a + b = -34". Why? How is this method? Lets factors be = (5x - a) (x - b) = 5x^2 -34x + 24 a + 5b = 34 ab = 24 Hence a=4, b=6 (5x - 4)(x - 6) are the factors.

VeritasPrepKarishma wrote:

TwoThrones wrote:

I was going to post this as an alternative, thank you for beating me to it. Although this method would take me about a minute and a half, it would be shorter than using the splitting method (I'm just not good at it).

Actually 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!

And as for '(I'm just not good at it)', it is just about a little bit of practice. I assume you know that you need to find the factors that add up to give the middle term and multiply to give the product of the constant term and co-efficient of x^2. What you may have problems with is splitting the middle term.

e.g. the question above: \(5x^2-34x+24\) If you need to factorize it, you need 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...

Re: Hard Factoring question [#permalink]
26 Feb 2011, 19:59

2

This post received KUDOS

Expert's post

1

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gmat1220 wrote:

Hi Pardon me. I don't know if you mean "a + b = -34". Why? How is this method? Lets factors be = (5x - a) (x - b) = 5x^2 -34x + 24 a + 5b = 34 ab = 24 Hence a=4, b=6 (5x - 4)(x - 6) are the factors.

There are many ways of arriving at the factors. One method is splitting the middle term: \(5x^2 -34x + 24 = 5x^2 -30x - 4x + 24 = 5x(x - 6) -4(x - 6) = (5x - 4)(x - 6)\)

or simply, when you split the middle term into 30 and 4, the factors are \(5x^2 -34x + 24 = 5(x - 30/5)(x - 4/5)\)

What you have done is fine too of course, though for most people, a + 5b = 34 is definitely more difficult to work with than a + b = 34. Hence, more often than not, I suggest splitting the middle term. Use whatever works for you. _________________

Re: Hard Factoring question [#permalink]
10 Sep 2014, 22:39

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