Find all School-related info fast with the new School-Specific MBA Forum

It is currently 02 Aug 2015, 06:53
GMAT Club Tests

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Hard Factoring question

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
avatar
Joined: 17 Aug 2010
Posts: 90
Followers: 1

Kudos [?]: 5 [0], given: 22

Hard Factoring question [#permalink] New post 09 Feb 2011, 23:47
GUYS, Any idea how to factor ? (i came across this in OG though have no idea how they factored it )

5x^2-34x+24
Manager
Manager
avatar
Joined: 27 Jul 2010
Posts: 197
Location: Prague
Schools: University of Economics Prague
Followers: 1

Kudos [?]: 24 [0], given: 15

GMAT ToolKit User
Re: Hard Factoring question [#permalink] New post 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:

5x^2-34x+24 -x +x +6 -6 =

\((5x^2 - 35x + 30) + (x-6) = 5(x^2 - 7x +6)+ (x-6) = 5(x-1)*(x-6) + (x-6)\)
_________________

You want somethin', go get it. Period!

Intern
Intern
avatar
Joined: 19 Mar 2010
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Hard Factoring question [#permalink] New post 10 Feb 2011, 00:49
\(5x^2 -34x + 24\) = \(5x^2 - 30x - 4x + 24 = (x - 6)(5x - 4)\)
Expert Post
Founder
Founder
User avatar
Affiliations: AS - Gold, HH-Diamond
Joined: 04 Dec 2002
Posts: 12855
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
Followers: 2793

Kudos [?]: 14076 [0], given: 3876

GMAT ToolKit User Premium Member CAT Tests
Re: Hard Factoring question [#permalink] New post 10 Feb 2011, 00:52
Expert's post
Please continue the discussion in the PS Forum.
_________________

Founder of GMAT Club

Just starting out with GMAT? Start here... | Want to know your GMAT Score? Try GMAT Score Estimator
Need GMAT Book Recommendations? Best GMAT Books

Co-author of the GMAT Club tests

Have a blog? Feature it on GMAT Club!

GMAT Club Premium Membership - big benefits and savings

Manager
Manager
avatar
Joined: 17 Aug 2010
Posts: 90
Followers: 1

Kudos [?]: 5 [0], given: 22

Hard Factoring question [#permalink] New post 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:

5x^2-34x+24 -x +x +6 -6 =

\((5x^2 - 35x + 30) + (x-6) = 5(x^2 - 7x +6)+ (x-6) = 5(x-1)*(x-6) + (x-6)\)


craky : Thanks for the shortcut, yet how you were to know to pick exactly 6 ? and not for example 4, or 2 or even 5? does that come from practice?

kamallohia wrote:
\(5x^2 -34x + 24\) = \(5x^2 - 30x - 4x + 24 = (x - 6)(5x - 4)\)

kamallohia thx . yet could you plx be so kind to show the full detailed factoring? how you arrived at the final answer? im having still some problems
Intern
Intern
avatar
Joined: 27 Jan 2011
Posts: 21
Followers: 0

Kudos [?]: 3 [0], given: 27

Re: Hard Factoring question [#permalink] New post 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!

More related info @: http://www.sosmath.com/algebra/quadrati ... rmula.html


tinki wrote:
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:

5x^2-34x+24 -x +x +6 -6 =

\((5x^2 - 35x + 30) + (x-6) = 5(x^2 - 7x +6)+ (x-6) = 5(x-1)*(x-6) + (x-6)\)


craky : Thanks for the shortcut, yet how you were to know to pick exactly 6 ? and not for example 4, or 2 or even 5? does that come from practice?

kamallohia wrote:
\(5x^2 -34x + 24\) = \(5x^2 - 30x - 4x + 24 = (x - 6)(5x - 4)\)

kamallohia thx . yet could you plx be so kind to show the full detailed factoring? how you arrived at the final answer? im having still some problems
3 KUDOS received
Director
Director
User avatar
Status: -=Given to Fly=-
Joined: 04 Jan 2011
Posts: 837
Location: India
Concentration: Leadership, Strategy
Schools: Haas '18, Kelley '18
GMAT 1: 650 Q44 V37
GMAT 2: 710 Q48 V40
GMAT 3: 750 Q51 V40
GPA: 3.5
WE: Education (Education)
Followers: 39

Kudos [?]: 157 [3] , given: 78

Re: Hard Factoring question [#permalink] New post 10 Feb 2011, 01:43
3
This post received
KUDOS
2
This post was
BOOKMARKED
You can use a method called "Splitting the middle term"

The main concept is:

for ax^2 + bx + c = 0

Find two numbers such that the product is ac and the sum is b

In this case, the two numbers are:
-30 and -4
-30 -4 = -34 = b and -30 x(-4) = 120 = ac

and then split the middle term using the two numbers. The split form of the main equation is:

5x^2 - 30x -4x +24 = 0

Now just group the terms as follows:
5x(x-6) - 4(x-6) = 0
(5x-4) (x-6) = 0

I hope the explanation helps :)
_________________

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

Useful Threads

1. How to Review and Analyze your Mistakes (Post by BB at GMAT Club)

2. 4 Steps to Get the Most out out of your CATs (Manhattan GMAT Blog)

My Experience With GMAT

1. From 650 to 710 to 750 - My Tryst With GMAT

2. Quest to do my Best - My GMAT Journey Log

Manager
Manager
avatar
Joined: 17 Aug 2010
Posts: 90
Followers: 1

Kudos [?]: 5 [0], given: 22

Re: Hard Factoring question [#permalink] New post 10 Feb 2011, 01:57
Entwistle wrote:
You can use a method called "Splitting the middle term"

The main concept is:

for ax^2 + bx + c = 0

Find two numbers such that the product is ac and the sum is b

In this case, the two numbers are:
-30 and -4
-30 -4 = -34 = b and -30 x(-4) = 120 = ac

and then split the middle term using the two numbers. The split form of the main equation is:

5x^2 - 30x -4x +24 = 0

Now just group the terms as follows:
5x(x-6) - 4(x-6) = 0
(5x-4) (x-6) = 0

I hope the explanation helps :)

Great shortcut
Thanks
+kudo
Math Forum Moderator
avatar
Joined: 20 Dec 2010
Posts: 2028
Followers: 140

Kudos [?]: 1182 [0], given: 376

Re: Hard Factoring question [#permalink] New post 10 Feb 2011, 03:26
The following method can also be used as the last resort;

One should be efficient in finding square roots of a number if following this method;

for any quadratic equation;
\(ax^2+bx+c=0\)

where,
a- coefficient of \(x^2\)
b- coefficient of \(x\)
c - constant

The roots are \(\alpha,\beta\);
\(\alpha,\beta=\frac{-b \pm sqrt{b^2-4ac}}{2a}\)

For the above equation;
\(5x^2-34x+24=0\)

\(5x^2+(-34)x+24=0\)

\(a=5\)
\(b=-34\)
\(c=24\)

\(\alpha=\frac{-(-34)+sqrt{(-34)^2-4*5*24}}{2*5}\)
\(\alpha=\frac{34+sqrt{1156-480}}{10}\)
\(\alpha=\frac{34+sqrt{676}}{10}\)
\(\alpha=\frac{34+26}{10}\)
\(\alpha=\frac{60}{10}\)
\(\alpha=6\)


\(\beta=\frac{34-26}{10}\)
\(\beta=\frac{8}{10}\)
\(\beta=\frac{4}{5}\)


We got two roots as; \(\alpha=6,\beta=\frac{4}{5}\)

\((x-\alpha)(x-\beta)=0\)
\((x-6)(x-\frac{4}{5})=0\)
_________________

~fluke

GMAT Club Premium Membership - big benefits and savings

Director
Director
User avatar
Status: -=Given to Fly=-
Joined: 04 Jan 2011
Posts: 837
Location: India
Concentration: Leadership, Strategy
Schools: Haas '18, Kelley '18
GMAT 1: 650 Q44 V37
GMAT 2: 710 Q48 V40
GMAT 3: 750 Q51 V40
GPA: 3.5
WE: Education (Education)
Followers: 39

Kudos [?]: 157 [0], given: 78

Re: Hard Factoring question [#permalink] New post 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

Useful Threads

1. How to Review and Analyze your Mistakes (Post by BB at GMAT Club)

2. 4 Steps to Get the Most out out of your CATs (Manhattan GMAT Blog)

My Experience With GMAT

1. From 650 to 710 to 750 - My Tryst With GMAT

2. Quest to do my Best - My GMAT Journey Log


Last edited by AmrithS on 10 Feb 2011, 04:14, edited 1 time in total.
Director
Director
User avatar
Status: -=Given to Fly=-
Joined: 04 Jan 2011
Posts: 837
Location: India
Concentration: Leadership, Strategy
Schools: Haas '18, Kelley '18
GMAT 1: 650 Q44 V37
GMAT 2: 710 Q48 V40
GMAT 3: 750 Q51 V40
GPA: 3.5
WE: Education (Education)
Followers: 39

Kudos [?]: 157 [0], given: 78

Re: Hard Factoring question [#permalink] New post 10 Feb 2011, 04:13
Trivia: The square root term is called the discriminant of a quadratic equation i.e. sq rt. (b^2 - 4ac)
_________________

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

Useful Threads

1. How to Review and Analyze your Mistakes (Post by BB at GMAT Club)

2. 4 Steps to Get the Most out out of your CATs (Manhattan GMAT Blog)

My Experience With GMAT

1. From 650 to 710 to 750 - My Tryst With GMAT

2. Quest to do my Best - My GMAT Journey Log

Senior Manager
Senior Manager
User avatar
Joined: 21 Sep 2010
Posts: 261
Followers: 5

Kudos [?]: 26 [0], given: 56

GMAT ToolKit User
Re: Hard Factoring question [#permalink] New post 11 Feb 2011, 10:11
fluke wrote:
The following method can also be used as the last resort;

One should be efficient in finding square roots of a number if following this method;

for any quadratic equation;
\(ax^2+bx+c=0\)

where,
a- coefficient of \(x^2\)
b- coefficient of \(x\)
c - constant

The roots are \(\alpha,\beta\);
\(\alpha,\beta=\frac{-b \pm sqrt{b^2-4ac}}{2a}\)

For the above equation;
\(5x^2-34x+24=0\)

\(5x^2+(-34)x+24=0\)

\(a=5\)
\(b=-34\)
\(c=24\)

\(\alpha=\frac{-(-34)+sqrt{(-34)^2-4*5*24}}{2*5}\)
\(\alpha=\frac{34+sqrt{1156-480}}{10}\)
\(\alpha=\frac{34+sqrt{676}}{10}\)
\(\alpha=\frac{34+26}{10}\)
\(\alpha=\frac{60}{10}\)
\(\alpha=6\)


\(\beta=\frac{34-26}{10}\)
\(\beta=\frac{8}{10}\)
\(\beta=\frac{4}{5}\)


We got two roots as; \(\alpha=6,\beta=\frac{4}{5}\)

\((x-\alpha)(x-\beta)=0\)
\((x-6)(x-\frac{4}{5})=0\)


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."

--T.S. Elliot

Expert Post
4 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 5746
Location: Pune, India
Followers: 1446

Kudos [?]: 7605 [4] , given: 186

Re: Hard Factoring question [#permalink] New post 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...
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

Manager
Manager
avatar
Joined: 11 Feb 2011
Posts: 80
Followers: 1

Kudos [?]: 2 [0], given: 0

Re: Hard Factoring question [#permalink] New post 11 Feb 2011, 14:48
5x^2 - 30x -4x+24= 5x-4 * x-6 = 0
x=6 or x=4/5
Director
Director
avatar
Status: Matriculating
Affiliations: Chicago Booth Class of 2015
Joined: 03 Feb 2011
Posts: 926
Followers: 13

Kudos [?]: 240 [0], given: 123

Reviews Badge
Re: Hard Factoring question [#permalink] New post 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...
Expert Post
2 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 5746
Location: Pune, India
Followers: 1446

Kudos [?]: 7605 [2] , given: 186

Re: Hard Factoring question [#permalink] New post 26 Feb 2011, 19:59
2
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 5712
Followers: 323

Kudos [?]: 63 [0], given: 0

Premium Member
Re: Hard Factoring question [#permalink] New post 10 Sep 2014, 22:39
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Re: Hard Factoring question   [#permalink] 10 Sep 2014, 22:39
    Similar topics Author Replies Last post
Similar
Topics:
6 Experts publish their posts in the topic Stuck between Hard and Very Hard Quant Questions pawanCEO 9 20 Apr 2015, 21:40
1 Experts publish their posts in the topic question on number of factors Vips0000 5 07 Nov 2012, 02:59
Quant Hard (800 level) Practice Questions - recommendation? abhicoolmax 1 21 May 2011, 08:38
12 Experts publish their posts in the topic Set T consists of all points (x,y) such that x^2+y^2 =1 tradinggenius 13 28 Nov 2010, 07:54
Couple of questions about factors and remainders shaselai 1 05 Apr 2010, 08:58
Display posts from previous: Sort by

Hard Factoring question

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.