GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 28 Feb 2020, 14:43 ### 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

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.  # Hard Factoring question

Author Message
TAGS:

### Hide Tags

Manager  Joined: 17 Aug 2010
Posts: 73

### Show Tags

2
GUYS, Any idea how to factor ? (i came across this in OG though have no idea how they factored it )

5x^2-34x+24
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10130
Location: Pune, India

### Show Tags

9
6
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

Director  Status: -=Given to Fly=-
Joined: 04 Jan 2011
Posts: 784
Location: India
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)

### Show Tags

4
2
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 _________________
##### General Discussion
Manager  Joined: 27 Jul 2010
Posts: 130
Location: Prague
Schools: University of Economics Prague

### Show Tags

1
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 =

$$(5x^2 - 35x + 30) + (x-6) = 5(x^2 - 7x +6)+ (x-6) = 5(x-1)*(x-6) + (x-6)$$
Intern  Joined: 19 Mar 2010
Posts: 1

### Show Tags

1
$$5x^2 -34x + 24$$ = $$5x^2 - 30x - 4x + 24 = (x - 6)(5x - 4)$$
Founder  V
Joined: 04 Dec 2002
Posts: 19077
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5

### Show Tags

Please continue the discussion in the PS Forum.
_________________
Manager  Joined: 17 Aug 2010
Posts: 73

### Show Tags

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  Joined: 27 Jan 2011
Posts: 13

### Show Tags

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
Manager  Joined: 17 Aug 2010
Posts: 73

### Show Tags

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
Retired Moderator Joined: 20 Dec 2010
Posts: 1514

### Show Tags

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;

$$ax^2+bx+c=0$$

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; $$\alpha=6,\beta=\frac{4}{5}$$

$$(x-\alpha)(x-\beta)=0$$
$$(x-6)(x-\frac{4}{5})=0$$
Director  Status: -=Given to Fly=-
Joined: 04 Jan 2011
Posts: 784
Location: India
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)

### Show Tags

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.
_________________

Originally posted by AmrithS on 10 Feb 2011, 04:12.
Last edited by AmrithS on 10 Feb 2011, 04:14, edited 1 time in total.
Director  Status: -=Given to Fly=-
Joined: 04 Jan 2011
Posts: 784
Location: India
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)

### Show Tags

Trivia: The square root term is called the discriminant of a quadratic equation i.e. sq rt. (b^2 - 4ac)
_________________
Manager  Joined: 21 Sep 2010
Posts: 247

### Show Tags

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;

$$ax^2+bx+c=0$$

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
Intern  Joined: 11 Feb 2011
Posts: 26

### Show Tags

5x^2 - 30x -4x+24= 5x-4 * x-6 = 0
x=6 or x=4/5
Director  Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 629

### Show Tags

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...
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10130
Location: Pune, India

### Show Tags

4
3
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

Non-Human User Joined: 09 Sep 2013
Posts: 14155

### Show Tags

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.
_________________ Re: Hard Factoring question   [#permalink] 27 Aug 2018, 16:33
Display posts from previous: Sort by

# Hard Factoring question  