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

 It is currently 06 Jul 2020, 01:46

### 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

# If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager
Joined: 23 Oct 2009
Posts: 62
If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is  [#permalink]

### Show Tags

12 Nov 2010, 03:06
6
52
00:00

Difficulty:

25% (medium)

Question Stats:

78% (02:27) correct 22% (02:48) wrong based on 478 sessions

### HideShow timer Statistics

If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is equivalent to which of the following?

A. (x^2) - x - 6
B. (x^3) + 3(x^2) + 3x
C. (x^3) - 25
D. (x^3) - 5(x^2) - 3x
E. (x^2) + x + 10

I know the problem can be solved by assuming x=2 and then checking which equation yields the correct answer (the correct answer is option A), but I want to understand how you would solve the equation through algebra as well.

Thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 64966
Re: Help with an Algebra PS question  [#permalink]

### Show Tags

12 Nov 2010, 03:31
12
9
rahul321 wrote:
If H = [(x^3) - 6(x^2) - x + 30] / (x-5) and x ≠(is not equal to) 5, then H is equivalent to which of the following?

A - (x^2) - x - 6

B - (x^3) + 3(x^2) + 3x

C - (x^3) - 25

D - (x^3) - 5(x^2) - 3x

E - (x^2) + x + 10

I know the problem can be solved by assuming x=2 and then checking which equation yields the correct answer (the correct answer is option A), but I want to understand how you would solve the equation through algebra as well.

Thanks!

Yes, substituting the values for x is probably the best way but if you want, algebraic approach is given below.

The point here is to factor out x-5 from nominator and then to reduce by it:

$$H=\frac{x^3-6x^2-x+ 30}{x-5}=\frac{x^3-5x^2-x^2-x+25+5}{x-5}=\frac{(x^3-5x^2)-(x^2-25)-(x-5)}{x-5}=\frac{x^2(x-5)-(x-5)(x+5)-(x-5)}{x-5}=$$
$$=\frac{(x-5)(x^2-(x+5)-1)}{x-5}=x^2-x-6$$.

_________________
Retired Moderator
Joined: 02 Sep 2010
Posts: 702
Location: London
Re: Help with an Algebra PS question  [#permalink]

### Show Tags

12 Nov 2010, 03:33
6
1
Given the above expression can be simplified into only a numerator terms means that the numerator must be factorable as (x-5)*Expression, where this Expression is the answer

Since we know this product has degree 3, the expression can at most have degree 2. So that rules out 3 options straight away.
Also the coefficient of x^0 in the product is +30, hence in the expression this must be -6 since -6*-5=30

From these two facts, answer must be (a)

Alternatively substitute x=0, which is the quickest way to get to (a)

Posted from my mobile device

Posted from my mobile device
_________________
##### General Discussion
Manager
Joined: 23 Oct 2009
Posts: 62
Re: Help with an Algebra PS question  [#permalink]

### Show Tags

12 Nov 2010, 03:58
1
Bunuel wrote:
rahul321 wrote:
If H = [(x^3) - 6(x^2) - x + 30] / (x-5) and x ≠(is not equal to) 5, then H is equivalent to which of the following?

A - (x^2) - x - 6

B - (x^3) + 3(x^2) + 3x

C - (x^3) - 25

D - (x^3) - 5(x^2) - 3x

E - (x^2) + x + 10

I know the problem can be solved by assuming x=2 and then checking which equation yields the correct answer (the correct answer is option A), but I want to understand how you would solve the equation through algebra as well.

Thanks!

Yes, substituting the values for x is probably the best way but if you want, algebraic approach is given below.

The point here is to factor out x-5 from nominator and then to reduce by it:

$$H=\frac{x^3-6x^2-x+ 30}{x-5}=\frac{x^3-5x^2-x^2-x+25+5}{x-5}=\frac{(x^3-5x^2)-(x^2-25)-(x-5)}{x-5}=\frac{x^2(x-5)-(x-5)(x+5)-(x-5)}{x-5}=$$
$$=\frac{(x-5)(x^2-(x+5)-1)}{x-5}=x^2-x-6$$.

Thank you Bunuel & shrouded1!! I knew factoring out (x-5) from the nominator was the key but couldn't figure out how to do that...

Substitution is clearly easier but knowing both methods helps.
Manager
Joined: 23 Oct 2009
Posts: 62
Re: Help with an Algebra PS question  [#permalink]

### Show Tags

12 Nov 2010, 04:04
1
rahul321 wrote:
If H = [(x^3) - 6(x^2) - x + 30] / (x-5) and x ≠(is not equal to) 5, then H is equivalent to which of the following?

A - (x^2) - x - 6

B - (x^3) + 3(x^2) + 3x

C - (x^3) - 25

D - (x^3) - 5(x^2) - 3x

E - (x^2) + x + 10

I know the problem can be solved by assuming x=2 and then checking which equation yields the correct answer (the correct answer is option A), but I want to understand how you would solve the equation through algebra as well.

Thanks!

For others who might come across this:

You can also do:

(x^3) - 5(x^2) - (x^2) - x + 30
(x^2)(x - 5) - (x^2 + x -30)
x^2(x - 5) - (x + 6) (x - 5)
(x-5) [x^2 - (x + 6)]
(x-5) (x^2 - x - 6)
Dividing by (x - 5) = x^2 - x - 6
Option A.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10634
Location: Pune, India
Re: Help with an Algebra PS question  [#permalink]

### Show Tags

12 Nov 2010, 06:27
3
rahul321 wrote:
If H = [(x^3) - 6(x^2) - x + 30] / (x-5) and x ≠(is not equal to) 5, then H is equivalent to which of the following?

A - (x^2) - x - 6

B - (x^3) + 3(x^2) + 3x

C - (x^3) - 25

D - (x^3) - 5(x^2) - 3x

E - (x^2) + x + 10

I know the problem can be solved by assuming x=2 and then checking which equation yields the correct answer (the correct answer is option A), but I want to understand how you would solve the equation through algebra as well.

Thanks!

If you know that an equation with degree 3 has a linear factor, it's pretty easy to find the quadratic factor.
Co-efficient of $$x^2$$ and the constant term can be found by observing the $$x^3$$ term and constant term on the left.

Attachment:

Ques.jpg [ 23.79 KiB | Viewed 8203 times ]

This is where you stop in your question because only one option has -6 as the constant term. But, if you do need to find the middle term, do the following. Let's find how we obtained the -x term on the left. For that, on the right, x got multiplied by -6 and the product (-6x) got added to the product of -5 and the missing middle term. To get -x, you must add 5x to -6x, therefore, the missing middle term must be -x.
so you get the quadratic factor as $$x^2 - x - 6$$
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Senior Manager
Joined: 25 Jul 2010
Posts: 290
Location: India
Concentration: Strategy, Technology
GMAT 1: 770 Q51 V46
GPA: 3.2
WE: Engineering (Computer Software)
Re: Help with an Algebra PS question  [#permalink]

### Show Tags

12 Nov 2010, 06:36
1
You could also multiply all the 4 options with (x-5) and see which give (x^3) - 6(x^2) - x + 30
Math Expert
Joined: 02 Sep 2009
Posts: 64966
Re: If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is  [#permalink]

### Show Tags

03 Jul 2013, 00:24
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Algebra: algebra-101576.html

DS Algebra Questions to practice: search.php?search_id=tag&tag_id=29
PS Algebra Questions to practice: search.php?search_id=tag&tag_id=50

Special algebra set: new-algebra-set-149349.html

_________________
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1706
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is  [#permalink]

### Show Tags

07 Oct 2014, 02:35
1
1
$$\frac{x^3 - 6x^2 - x + 30}{x-5} = H$$

$$x^3 - 6x^2 - x + 30 = (x-5) * H$$

Maximum power of x on RHS is 3; power of x in (x-5) is 1, so power of x in H should be 2

So, options B, C & D can be ignored

Working with Option A

$$(x-5) * (x^2 - x - 6) = x^3 - x^2 - 6x - 5x^2 + 5x + 30 = x^3 - 6x^2 - x + 30$$

Intern
Joined: 23 Dec 2014
Posts: 47
Re: If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is  [#permalink]

### Show Tags

02 Jan 2015, 04:04
anon321 wrote:
If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is equivalent to which of the following?

A. (x^2) - x - 6
B. (x^3) + 3(x^2) + 3x
C. (x^3) - 25
D. (x^3) - 5(x^2) - 3x
E. (x^2) + x + 10

I know the problem can be solved by assuming x=2 and then checking which equation yields the correct answer (the correct answer is option A), but I want to understand how you would solve the equation through algebra as well.

Thanks!

I used x=1 and got the right answer. Is it a good way to answer this type of question or algebra process is better?
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 17021
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is  [#permalink]

### Show Tags

02 Jan 2015, 12:52
1
Hi Salvetor,

GMAT questions are usually written in such a way that you can get to the correct answer using a variety of approaches. Thus, it can be a tricky argument to ask if one approach is "best." On Test Day, you have a number of different goals:

1) Get the immediate question correct, if you can.
2) Answer it in a reasonable amount of time (but don't spend too much time).
3) Get the highest score that you can in the OVERALL section.

I'm a BIG fan of TESTing VALUES (the approach that you used), but I also know that sometimes "doing math", using Number Properties, TESTing THE ANSWERS, etc. might be fastest/best depending on the situation. This is all meant to say that by practicing more than one approach, you will have the flexibility to choose how to approach each prompt (instead of getting "stuck" and having just one option which might take too long to implement) and a greater chance to score at a higher level.

GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at: Rich.C@empowergmat.com

The Course Used By GMAT Club Moderators To Earn 750+

souvik101990 Score: 760 Q50 V42 ★★★★★
ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
Intern
Joined: 23 Dec 2014
Posts: 47
Re: If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is  [#permalink]

### Show Tags

08 Jan 2015, 14:22
EMPOWERgmatRichC wrote:
Hi Salvetor,

GMAT questions are usually written in such a way that you can get to the correct answer using a variety of approaches. Thus, it can be a tricky argument to ask if one approach is "best." On Test Day, you have a number of different goals:

1) Get the immediate question correct, if you can.
2) Answer it in a reasonable amount of time (but don't spend too much time).
3) Get the highest score that you can in the OVERALL section.

I'm a BIG fan of TESTing VALUES (the approach that you used), but I also know that sometimes "doing math", using Number Properties, TESTing THE ANSWERS, etc. might be fastest/best depending on the situation. This is all meant to say that by practicing more than one approach, you will have the flexibility to choose how to approach each prompt (instead of getting "stuck" and having just one option which might take too long to implement) and a greater chance to score at a higher level.

GMAT assassins aren't born, they're made,
Rich

Thanks for the explanation. And special thanks for this quote " GMAT assassins aren't born, they're made"
Intern
Joined: 07 Jul 2014
Posts: 2
Re: If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is  [#permalink]

### Show Tags

30 Jan 2015, 18:56
If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is equivalent to which of the following?

A. (x^2) - x - 6
B. (x^3) + 3(x^2) + 3x
C. (x^3) - 25
D. (x^3) - 5(x^2) - 3x
E. (x^2) + x + 10

You also can do this;
A. : (x^2) - x - 6 *(x-5) >>> equal to (x^3) - 6(x^2) - x + 30] CORRECT
B. : (x^3) + 3(x^2) + 3x *(x-5) >>> not equal to (x^3) - 6(x^2) - x + 30] INCORRECT
C. : (x^3) - 25 *(x-5) >>> not equal to (x^3) - 6(x^2) - x + 30] INCORRECT
D. : (x^3) - 5(x^2) - 3x *(x-5) >>> not equal to (x^3) - 6(x^2) - x + 30] INCORRECT
E. : (x^2) + x + 10 *(x-5) >>> not equal to (x^3) - 6(x^2) - x + 30] INCORRECT
Intern
Joined: 12 Oct 2015
Posts: 2
Re: If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is  [#permalink]

### Show Tags

15 Mar 2016, 09:53
simply put x=o in all equations ..
RSM Erasmus Moderator
Joined: 26 Mar 2013
Posts: 2470
Concentration: Operations, Strategy
Schools: Erasmus
Re: If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is  [#permalink]

### Show Tags

05 May 2017, 17:59
1
anon321 wrote:
If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is equivalent to which of the following?

A. (x^2) - x - 6
B. (x^3) + 3(x^2) + 3x
C. (x^3) - 25
D. (x^3) - 5(x^2) - 3x
E. (x^2) + x + 10

Let x=0 ...... Then H = 30/-5 =-6

Apply in the answer choices:

A) -6 .........Keep

B) 0 ...........Eliminate

C) -5...........Eliminate

D) 0 ...........Eliminate

E) 10..........Eliminate

Intern
Joined: 25 May 2020
Posts: 15
Re: If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is  [#permalink]

### Show Tags

31 May 2020, 01:09
As we see, there are only variables in the question and in the answer choices. We can TEST some values and check the options.
Let's assume x = 6 as the denominator will be positive.
Substituting x = 6 in the equation we get

$$[ \frac{((6^3) - 6(6^2) - 6 + 30) }{ (6-5)} ]$$

= $$[ \frac{( - 6 + 30) }{ 1 } ]$$

= 24

Now, let's check the options.

A. $$(x^2) - x - 6 = (6^2) - 6 - 6 = 36 - 12 = 24$$. This matches, but let's check other options as well.

B. $$(x^3) + 3(x^2) + 3x = (6^3) + 3(6^2) + 3 \times 6$$. everything is added, will result in a very big number. Eliminate

C. $$(x^3) - 25 = (6^3) - 25 = 216 - 25$$. No need to calculate. Very big than 24. Eliminate

D. $$(x^3) - 5(x^2) - 3x = (6^3) - 5(6^2) - 3 \times 6 = 216 - 5 \times 36 - 18 = 216 - 180 - 18 = 36 - 18 = 18$$. Eliminate.

E. $$(x^2) + x + 10 = (6^2) + 6 + 10$$. All terms added, will be greater than 24. Eliminate.

Hence, our option becomes A.

OA, A
anon321 wrote:
If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is equivalent to which of the following?

A. (x^2) - x - 6
B. (x^3) + 3(x^2) + 3x
C. (x^3) - 25
D. (x^3) - 5(x^2) - 3x
E. (x^2) + x + 10

I know the problem can be solved by assuming x=2 and then checking which equation yields the correct answer (the correct answer is option A), but I want to understand how you would solve the equation through algebra as well.

Thanks!
Re: If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is   [#permalink] 31 May 2020, 01:09

# If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne