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

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

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New post 12 Nov 2010, 03:06
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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!
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New post 12 Nov 2010, 03:31
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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:

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

Answer: A.
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Re: Help with an Algebra PS question  [#permalink]

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New post 12 Nov 2010, 03:33
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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)

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Re: Help with an Algebra PS question  [#permalink]

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New post 12 Nov 2010, 03:58
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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:

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

Answer: A.


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. :-D
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Re: Help with an Algebra PS question  [#permalink]

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New post 12 Nov 2010, 04:04
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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.
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Re: Help with an Algebra PS question  [#permalink]

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New post 12 Nov 2010, 06:27
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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
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 \(x^2 - x - 6\)
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Re: Help with an Algebra PS question  [#permalink]

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New post 12 Nov 2010, 06:36
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You could also multiply all the 4 options with (x-5) and see which give (x^3) - 6(x^2) - x + 30 :)
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Re: If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is  [#permalink]

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

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New post 07 Oct 2014, 02:35
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\(\frac{x^3 - 6x^2 - x + 30}{x-5} = H\)

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

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

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

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

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New post 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?
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Re: If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is  [#permalink]

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New post 02 Jan 2015, 12:52
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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.

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

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New post 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" :)
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Re: If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is  [#permalink]

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

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New post 15 Mar 2016, 09:53
simply put x=o in all equations ..
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Re: If H = [(x^3) - 6(x^2) - x + 30] / (x-5)and x ≠ 5, then H is  [#permalink]

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New post 05 May 2017, 17:59
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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

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

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New post 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!
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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

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