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

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.

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.

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.


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 \(x^2 - x - 6\)

You could also multiply all the 4 options with (x-5) and see which give (x^3) - 6(x^2) - x + 30

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

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

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

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

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?

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"

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

simply put x=o in all equations ..

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

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

H=(x3−6x2−x+30)/(x−5)
H(x-5)=(x3−6x2−x+30)

Let H = Ax2+Bx+C

Then, (Ax2+Bx+C)(x-5)=(x3−6x2−x+30)
=>Ax3-5Ax2+Bx2-5Bx+Cx-5C=x3−6x2−x+30
Comparing coefficients,
=> A = 1
=> -5A + B = -6
=> -5*1+B=-6
=> B =-1
=>-5C = 30
=> C = -6

Hence H = x2 -x -6 i.e. A

well, just put x=0

The main equation boils down to -6.

Only A = -6

The rest of the calculations are pretty fast because 0 eliminates the variables.

S answer is (A)