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805+ (Hard)|   Algebra|   Inequalities|      
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How many non-positive integers satisfy the inequality (9 x^2)(x +1) Updated on: 31 Jan 2026, 01:15
Originally posted by EgmatQuantExpert on 11 Jan 2019, 06:59.
Last edited by Bunuel on 31 Jan 2026, 01:15, edited 2 times in total.
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C
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95% (hard)

Question Stats:

30% (02:10) correct 70% (02:26) wrong based on 437 sessions

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How many non-positive integers satisfy the inequality \(\frac{(9 – x^2)(x +1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0\)?

A. 1
B. 2
C. 3
D. 4
E. Infinite

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e-GMAT Question of the Week #31

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C
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How many non-positive integers satisfy the inequality (9 x^2)(x +1) 14 Jan 2019, 09:59
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How many non-positive integers satisfy the inequality \(\frac{(9 – x^2)(x +1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0\)?

    A. 1
    B. 2
    C. 3
    D. 4
    E. Infinite

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Solution attached

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How many non-positive integers satisfy the inequality (9 x^2)(x +1) 14 Jan 2019, 14:20
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hi chetan2u Pls please help me out with solving this using wavy line method
I have tried but i still keep getting B as my answer
Thanks
Also in the Explanation in the picture the writer wrote (9-x)^2 but the question says (9-x^2) they are not the same thing
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How many non-positive integers satisfy the inequality (9 x^2)(x +1) 14 Jan 2019, 19:50
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hi chetan2u Pls please help me out with solving this using wavy line method
I have tried but i still keep getting B as my answer
Thanks
Also in the Explanation in the picture the writer wrote (9-x)^2 but the question says (9-x^2) they are not the same thing
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\(\frac{(9 – x^2)(x +1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0\)...

Hi..

You can get your answer by just looking at the extreme values of X for which the equation becomes 0..
Here the values will be 3 and -3, but since we are looking at only non positive integers, see the effect on the entire equation beyond x less than -3..
So substitute X as -4..
\(\frac{(9 – (-4)^2)(-4+1)(-4 + 2)^2}{(-4+ 3)(-4– 2)} ≤ 0......\frac{(-7)*(-3)*4}{(-1)*(-6)}=14>0\)..
So beyond x<-3, the equation will be positive, so discard all the values..

Now, we are left with 0,-1,-2,-3
Discard -3 as the denominator becomes 0 and our equation becomes undefined..
-1 and -2 will give you a 0 as value as numerator has (x+1) and (x+2) in numerator, so ok.
When x is 0, term becomes 9*1*2/(3*(-2))=-3, so ok..

Therefore values will be 0, -1, -2--------3 values

C
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How many non-positive integers satisfy the inequality (9 x^2)(x +1) 14 Jan 2019, 21:21
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X cannot be equals to -3 or 2 since numerator is (3+X)(X-2).
Non positive integer starts from 0 all the way to negative infinity.
So just substitute X=0, X=-1, X=-2, X=-4 into the equation and see if the answer is a non positive value.
When X=0, the equation is negative.
When X=-1, the equation is negative.
When X=-2, the equation is negative.
When X>=-4, the equation is positive.
C.
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How many non-positive integers satisfy the inequality (9 x^2)(x +1) 15 Jan 2019, 01:10
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Thanks so much chetan2u

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How many non-positive integers satisfy the inequality (9 x^2)(x +1) 16 Jan 2019, 21:03
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Solution


Given:
    • \(\frac{(9 – x^2)(x +1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0\)

To find:
    • The number of non-positive integer values of x that satisfy the given inequality

Approach and Working:
    • \(\frac{(9 – x^2)(x +1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0\)

The denominator must not be = 0. Thus, x ≠ {-3, 2}
    • \(\frac{(3 – x)(3 + x)(x + 1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0\)

We can cancel out the common term (x + 3) in both numerator and denominator
    • \(\frac{(3 – x)(x + 1)(x + 2)^2}{(x – 2)} ≤ 0\)

And, we know that \((x + 2)^2\) is always ≥ 0. It is equal to 0, when x = -2
    • Thus, if x = -2, the inequality will be 0
    • So, we get, \(\frac{(3 – x)(x + 1)}{(x – 2)} ≤ 0\)

Now, if we multiply both numerator and denominator by (x – 2), we get,
    • \(\frac{(3 – x)(x + 1)(x - 2)}{(x – 2)^2} ≤ 0\)
      o Implies, (3 – x)(x + 1)(x - 2) ≤ 0
      o Multiplying by -1 on both sides, we get, (x - 3)(x + 1)(x - 2) ≥ 0

    • The zero points of the above inequality are x = {-1, 2, 3}

Let’s represent this on a number line and identify the regions, where this expression will give a non-negative value.



    • From the above diagram, we can see that (x - 3)(x + 1)(x - 2) ≥ 0 in the regions x ≥ 3 and -1 ≤ x ≤ 2.
    • But, we are asked for non-positive values of x, they are x = {-1, 0}

Therefore, the non-positive values of x, which satisfy the inequality, \(\frac{(9 – x^2)(x +1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0\) are x = {-2, -1, 0}

Hence the correct answer is Option C.

Answer: C

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How many non-positive integers satisfy the inequality (9 x^2)(x +1) 30 Jan 2026, 23:15
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Hi Sir,


How this solution is possible -1<=x<=2. How x=2 is possible ? when x=2 then denominator will be zero


EgmatQuantExpert

Solution


Given:

• \(\frac{(9 – x^2)(x +1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0\)

To find:

• The number of non-positive integer values of x that satisfy the given inequality

Approach and Working:

• \(\frac{(9 – x^2)(x +1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0\)

The denominator must not be = 0. Thus, x ≠ {-3, 2}

• \(\frac{(3 – x)(3 + x)(x + 1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0\)

We can cancel out the common term (x + 3) in both numerator and denominator

• \(\frac{(3 – x)(x + 1)(x + 2)^2}{(x – 2)} ≤ 0\)

And, we know that \((x + 2)^2\) is always ≥ 0. It is equal to 0, when x = -2

• Thus, if x = -2, the inequality will be 0
• So, we get, \(\frac{(3 – x)(x + 1)}{(x – 2)} ≤ 0\)

Now, if we multiply both numerator and denominator by (x – 2), we get,
    • \(\frac{(3 – x)(x + 1)(x - 2)}{(x – 2)^2} ≤ 0\)
    • o Implies, (3 – x)(x + 1)(x - 2) ≤ 0o Multiplying by -1 on both sides, we get, (x - 3)(x + 1)(x - 2) ≥ 0
    • The zero points of the above inequality are x = {-1, 2, 3}

Let’s represent this on a number line and identify the regions, where this expression will give a non-negative value.




• From the above diagram, we can see that (x - 3)(x + 1)(x - 2) ≥ 0 in the regions x ≥ 3 and -1 ≤ x ≤ 2.
• But, we are asked for non-positive values of x, they are x = {-1, 0}

Therefore, the non-positive values of x, which satisfy the inequality, \(\frac{(9 – x^2)(x +1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0\) are x = {-2, -1, 0}

Hence the correct answer is Option C.

Answer: C

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How many non-positive integers satisfy the inequality (9 x^2)(x +1) Updated on: 03 Feb 2026, 04:29
Originally posted by Bunuel on 31 Jan 2026, 01:19.
Last edited by Bunuel on 03 Feb 2026, 04:29, edited 1 time in total.
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arunbhati
Hi Sir,


How this solution is possible -1<=x<=2. How x=2 is possible ? when x=2 then denominator will be zero



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You are right, x cannot be 2. The solution to \(\frac{(9 – x^2)(x + 1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0\) is \(-1 ≤ x < 2\) or \(x ≥ 3\) or \(x = -2\).
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How many non-positive integers satisfy the inequality (9 x^2)(x +1) Updated on: 03 Feb 2026, 04:31
Originally posted by arunbhati on 03 Feb 2026, 02:30.
Last edited by Bunuel on 03 Feb 2026, 04:31, edited 1 time in total.
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Hi Bunuel,

if X is not equal to 2 then it should have only two non-postive solution that is -1 and 0
Bunuel


You are right, x cannot be 2. The solution to \(\frac{(9 – x^2)(x + 1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0\) is \(-1 ≤ x < 2\) or \(x ≥ 3\) or \(x = -2\) .
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How many non-positive integers satisfy the inequality (9 x^2)(x +1) 03 Feb 2026, 04:31
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arunbhati
Hi Bunuel,

if X is not equal to 2 then it should have only two non-postive solution that is -1 and 0

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The solution to \(\frac{(9 – x^2)(x + 1)(x + 2)^2}{(x + 3)(x – 2)} ≤ 0\) is \(-1 ≤ x < 2\) or \(x ≥ 3\) or \(x = -2\).

So, non-positive solutions are -2, -1, and 0.
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