m is the smallest positive integer such that for any integer n ≥ m

Question

m  is the smallest positive integer such that for any integer  n≥m , the quantity  n^3–7n^2+11n–5  is positive. What is the value of  m ?

A. 4
B. 5
C. 8
D. 6
E. 11

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dips1122 · Accepted Answer

Quickly check for options. If we substitute n for 5, the equation = 0. But we need it to be +ve for all n>=m. At m=5, the value is 0. So for m>5, it will be positive. Smallest value of m = 6. IMO D

shameekv1989 · Answer

The equation  n^3–7n^2+11n–5  = 0 at n = 5

Therefore for the expression to be positive n>5. 
Thus the smallest positive integer for which  n >= m  m has to be 5

Answer - B

Archit3110 · Answer

re write given expression as 
  n^3–7n^2+11n–5  
  (n^2(n-7)+11n–5)

plugin values of options since  n≥m 
at a=4 we get -ve 
at a=5 we get 0 
at a=6 we get +ve value
IMO D; sufficient

Arvind90 · Answer

why cant m be any positive integer value less than 6? the equation states n>=m.
We know that the smallest value of n is 6 as the cubic equation is positive for n=6.

n is  greater than or equal to  m.

so then, m can be 5 or 4?

Can anyone explain what am i missing here?

Dream2020 · Answer

Is there any other way to solve this problem without resorting to Trail and error? It is very time consuming.

DippyDiva · Answer

Factorize the cubic equation to (n-1)(n-1)(n-5) which is positive for all integers greater than 5. So the smalleo value of n turns out to be 6 which equals m

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Kalpit1212 · Answer

can you please tell how did you factorize a cubic polynomial?

DippyDiva · Answer

You normally look for one number when substituted gives 0. In this case finding that one number was easy. So once you get '1' as a root n-1 becomes a factor. Divide the cubic by n-1 to get a quadratic that can easily be factorized.

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RastogiSarthak99 · Answer

This is solveable for when n = m and replace n for any of the values mentioned in the list.

Bunuel  correct me if I am wrong but in GMAT options are listed in ascending or descending order of value right? This list here seems random. TIA.

Chacotay · Answer

One way to factorize polynomials is to use Rene Descartes Rational Root Theorem. In our case the factors of -5 are +1,-1,+5,-5 and the factors of the leading term 1 are +1,-1. Therefore, the rational possible roots of the polynomial are : +1,-1,+5,-5.
By checking it is obvious that +1 is a root of the polynomial and by a long division we find that 
n^3 – 7n^2 + 11n - 5 = (n^2 – 6n + 5)(n – 1) = (n - 5)(n – 1)^2
However, (n – 1)^2 always positive. So, n – 5 > 0 => n > 5 
Solution is n = 6 => D

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m is the smallest positive integer such that for any integer n ≥ m

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m is the smallest positive integer such that for any integer n ≥ m

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