This is a question from Polynomails that appeared in the CAT 2018 Question Paper. There are essentially two ways to solve the question. Try it yourself before looking at the video explanation where both the ways to solve this questionare explained.

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Question 14 : The smallest integer n such that n^{3} - 11n^{2} + 32n - 28 ＞ 0 is [TITA]

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We have to find the smallest integer n such that n^{3} - 11n^{2} + 32n - 28 > 0

For this we can assume some values for n, such that n = 10 , 9 , 8 ,.. so on and can find the smallest integer

n^{3} - 11n^{2} + 32n - 28 > 0

Let us now assume that n = 10

1000 – 1100 + 320 - 28 > 0

192 > 0 Hence this n = 10 works out

Similarly n = 9 , 8 also works so we can try out with n = 8 and if it works we can try for lower numbers otherwise we can try with 9

n^{3} - 11n^{2} + 32n - 28 > 0

512 – 704 + 256 – 28 > 0

36 > 0

Since n = 8 works lets check for n = 7

343 – 539 + 224 – 28 = 0

Hence the smallest integer was found to be 8

The other way of thinking about it can be by substituting it with smaller numbers and checking if something can factorize it

So \\frac{P(x)}{x-a}) where remainder = 0

Let's substitute n = 1 with n^{3} - 11n^{2} + 32n - 28

1 – 11 + 32 – 28 = -6 this doesn’t work and now we can substitute n = 2 and check it out

8 – 44 – 64 – 28 = 0

So this number n^{3} - 11n^{2} + 32n - 28 is a multiple of n-2

(n-2)(n^{2} – 9n + 14)

(n – 2)(n - 2)(n - 7) > 0

This will be greater than 0 when n = 7 and when n is between 2 and 7 it will be negative

When n = 1 ,

1 - 11 + 32 - 28 = -6 so this doesn’t work

Hence the smallest integer which will work is n = 8

The question is **"The smallest integer n such that n ^{3} - 11n^{2} + 32n - 28 ＞ 0 is [TITA]"**

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