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

Valid until 28th Sep

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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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