1. The smallest integer $$n$$ such that $$n^3-11n^2+32n-28>0$$ is





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  • By: anil on 05 May 2019 02.28 am
    We can see that at n = 2, $$n^3-11n^2+32n-28=0$$ i.e. (n-2) is a factor of $$n^3-11n^2+32n-28$$ $$dfrac{n^3-11n^2+32n-28}{n-2}=n^2-9n+14$$ We can further factorize n^2-9n+14 as (n-2)(n-7). $$n^3-11n^2+32n-28=(n-2)^2(n-7)$$ $$Rightarrow$$ $$n^3-11n^2+32n-28>0$$ $$Rightarrow$$ $$(n-2)^2(n-7)>0$$
    Therefore, we can say that n-7>0 Hence, n$$_{min}$$ = 8
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