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let $P(x)=ax^3+bx^2+cx+d,a,b,c,d\in R$, such that $$\min{\{d,b+d\}}>\max{\{|c|,|a+c|\}}$$

show that $P(x)=0$ have no real roots in $[-1,1]$

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You have left out two key pieces of information in the question: the context in which you encountered the question (what class, what book), and what you have tried already. Separately, questions that are phrased in the imperative ("Show that") are often considered impolite, because they can come across as orders rather than as questions. – Carl Mummert Jun 26 '13 at 2:15
What's the close votes for?? OP has numerous questions, and contributed answers to non-trivial questions. I think the phrasing has more to do with English not being OP's first language. – Calvin Lin Jun 26 '13 at 4:01
@Calvin Lin: the number of close votes matches the number of upvotes on my previous comment, if you count the author (me) as an upvote. This particular question could be substantially improved. – Carl Mummert Jun 26 '13 at 11:05
up vote 4 down vote accepted

$P(-1)=-a+b-c+d$ and $P(1) = a+b+c+d$ so $P(-1)\times P(1)= -a^2+b^2-c^2+d^2-2ac+2bd=-(a+c)^2+(b+d)^2$ Since $Min\{d,b+d\}>Max${|c|,|a+c|}, we have $P(-1)\times P(1)>0$ and we have, granted by Bolzano's Theorem, that the number of roots of $P(x)=0$ in $[-1,1]$ is even, that means that $P(x)$ has $0$ or $2$ roots in $[-1,1]$.

Now uses the Cardano's formula to show that it cant have 3 distinct real roots, wich means that 2 roots are complex number and only one is real and this one cant be in $[-1,1]$

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Does this account for the possibility that it has a double root in $[-1,1]$? – Antonio Vargas Jun 26 '13 at 16:46

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