How exactly can I find no. of rational roots, not only for this but similar type of questions?
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5Are you familiar with the rational root theorem? https://en.wikipedia.org/wiki/Rational_root_theorem – erfink Apr 09 '17 at 08:58
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Hint: By rational root theorem, the numerator of any rational root (in simplest form) divides the constant, and the denominator divides the coefficient of the highest power. In this case, they are $-1$ and $1$ respectively. – Sarvesh Ravichandran Iyer Apr 09 '17 at 08:58
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Possible duplicate of Rational Root theorem issue – mlc Apr 09 '17 at 09:37
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1@mlc , It isn't a duplicate of this question, since she was not knowing what rational root theorem is, how can she had a issue with that? – Jaideep Khare Apr 09 '17 at 09:58
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@Jaideep Khare: point taken. – mlc Apr 09 '17 at 10:29
2 Answers
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We will prove that this equation has no rational roots by contradiction.
Suppose $x= \dfrac pq$ (Rational) is a root of this equation , where $\gcd (p,q) =1$
Now put the value of $x$ in the equation :
$$ \Big (\frac{p}{q} \Big )^3 - 3 \Big (\frac{p}{q} \Big ) -1=0 \implies p^3-3pq^2-q^3=0$$
Now take a look at the following cases :
Case 1 : If $p$ is even $\implies$ $q$ is also even. Not possible , since $\gcd (p,q) =1$
Case 2 : If $p$ is odd, and $q$ is also odd , This expression can never be zero. (odd $-$ odd $-$ odd $\neq 0$)
Case 3 : If $p$ is odd, and $q$ is even, again, this expression can never be zero. (odd $-$ even $-$ even $\neq 0$)
Our assumption was wrong.
Therefore, this equation has no rational root.
Jaideep Khare
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Thanks I get it now, btw can you help me by telling how exactly I can write mathematical equations here? – Iti Shree Apr 09 '17 at 09:11
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@ItiShree Put a dollar sign in beginning and one in the end to write in $LaTeX$ (i.e. coding language for writing math ) . You can click 'edit' on your question (or my answer) to see that how coding is done.For further help : See this https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Jaideep Khare Apr 09 '17 at 09:17
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There is a theorem which states that all and only rational roots of a polynomial with integer coefficients are to be searched among the ones of the form $\frac ab$ where $a\in\Bbb Z$ divides the known term and $b\in\Bbb Z$ (obviously $b\neq0$) divides the coefficient of the highest power.
So in your case all and only rational roots could be $\pm1$, but plugging them into your polynomial, you'll easily see that none of them is a root.
Joe
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