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For the cubic equation, $2x^3 -9x^2 +12x -3 =0$ the real root is $$\frac{1}{2}(3 - \frac{1}{(3-2\sqrt{2})^\frac{1}{3}}-(3-2\sqrt{2})^\frac{1}{3})$$ according to Wolfram Alpha. But in High School Mathematics I learnt that

When $p(x) = ax^3 + bx^2 +cx+d$ then possible zeroes $= \frac{\text{factors of d}}{\text{factors of a}}$.

According to this the only possibilities are rational zeroes: $±\frac{1}{2}$, $±\frac{3}{2}$, $±1$, $±3$. Why then do we get irrational zeroes? In this kind of problems, how can we derive answers without using calculator(I also mean without using cubic formula)?

S Das
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2 Answers2

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The result you have learned is called the Rational Root Theorem. In the case of $2x^3-9x^2+12x-3$ we see that there are no rational roots. Of course, every cubic has a real root (which is not rational in this case) by the mean value theorem, without using the cubic formula.

Dietrich Burde
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  • I don't understand it quite well. But does it mean that if $f(k).f(k+1) < 0$, at least one root lies in the interval $(k,k+1)$ which in this case is $(0,1)$? Can we go beyond this? – S Das Jul 07 '21 at 09:31
  • Yes, the mean value theorem tells you that, see this post. What exactly don't you understand quite well? – Dietrich Burde Jul 07 '21 at 10:07
  • Yes, I understand that every cubic equation has a real root. But using IVT I can only find the range in which the roots can lie. I want to know how I can calculate the roots precisely? – S Das Jul 08 '21 at 00:20
  • You can use Cardano's formula, or a numerical way to compute it. Have a look at the posts about it at this site. – Dietrich Burde Jul 08 '21 at 08:24
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The proposition states that if $p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots + a_1x+a_0$ is a polynomial with integer coefficient and if $r/s \in \mathbb{Q}$ is a root of $p(x)$, then $r \mid a_0 $ and $s \mid a_n$.

Galois Theory is a powerful tool for determining the solvability of polynomials using elementary algebraic operation: summation, multiplication, division and roots. Also Cardano's formula is a nice approach to cubics.

Masoud
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