Find the rational roots of $x^{3}-{2x^{2}\over 3}+3x-2$

Question

I need to find the rational roots of $$x^{3}-{2x^{2}\over 3}+3x-2$$ I thought about using descartes´ rational root theorem but I need to have integers as coefficients of my polynomial: can I work with this polynomial$$3x^{3}-2x^{2}+9x-6$$ ? and use the theorem to find the roots? I will appreciate your help

Answer 1

Technically the two polynomials: $$ p(x) = x^3 - \frac{2x^2}{3} + 3x - 2 \\ q(x) = 3x^3 - 2x^2 + 9x - 6 $$ are entirely different functions, however in the perspective of roots, they clearly have the same roots. Let $\alpha$ be a root of $q(x)$. Then we have that $$ 0 = q(\alpha) = 3\alpha^3 - 2\alpha^2 + 9\alpha - 6\alpha \implies 0 = \frac{0}{3} = \alpha^3 - \frac{2\alpha^2}{3} + 3\alpha - 2 \implies p(\alpha)= 0 $$ which means then that $\alpha$ is a root of $p(x)$. Likewise one can show if $\alpha$ is a root of $p(x)$ then it's a root of $q(x)$. If you're curious about this sort of process you can look into the content of a polynomial and primitive polynomials and Guass' Lemma.

Now for your problem we can factor $q(x)$ using the rational roots theorem like so:

We are looking for elements $\alpha = \frac{p}{q}$ where $p \mid -6$ and $q \mid 3$. Clearly $q = \pm3 \lor q = \pm1$ and $p = \pm6 \lor p = \pm3 \lor p \mid \pm2 \lor p \mid \pm 1$. I'm sure from here you see how to check the cases and find the roots. If you need help let me know, but I found while learning these things it was good to try and do them by myself entirely.

Answer 2

Factorise

$3x^3-2x^2+9x-6=0 \implies (3x-2)x^2+3(3x-2)=0 \implies (3 x-2) (x^2+3) = 0$

For the original

$x^3-\frac 2 3x^2+3x-2=0 \implies (x-\frac 2 3)x^2+3(x-\frac 2 3)=0 \implies (x-\frac 2 3) (x^2+3) = 0$

Same roots. $\checkmark$

Answer 3

$$x^3-\frac{2x^2}{3}+3x-2=0$$ $$3x^3-2x^2+9x-6=0$$ Possible roots from rational root theorem: $\pm\{1, \ 2, \ 3, \ 6, \ \frac 13, \ \frac 23\}$

After trying the possible roots, I found out that $x=\frac 23$ is a root (it makes the equation equal to $0$). Using long division, you can find out that the other factor is $x^2+3$, which does not have any real roots. $$3x^3-2x^3+9x-6=\left(x-\frac 23\right)\left(x^2+3\right)$$ The only rational root is: $$\color{green}{x=\frac 23}$$ The reason why you can work with $3x^3-2x^2+9x-6$ is because it has the same roots as the original. You actually multiply both sides by $3$ to get $3x^3-2x^2+9x-6$. $0\times 3=0$, so $3x^3-2x^2+9x-6$ still equals $0$.