Find the zero of the polynomial

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Find the zero of the polynomial $f(x) = x^3 - 5x^2- 2x + 24,$ if it is given that the product of the zeroes is 12.

Thanks in advance.

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This could be solved by Bezout theorem if it was all ready told that roots are integers,other than that you could have tried plugging $(1,12),(2,6),(3,4)$ as the solution since product must be $12$ $$\frac{(x^3-5x^2-2x+24)}{x+2}=x^2-7x+12\\x_{1,2}=\frac{7\pm\sqrt{49-48}}{2}\\x_1=4,x_2=3,x_3=-2\\$$ Clearly product is $-24$ not $12$,maybe it was about positive roots?

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There is a simple way to check for rational roots of a polynomial with integer coefficients.

Let $P(x) = a_nx^n + ... + a_1x+a_0$. If $x=\frac{p}{q}$ is a root, with coprime $p$ and $q$, then

$$a_np^n+a_{n-1}p^{n-1}q + ... + a_1pq^{n-1} + a_0q^n = 0$$

Thus, $q$ must be a divisor of $a_n$ and $p$ a divisor of $a_0$.

Here, $a_n=1$ so any rational root must be an integer, and $a_0=24$, so if there is any integer root, it must be a divisor of $24$.

There are only a few possibilities: $1,2,3,4,6,8,12,24$, with plus or minus sign. So only a few checks to do. Once you get one root, you can reduce the degree by polynomial division.


As mentioned in the comments, here there is a faster way: with the extra hint that two of the three roots have product 12, since the product of the three roots is $-24$, that means the third root must be $-2$, so you can divide your polynomial by $x+2$ and find the two other either by the same reasoning, or by the quadratic formula.