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.
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.
On
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.
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?