Problem with the rational root theorem

Question

Consider this polynomial: $f(x)=(2x+5)(x-3)(x+8/3)=0$. Then $f(x)=2x^3+...+(-40)$

Here is a list of all factors of $40$ and $2$:

$40$: $±1$, $±2$, $±4$, $±5$, $±8$, $±10$, $±20$

$2$: $±2$, $±1$

Now, $3$ is clearly a root, but there is no combination between factors of $40$ divided by factors of $2$ that gives $3$ as a result. What am I doing wrong?

Answer 1

The problem is that for the rational root theorem to work, we need all the coefficients to be integral.

Expanding, we have

$$f(x)=(2x+5)(x-3)(x+8/3)=2x^3+\frac{13}{3}x^2-\frac{53}{3}x-40$$

Now notice that the solutions to $f(x)=0$ are the same as the solutions to $3f(x)=0$, so we can safely multiply through by $3$ to obtain the equation

$$6x^3+13x^2-53x-120=0$$

Now because all of the coefficients are integral, we can apply the rational root theorem and find that $3$ may be a root (since $3\mid120$).

Answer 2

You don't have all integer coefficients of your middle terms in the expanded polynomial. Multiplying through by three will get rid of fractional coefficients and give you the possibility of 3 being a root.

The rational root theorem requires all integer coefficients.