Verify my answer: Could $f(x) = 2x^4-7x^3-13x^2+68x-30$ have a rational zero between $5$ and $6$?

Question

I would like someone to verify my answer to the following question:

Let $f(x) = 2x^4-7x^3-13x^2+68x-30$. Could $f$ have a rational zero between $5$ and $6$? If so, what? If not, tell why not.

My answer:

$f(5) = 360$, $f(6) = 990$

Therefore, $f$ does not have a rational zero between $5$ and $6$ because one of the corollaries of Intermediate value theorem states the following:

If a continuous function has values of opposite sign inside an interval, then it has root in that interval.

But in this case, the signs of the values did not change and therefore, again, $f$ does not have a rational zero.

Is my argument correct? Please send me your suggestions.

Answer 1

You are right in thinking that if you could prove $f$ has no roots in the interval $(5, 6)$ then surely $f$ would have no rational roots in that interval. But if $f$ had a real, irrational root in $(5, 6)$ then you'd be stuck. For example, the slightly modified polynomial $$g(x) = 2x^4 - 7x^3 - \color{red}{31}x^2 + 68x - 30$$ has $g(5) = -90$ and $g(6) = 342$, so by IVT there is a root between $5$ and $6$ (it's $5.312...$). How can we tell if it's rational or not? Here's a technique that answers the question without appealing to continuity.

The rational root theorem says that if $f$ is a polynomial with integer coefficients and if $f(r/s) = 0$ for some rational number $r/s$ in lowest terms, then $r$ divides the constant term of $f$ and $s$ divides the leading term of $f$. Thus, any rational root of your $f$ has to have $r$ a divisor of $30$ and $s$ a divisor of $2$. So, $s \in \{1, 2\}$ and $r \in \{1, 2, 3, 5, 6, 10, 15, 30\}$. Forming all possible combinations $r/s$ we see that none of them lies between $5$ and $6$. So $f$ has no rational root there. (If "between" is supposed to include the endpoints, then, as you have already checked, $f(5)$ and $f(6)$ are not zero.)

Answer 2

$f(5)>0$ and $f(6)>0$, you cannot conclude anything from that, as it is stated that it works only if they have opposite signs.

You have to provide information about the monotony of the function.

Answer 3

Your argument isn't exactly correct.

Logically speaking, the IVT states $$\text{Continuous + Values of Opposite Sign in Interval} \implies \text{Root in Interval}$$ This does NOT imply that $$\text{Continuous + No Values of Opposite Sign in Interval} \implies \text{No Root in Interval}$$ Actually, for your own reference, the contrapositive would be true: $$\text{No Root in Interval} \implies \text{at least one of not Continuous or No Values of Opposite Sign in Interval}\text{.}$$ In general, if $P \implies Q$, then $\text{not }Q \implies \text{not }P$.

Particularly for this question, you've shown that $f(5)$ and $f(6)$ are both positive. But it is possible for a continuous function with these values to still hit $0$ in $(5, 6)$. As an extreme example:

and intuitively, $f$ is continuous.

What you need to show is that $f$ is monotonically increasing in $[5, 6]$ to claim that it never hits $0$ in $[5, 6]$:

Answer 4

$$f(x) = 2x^4-7x^3-13x^2+68x-30$$

By the rational root theorem, all rational roots, if they exist, would have numerators factors of $30$, and denominators factors of $2$. This is often written as $$\pm \dfrac{\{30,1,15,2,5,3\}}{\{2,1\}}$$. Check if there are any between $5$ and $6$.

Answer 5

But in this case, the signs of the values did not change and therefore, again, $f$ does not have a rational zero.

All you can derive from $f(5),f(6)$ being both positive is that the equation has an even number of roots in $(5,6)$ (including possibly none).

You could use the rational root theorem and check how many rationals between $5$ and $6$ can be written as $p/q$ with $p \mid 30$ and $q \mid 2$, then try each of them to see whether it's a root.

Or, since this looks suspiciously similar to Finding the zeroes of the function, you already know that there are only two real roots, and they are given by the equation $2 x^2 + 5x - 3 = 0$.