according to the rational root theorem, $-\frac{7}{8}$ is a potential rational root of which…

Explanation:

Step1: Recall Rational Root Theorem

The Rational Root Theorem states that if a polynomial function $f(x)=a_nx^n + a_{n - 1}x^{n-1}+\cdots+a_1x + a_0$ has integer coefficients, then a potential rational root is of the form $\frac{p}{q}$, where $p$ is a factor of the constant term $a_0$ and $q$ is a factor of the leading - coefficient $a_n$. Here, the potential root is $-\frac{7}{8}$, so $p=\pm7$ and $q = \pm8$.

Step2: Check leading - coefficient and constant term for each function

For $f(x)=24x^7 + 3x^6+4x^3 - x - 28$

The leading - coefficient $a_n = 24$ and the constant term $a_0=-28$. The factors of $24$ are $\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12,\pm24$ and the factors of $28$ are $\pm1,\pm2,\pm4,\pm7,\pm14,\pm28$. Since $q = 8$ is a factor of $24$ and $p = 7$ is a factor of $28$, $-\frac{7}{8}$ is a potential rational root.

For $f(x)=28x^7 + 3x^6+4x^3 - x - 24$

The leading - coefficient $a_n = 28$ and the constant term $a_0=-24$. The factors of $28$ are $\pm1,\pm2,\pm4,\pm7,\pm14,\pm28$ and the factors of $24$ are $\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12,\pm24$. Since $q = 8$ is not a factor of $28$, $-\frac{7}{8}$ is not a potential rational root.

For $f(x)=30x^7 + 3x^6+4x^3 - x - 56$

The leading - coefficient $a_n = 30$ and the constant term $a_0=-56$. The factors of $30$ are $\pm1,\pm2,\pm3,\pm5,\pm6,\pm10,\pm15,\pm30$ and the factors of $56$ are $\pm1,\pm2,\pm4,\pm7,\pm8,\pm14,\pm28,\pm56$. Since $q = 8$ is not a factor of $30$, $-\frac{7}{8}$ is not a potential rational root.

For $f(x)=56x^7 + 3x^6+4x^3 - x - 30$

The leading - coefficient $a_n = 56$ and the constant term $a_0=-30$. The factors of $56$ are $\pm1,\pm2,\pm4,\pm7,\pm8,\pm14,\pm28,\pm56$ and the factors of $30$ are $\pm1,\pm2,\pm3,\pm5,\pm6,\pm10,\pm15,\pm30$. Since $q = 8$ is not a factor of $30$, $-\frac{7}{8}$ is not a potential rational root.

Answer:

$f(x)=24x^7 + 3x^6+4x^3 - x - 28$