1. **Problem Statement:** Find all possible rational zeros of the polynomial $x^3 - 5x + 4$ using the Rational Root Theorem and then find the actual rational zeros. 2. **Rational Root Theorem:** Possible rational zeros are of the form $\pm \frac{p}{q}$, where $p$ divides the constant term and $q$ divides the leading coefficient. 3. For $x^3 - 5x + 4$, the constant term is 4 and the leading coefficient is 1. 4. Possible values of $p$ (factors of 4): $\pm 1, \pm 2, \pm 4$. 5. Possible values of $q$ (factors of 1): $\pm 1$. 6. Therefore, possible rational zeros are $\pm 1, \pm 2, \pm 4$. 7. **Test each candidate by substitution:** - For $x=1$: $1^3 - 5(1) + 4 = 1 - 5 + 4 = 0$ so $x=1$ is a zero. - For $x=-1$: $(-1)^3 - 5(-1) + 4 = -1 + 5 + 4 = 8 \neq 0$. - For $x=2$: $2^3 - 5(2) + 4 = 8 - 10 + 4 = 2 \neq 0$. - For $x=-2$: $(-2)^3 - 5(-2) + 4 = -8 + 10 + 4 = 6 \neq 0$. - For $x=4$: $4^3 - 5(4) + 4 = 64 - 20 + 4 = 48 \neq 0$. - For $x=-4$: $(-4)^3 - 5(-4) + 4 = -64 + 20 + 4 = -40 \neq 0$. 8. Since $x=1$ is a root, divide the polynomial by $x-1$ using synthetic division: $$\begin{array}{r|rrrr} 1 & 1 & 0 & -5 & 4 \\ & & 1 & 1 & -4 \\ \hline & 1 & 1 & -4 & 0 \end{array}$$ 9. The quotient is $x^2 + x - 4$. 10. Solve $x^2 + x - 4 = 0$ using the quadratic formula: $$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-4)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2}$$ 11. These roots are irrational, so the only rational zero is $x=1$. **Final answer:** Possible rational zeros: $\pm 1, \pm 2, \pm 4$. Actual rational zero: $x=1$.