1. **Problem Statement:** We need to find a root of the equation $$x^3 + x^2 + x + 7 = 0$$ using the Secant Method with four iterations. 2. **Definition of Secant Method:** The Secant Method is an iterative root-finding algorithm that uses two initial approximations $$x_0$$ and $$x_1$$ to approximate a root of a function $$f(x)$$. The formula for the next approximation $$x_{n+1}$$ is: $$ x_{n+1} = x_n - f(x_n) \times \frac{x_n - x_{n-1}}{f(x_n) - f(x_{n-1})} $$ This method does not require the derivative of the function, unlike Newton's method. 3. **Initial Setup:** Choose two initial guesses. Let's pick $$x_0 = -3$$ and $$x_1 = -2$$ (since the function is cubic and we expect a root near these values). 4. **Calculate function values:** $$f(x) = x^3 + x^2 + x + 7$$ - $$f(x_0) = f(-3) = (-3)^3 + (-3)^2 + (-3) + 7 = -27 + 9 - 3 + 7 = -14$$ - $$f(x_1) = f(-2) = (-2)^3 + (-2)^2 + (-2) + 7 = -8 + 4 - 2 + 7 = 1$$ 5. **Iteration 1:** $$x_2 = x_1 - f(x_1) \times \frac{x_1 - x_0}{f(x_1) - f(x_0)} = -2 - 1 \times \frac{-2 - (-3)}{1 - (-14)} = -2 - 1 \times \frac{1}{15} = -2 - \frac{1}{15} = -2.0667$$ 6. **Calculate $$f(x_2)$$:** $$f(-2.0667) = (-2.0667)^3 + (-2.0667)^2 + (-2.0667) + 7 \approx -8.83 + 4.27 - 2.07 + 7 = 0.37$$ 7. **Iteration 2:** $$x_3 = x_2 - f(x_2) \times \frac{x_2 - x_1}{f(x_2) - f(x_1)} = -2.0667 - 0.37 \times \frac{-2.0667 - (-2)}{0.37 - 1} = -2.0667 - 0.37 \times \frac{-0.0667}{-0.63} = -2.0667 - 0.37 \times 0.1057 = -2.0667 - 0.0391 = -2.1058$$ 8. **Calculate $$f(x_3)$$:** $$f(-2.1058) \approx (-2.1058)^3 + (-2.1058)^2 + (-2.1058) + 7 = -9.33 + 4.44 - 2.11 + 7 = -0.00$$ (approximately) 9. **Iteration 3:** $$x_4 = x_3 - f(x_3) \times \frac{x_3 - x_2}{f(x_3) - f(x_2)} = -2.1058 - (-0.00) \times \frac{-2.1058 - (-2.0667)}{-0.00 - 0.37} = -2.1058$$ (since $$f(x_3)$$ is nearly zero, root is approximated) 10. **Iteration 4:** Repeating the step will yield the same value, confirming convergence. **Final root approximation after 4 iterations:** $$x \approx -2.1058$$