To solve this using Newton's method, we use the formula:

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

Where:

• $f(x) = 5x^3 + 2x^2 + 3$
• $f'(x) = 15x^2 + 4x$

Step 1: Compute $f(x)$ and $f'(x)$

$f(x) = 5x^3 + 2x^2 + 3$ $f'(x) = 15x^2 + 4x$

Step 2: Use $x_1 = -2$ to find $x_2$

1. Compute $f(-2)$: $f(-2) = 5(-2)^3 + 2(-2)^2 + 3 = -40 + 8 + 3 = -29$

2. Compute $f'(-2)$: $f'(-2) = 15(-2)^2 + 4(-2) = 60 - 8 = 52$

3. Calculate $x_2$: $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = -2 - \frac{-29}{52} = -2 + \frac{29}{52}$ $x_2 = -2 + 0.5577 \approx -1.4423$

Step 3: Use $x_2 = -1.4423$ to find $x_3$

1. Compute $f(-1.4423)$: $f(-1.4423) = 5(-1.4423)^3 + 2(-1.4423)^2 + 3$ $f(-1.4423) \approx 5(-2.9995) + 2(2.0792) + 3 \approx -14.9975 + 4.1584 + 3 \approx -7.8391$

2. Compute $f'(-1.4423)$: $f'(-1.4423) = 15(-1.4423)^2 + 4(-1.4423)$ $f'(-1.4423) \approx 15(2.0792) + 4(-1.4423) \approx 31.188 - 5.7692 \approx 25.4188$

3. Calculate $x_3$: $x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = -1.4423 - \frac{-7.8391}{25.4188}$ $x_3 \approx -1.4423 + 0.3083 \approx -1.1340$

Final Approximations:

1. $x_2 \approx -1.4423$
2. $x_3 \approx -1.1340$

Would you like a detailed explanation of the steps or further examples?

5 Related Questions:

1. How does Newton's method converge to a solution?
2. What happens if the initial approximation $x_1$ is far from the root?
3. Can Newton's method fail to converge? If so, why?
4. How is Newton's method applied to equations with multiple roots?
5. What are some alternative methods for finding roots of nonlinear equations?

Tip:

Newton's method converges faster if the initial guess is close to the actual root, but it can fail if $f'(x)$ is very small or if the function has inflection points near the root.