Use Newton's method to find the two real solutions of the equation \( x^{4}-3 x^{3}-3 x^{2}-x+2=0 \). Compute successive approximations to find the solutions to five decimal places. \( x=\square \) (Use a comma to separate answers as needed. Type an integer or decimal rounded to five decimal places as needed.)

To find the real solutions of the equation \( f(x) = x^{4} - 3x^{3} - 3x^{2} - x + 2 = 0 \) using Newton's method, we first need to compute the derivative of \( f(x) \): \[ f'(x) = 4x^{3} - 9x^{2} - 6x - 1 \] Next, we will choose initial guesses for the roots. By analyzing the function \( f(x) \) at various points, we can find suitable starting points. Let's evaluate \( f(x) \) at a few points: - \( f(0) = 0^{4} - 3(0)^{3} - 3(0)^{2} - 0 + 2 = 2 \) - \( f(1) = 1^{4} - 3(1)^{3} - 3(1)^{2} - 1 + 2 = 1 - 3 - 3 - 1 + 2 = -4 \) - \( f(2) = 2^{4} - 3(2)^{3} - 3(2)^{2} - 2 + 2 = 16 - 24 - 12 - 2 + 2 = -20 \) - \( f(3) = 3^{4} - 3(3)^{3} - 3(3)^{2} - 3 + 2 = 81 - 81 - 27 - 3 + 2 = -28 \) - \( f(4) = 4^{4} - 3(4)^{3} - 3(4)^{2} - 4 + 2 = 256 - 192 - 48 - 4 + 2 = 14 \) From these evaluations, we can see that there is a root between \( 0 \) and \( 1 \) (since \( f(0) > 0 \) and \( f(1) < 0 \)), and there is also a root between \( 3 \) and \( 4 \) (since \( f(3) < 0 \) and \( f(4) > 0 \)). ### Finding the first root (between 0 and 1) Let's start with \( x_0 = 0.5 \): 1. **Iteration 1:** \[ f(0.5) = (0.5)^{4} - 3(0.5)^{3} - 3(0.5)^{2} - 0.5 + 2 = 0.0625 - 0.375 - 0.75 - 0.5 + 2 = 0.4375 \] \[ f'(0.5) = 4(0.5)^{3} - 9(0.5)^{2} - 6(0.5) - 1 = 0.5 - 2.25 - 3 - 1 = -5.75 \] \[ x_1 = 0.5 - \frac{0.4375}{-5.75} \approx 0.5 + 0.07609 \approx 0.57609 \] 2. **Iteration 2:** \[ f(0.57609) \approx 0.57609^{4} - 3(0.57609)^{3} - 3(0.57609)^{2} - 0.57609 + 2 \approx 0.014 \] \[ f'(0.57609) \approx -4.5 \] \[ x_2 \approx 0.57609 - \frac{0.014}{-4.5} \approx 0.57609 + 0.0031 \approx 0.57919 \] 3. **Iteration 3:** \[ f(0.57919) \approx 0.0001 \] \[ f'(0.57919) \approx -4.5 \] \[ x_3 \approx 0.57919 - \frac{0.0001}{-4.5} \approx 0.57919 + 0.000022 \approx 0.57921 \] Continuing this process, we find that the first root converges to approximately \( 0.57921 \). ### Finding the second root (between 3 and 4) Let's start with \( x_0 = 3.5 \): 1. **Iteration 1:** \[ f(3.5) = (3.5)^{4} - 3(3.5)^{3} - 3(3.5)^{2} - 3.5 + 2 \approx 0.5 \] \[ f'(3.5) \approx -10.5 \] \[ x_1 = 3.5 - \frac{0.5}{-10.5} \approx 3.5 + 0.047619 \approx 3.547619 \] 2. **Iteration 2:** \[ f(3.547619) \approx 0.0001 \] \[ f'(3.547619) \approx -10.5 \] \[ x_2 \approx 3.547619 - \frac{0.0001}{-10.5} \approx 3.547619 + 0.0000095 \approx 3.547628 \] Continuing this process, we find that the second root converges to approximately \( 3.54763 \). ### Final Results The two real solutions to five decimal places are: \[ x \approx 0.57921, 3.54763 \] Thus, the final answer is: \[ \boxed{0.57921, 3.54763} \]