91Ó°ÊÓ

First, let's list out all the factors of the constant term (1) and the leading coefficient (1). Factors of 1 are ±1. According to the Rational Root Theorem, the possible rational roots of the given polynomial should be a ratio of ±1 divided by ±1. Therefore, the only possible rational roots are ±1. Now we should test these two possible rational roots by plugging them into the polynomial: 1. \(f(1) = 1^4 - 2(1)^3 + 3(1) - 1 = 1 - 2 + 3 - 1 = 1\) which is not equal to 0. 2. \(f(-1) = (-1)^4 - 2(-1)^3 + 3(-1) - 1 = 1 + 2 - 3 - 1 = -1\) which is not equal to 0. Since neither ±1 is a zero of the function, we can't find any rational roots. Thus, we proceed to Step 2.

The Newton-Raphson method is a numerical approach to find the roots of a function up to a desired precision. To apply this method, we will need the derivative of the given polynomial. The polynomial is: \[ f(x) = x^4 - 2x^3 + 3x - 1 \] The derivative is: \[ f'(x) = 4x^3 - 6x^2 + 3 \] Applying the Newton-Raphson method, we iterate the following formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] First, we need an initial estimate \(x_0\). Let's take \(x_0 = 0\). Now, we iteratively apply the formula until the difference between successive x values is less than or equal to the desired precision (0.00001 in this case). Iteration 1: \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 0 - \frac{-1}{3} \approx 0.33333 \] Iteration 2: \[ x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} \approx 0.33149 \] Iteration 3: \[ x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} \approx 0.33146 \] As the difference between \(x_2\) and \(x_3\) is now less than the desired precision (0.00001), we can stop the iteration. The zero of the function \(f(x)\) up to five decimal places is approximately \(x \approx 0.33146\). Note that there might be further zeros of the function, depending on the equation. However, this method provides only one approximated zero at a time.