positive real root of the equation using Horner's Method

Question

obtain a positive real root of the equation $$x^3-3x^2+2.5=0$$ that lines between $1$ and $2$ using Horner's method.

Answer 1

Use Horner's Method to find a positive real root of:

$$\tag 1 x^3-3x^2+2.5 = 0$$

We re-write $(1)$ as:

$$\tag 2 2x^3-6x^2+5 = 0$$

I will assume you have access to Horner's algorithm and the intermediate values below are from the synthetic division steps.

• $x_0 = 1, a_3 = 2, a_2 = -6, a_1 = 0, a_0 = 5$
• $b_3 = 2, b_2 = -4, b_1 = -4, b_0 = 1 = P(1)$
• $b'_2 = 2, b'_1 = -2, b'_0 = -6 = Q(1) = P'(1)$

$$x_1= x_0 - \dfrac{P(x_0)}{P'(x_0)} = 1 - \dfrac{1}{-6} = \dfrac{7}{6} = 1.1666$$

Repeat process with $x_1 = 1.1666$, so we get:

$$x_2= x_1 - \dfrac{P(x_1)}{P'(x_1)} = 1.1666 - \dfrac{0.0081}{-6.80214} = 1.16779$$

Repeat this process a couple of more times to improve the accuracy since it is easy to do by hand.

The actual root (we already have two digit accuracy from above) for comparison purposes is:

$$x = 1.16825440178103$$