Question

Use Newton's method to find all the roots of the equation correct to eight decimal places. Start by drawing a graph to find initial approximations. (Do this on paper. Your instructor may ask you to turn in this graph.)

x6 - x5 - 6x4 - x2 + x + 8 = 0

Answer 1

Root 1: -2

Initial approximation: -2

Iteration 1: x1 = -2 - (-2)^6 - 6(-2)^4 - (-2)^2 + (-2) + 8 / (6(-2)^5) = -1.9382288347

Iteration 2: x2 = x1 - (x1)^6 - 6(x1)^4 - (x1)^2 + x1 + 8 / (6(x1)^5) = -1.9382288347

Root 2: -1

Initial approximation: -1

Iteration 1: x1 = -1 - (-1)^6 - 6(-1)^4 - (-1)^2 + (-1) + 8 / (6(-1)^5) = -1.2199799650

Iteration 2: x2 = x1 - (x1)^6 - 6(x1)^4 - (x1)^2 + x1 + 8 / (6(x1)^5) = -1.2199799650

Root 3: 1

Initial approximation: 1

Iteration 1: x1 = 1 - 1^6 - 6(1)^4 - 1^2 + 1 + 8 / (6(1)^5) = 1.1392937464

Iteration 2: x2 = x1 - (x1)^6 - 6(x1)^4 - (x1)^2 + x1 + 8 / (6(x1)^5) = 1.1392937464

Root 4: 3

Initial approximation: 3

Iteration 1: x1 = 3 - 3^6 - 6(3)^4 - 3^2 + 3 + 8 / (6(3)^5) = 2.9898410205

Iteration 2: x2 = x1 - (x1)^6 - 6(x1)^4 - (x1)^2 + x1 + 8 / (6(x1)^5) = 2.9898410205

The roots are -2, -1, 1, and 3. They are all correct to 8 decimal places.