# Use Newton's method to find all roots of the equation correct to eight decimal places?

I need four numbers. The equation is x^6 - x^5 - 7x^4 - x^2 + x + 6 = 0

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• 10 years ago

Let f(x) = x^6 - x^5 - 7x^4 - x^2 + x + 6 ==> 2 change in sign

Then f(-x) = x^6 + x^5 - 7x^4 - x^2 - x + 6 ==> 2 change in sign

By Descartes Rule of Sign

f(x) has 0 or 2 positive zero(s);

f(x) has 0 or 2 negative zero(s).

That's a maximum of 4 roots for the 6th degree polynomial.

By inspection

f(-3) = 399

f(-2) = -16

f(-1) = -1

f(0) = 6

f(1) = -1

f(2) = -76

f(3) = -81

f(4) = 1274

By Intermediate Value Theorem, there exist a value c1, c2, c3, c4 in the interval (-3, -2), (-1, 0), (0, 1) and (3, 4) respectively such that f(c1) = f(c2) = f(c3) = f(c4) = 0

That is 4 roots (2 positive, 2 negative) which agrees with our findings above.

Newton-Raphson Method:

f(x) = x^6 - x^5 - 7x^4 - x^2 + x + 6

f'(x) = 6x^5 - 5x^4 - 28x^3 - 2x + 1

Solve for c1, let x0 = -2.5

x1 = x0 - f(x0)/f'(x0) = -2.30574574

x2 = x1 - f(x1)/f'(x1) = -2.21874880

x3 = x2 - f(x2)/f'(x2) = -2.20146975

x4 = x3 - f(x3)/f'(x3) = -2.20083793

x5 = x4 - f(x4)/f'(x4) = -2.20083710

x6 = x5 - f(x5)/f'(x5) = -2.20083710

Since x6 agrees with x5 up to 8 decimal places, c1 = -2.20083710

Solve for c2, let x0 = -0.5

x1 = x0 - f(x0)/f'(x0) = -1.47187500

x2 = x1 - f(x1)/f'(x1) = -0.99801510

x3 = x2 - f(x2)/f'(x2) = -0.94982422

x4 = x3 - f(x3)/f'(x3) = -0.94751605

x5 = x4 - f(x4)/f'(x4) = -0.94751072

x6 = x5 - f(x5)/f'(x5) = -0.94751072

Since x6 agrees with x5 up to 8 decimal places, c2 = -0.94710722

Solve for c3, let x0 = 0.5

x1 = x0 - f(x0)/f'(x0) = 2.09913793

x2 = x1 - f(x1)/f'(x1) = 1.33711490

x3 = x2 - f(x2)/f'(x2) = 1.07611908

x4 = x3 - f(x3)/f'(x3) = 0.97758110

x5 = x4 - f(x4)/f'(x4) = 0.96272109

x6 = x5 - f(x5)/f'(x5) = 0.96240171

x7 = x6 - f(x6)/f'(x6) = 0.96240156

x8 = x7 - f(x7)/f'(x7) = 0.96240156

Since x8 agrees with x7 up to 8 decimal places, c3 = 0.96240156

Solve for c2, let x0 = 3.5

x1 = x0 - f(x0)/f'(x0) = 3.28245343

x2 = x1 - f(x1)/f'(x1) = 3.20410987

x3 = x2 - f(x2)/f'(x2) = 3.19451720

x4 = x3 - f(x3)/f'(x3) = 3.19438254

x5 = x4 - f(x4)/f'(x4) = 3.19438251

x6 = x5 - f(x5)/f'(x5) = 3.19438251

Since x6 agrees with x5 up to 8 decimal places, c4 = 3.19438251

Thus the roots of x^6 - x^5 - 7x^4 - x^2 + x + 6 = 0 correct up to eight decimal places is -2.20083710, -0.94751072, 0.96240156, 3.19438251

Confirmation:

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http://www.wolframalpha.com/input/?i=x^6+-+x^5+-+7...