First draw the graph using a calculator or computer

Approximate the four roots (where the graph cuts the x axis) giving something like

`x= -1.8,-1.1,1.1,2.8`

Newton's method involves refining our estimates using the recursive formula

`x_(n+1) = x_n - f(x_n)/(f'(x_n))`

Now, we have `f(x) = x^6-x^5 -5x^4 -x^2+x+6`

`implies` ...

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First draw the graph using a calculator or computer

Approximate the four roots (where the graph cuts the x axis) giving something like

`x= -1.8,-1.1,1.1,2.8`

Newton's method involves refining our estimates using the recursive formula

`x_(n+1) = x_n - f(x_n)/(f'(x_n))`

Now, we have `f(x) = x^6-x^5 -5x^4 -x^2+x+6`

`implies` `f'(x) = 6x^5 -5x^4 -20x^3 -2x +1`

For each initial estimate we get a sequence of improving approximations:

1) ` `x0 -1.8

x1 -1.76907576

x2 -1.76617802

x3 -1.76615375

x4 -1.76615375

2) x0 -1.1

x1 -1.08064658

x2 -1.08057763

x3 -1.08057763

3) x0 1.1

x1 1.05017360

x2 1.04708065

x3 1.04706901

x4 1.04706901

4) x0 2.8

x1 2.78786233

x2 2.78761037

x3 2.78761026

x4 2.78761026

**Therefore we have real 4 solutions (the other two are complex)**

`x = -1.76615375`, `-1.08057763`, `1.04706901` and `2.78761026`