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`