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

Question

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

x^5 - x^4 - 5x^3 - x^2 + 6x + 3 = 0

Answer 1

From a graph we can see that there are three roots, at roughly -0.6, 1.2, 2.7.
root 1:
. . . x . . . . . . . . . . / . . . f(x) . . . . . . . / . . f'(x) . . .
-0.6 . . . . . . . . . . . . . . -0.08736 . . . . . 3.312
-0.573623188 . . . 2.576×10^-3. . . . 3.5079
-0.574357735 . . . 2.019×10^-6 . . . 3.5024
-0.574358311 . . . 1.243×10^-12 . . 3.5024

So root 1 is -0.57435831 to 8 d.p. (next correction term is approximately 4×10^-13, so this estimate is actually correct to 12 d.p.)

root 2:
. . . x . . . . . . . . . . / . . . f(x) . . . . . . . / . . f'(x) . . .
1.2 . . . . . . . . . . . . . . . 0.53472 . . . . . -14.544
1.236765677 . . . . -0.013766 . . . . -15.2861
1.235865124 . . . -7.961×10^-6 . . . -15.2684
1.235864603 . . . -2.672×10^-12 . . -15.2684
So root 2 is 1.23566460 to 8 d.p.

root 3:
. . . x . . . . . . . . . . / . . . f(x) . . . . . . . / . . f'(x) . . .
2.7 . . . . . . . . . . . . . . 3.83997 . . . . . 78.2385
2.650919688 . . . . 0.262128 . . . . . 67.6915
2.647047297 . . . 1.5467×10^-3 . . 66.89345
2.647024174 . . . 5.4926×10^-8 . . 66.88870
2.647024173 . . . -7.105×10^-15 . . 66.88870
So root 3 is 2.64702417 to 8 d.p.

Answer 2

Why can't you do it? Eh?