Question

the solution set for the following
1.X3+x2-6x-10=0
2.2x2-4x-4=0
3.x4-x3-4x2+x+1=0

Answer 1

1. x^3 + x^2 - 6x - 10 = 0
let f(x) = x^3 + x^2 - 6x - 10
f(1) = ... ≠ 0
f(-1) = ... ≠ 0
f(2) = 8 + 4 - 12 - 10 ≠ 0
f(-2) = -8 + 4 + 12 - 10 ≠ 0
f(5) ≠0, f(-5) ≠ 0
all rational possiblilities have been exhausted, tough from here on in.
Wolfram gave me this ....
http://www.wolframalpha.com/input/?i=x%5E3+%2B+x%5E2+-+6x+-+10+%3D+0
x = appr 2.6631, and 2 complex roots

2. I will use completing the square ...
could use the formula, but that would be harder.
first divide by 2
-----> x^2 - 2x = 2
x^2 - 2x + 1 = 2+1
(x-1)^2 = 3
x-1 = ± √3
x = 1 ± √3

3. let f(x) = x^4 - x^3 - 4x^2 + x + 1 = 0
WOW, even nastier than the first one

Wolfram says this
http://www.wolframalpha.com/input/?i=x%5E4+-+x%5E3+-+4x%5E2+%2B+x+%2B+1+%3D+0

What method was suggested to you to solve these?
Have you learned Newton's Method ?