Question

Find all the real solutions of x^3 - x^2 - 3x - 1 =0

Answer 1

f(-1) = -1^3 - (-1)^2 -3(-1) - 1 = = -1 - 1 + 3 - 1 = 0
so one solution is x = -1
and (x + 1) is a factor of the function

Answer 2

Dividing:

x^2 - 2x - 1
------------------
x+1) x^3 - x^2 - 3x - 1
x^3 + x^2
--------
-2x^2 - 3x
-2x^2 - 2x
----------
-x - 1
-x - 1
------

so f(x) = (x +1)(x^2 - 2x -1 )

Answer 3

to get the other roots

solve (x^2 - 2x - 1) = 0

Answer 4

x^3 - x^2 - 3x - 1 =0
f(x)=x^3 - x^2 - 3x - 1
now you have to take a value of x such that f(x)=0; if x=-1 then f(x)=0;lets see
f(-1)=(-1)^3-(-1)^2-3(-1)-1
=-1-1+3-1=-3+3=0
as for x=-1 f(x) is 0 so (x-(-1))=(x+1) is one of the factor of the equation
so now you can write
x^3 - x^2 - 3x - 1 =0;
=>x^2(x+1)-2x(x+1)-1(x+1)=0; [to be sure you can back calculate it]
=>(x+1)(x^2-2x-1)=0
so x+1=0 or x=-1
and x^2-2x-1=0 or x=[-2pm sqrt{(-2)^2-4*1*(-1)}\2*1] [pm means plus minus]
or x=1+sqrt (2 )or x= 1-sqrt (2)

ans: x=-1; 1+sqrt (2 ); x= 1-sqrt (2)