x = -1
x =(1-√5)/-2= 0.618
x =(1+√5)/-2=-1.618
Step  1  :
Equation at the end of step  1  :
 0 -  (((x3) +  2x2) -  1)  = 0 Â
Step  2  : Â
Step  3  :
Pulling out like terms :
3.1 Â Â Pull out like factors :
 -x3 - 2x2 + 1  =  -1 • (x3 + 2x2 - 1) Â
3.2 Â Â Find roots (zeroes) of : Â Â Â F(x) = x3 + 2x2 - 1
Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  x  which can be expressed as the quotient of two integers
The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q  then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient
In this case, the Leading Coefficient is  1  and the Trailing Constant is  -1.
The factor(s) are:
of the Leading Coefficient : Â 1
of the Trailing Constant : Â 1
Let us test
 P   Q   P/Q   F(P/Q)   Divisor
   -1    1     -1.00     0.00    x + 1 Â
   1    1     1.00     2.00   Â
The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms
In our case this means that
 x3 + 2x2 - 1 Â
can be divided with  x + 1 Â
Polynomial Long Division :
3.3 Â Â Polynomial Long Division
Dividing : Â x3 + 2x2 - 1 Â
               ("Dividend")
By     :   x + 1   ("Divisor")
dividend   x3  +  2x2    -  1 Â
- divisor  * x2   x3  +  x2     Â
remainder     x2    -  1 Â
- divisor  * x1     x2  +  x   Â
remainder      -  x  -  1 Â
- divisor  * -x0      -  x  -  1 Â
remainder         0
Quotient : Â x2+x-1 Â Remainder: Â 0 Â
Trying to factor by splitting the middle term
3.4   Factoring  x2+x-1 Â
The first term is,  x2  its coefficient is  1 .
The middle term is,  +x  its coefficient is  1 .
The last term, "the constant", is  -1 Â
Step-1 : Multiply the coefficient of the first term by the constant  1 • -1 = -1 Â
Step-2 : Find two factors of  -1  whose sum equals the coefficient of the middle term, which is  1 .
   -1   +   1   =   0 Â
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step  3  :
 (-x2 - x + 1) • (x + 1)  = 0 Â
Step  4  :
Theory - Roots of a product :
4.1 Â Â A product of several terms equals zero. Â
When a product of two or more terms equals zero, then at least one of the terms must be zero. Â
We shall now solve each term = 0 separately Â
In other words, we are going to solve as many equations as there are terms in the product Â
Any solution of term = 0 solves product = 0 as well.
Parabola, Finding the Vertex :
4.2    Find the Vertex of  y = -x2-x+1
For any parabola,Ax2+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is  -0.5000 Â
Plugging into the parabola formula  -0.5000  for  x  we can calculate the  y -coordinate : Â
 y = -1.0 * -0.50 * -0.50 - 1.0 * -0.50 + 1.0
or  y = 1.250
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : Â y = -x2-x+1
Axis of Symmetry (dashed) Â {x}={-0.50} Â
Vertex at  {x,y} = {-0.50, 1.25} Â
x -Intercepts (Roots) :
Root 1 at  {x,y} = { 0.62, 0.00} Â
Root 2 at  {x,y} = {-1.62, 0.00} Â
Solve Quadratic Equation by Completing The Square
4.3   Solving  -x2-x+1 = 0 by Completing The Square .
Multiply both sides of the equation by  (-1)  to obtain positive coefficient for the first term:
x2+x-1 = 0  Add  1  to both side of the equation :
 x2+x = 1
Now the clever bit: Take the coefficient of  x , which is  1 , divide by two, giving  1/2 , and finally square it giving  1/4 Â
Add  1/4  to both sides of the equation :
 On the right hand side we have :
 1  +  1/4   or,  (1/1)+(1/4) Â
 The common denominator of the two fractions is  4  Adding  (4/4)+(1/4)  gives  5/4 Â
 So adding to both sides we finally get :
 x2+x+(1/4) = 5/4
Adding  1/4  has completed the left hand side into a perfect square :
 x2+x+(1/4)  =
 (x+(1/2)) • (x+(1/2))  =
 (x+(1/2))2
Things which are equal to the same thing are also equal to one another. Since
 x2+x+(1/4) = 5/4 and
 x2+x+(1/4) = (x+(1/2))2
then, according to the law of transitivity,
 (x+(1/2))2 = 5/4
We'll refer to this Equation as  Eq. #4.3.1 Â
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
 (x+(1/2))2  is
 (x+(1/2))2/2 =
 (x+(1/2))1 =
 x+(1/2)
Now, applying the Square Root Principle to  Eq. #4.3.1  we get:
 x+(1/2) = √ 5/4
Subtract  1/2  from both sides to obtain:
 x = -1/2 + √ 5/4
Since a square root has two values, one positive and the other negative
 x2 + x - 1 = 0
 has two solutions:
 x = -1/2 + √ 5/4
 or
 x = -1/2 - √ 5/4
Note that  √ 5/4 can be written as
 √ 5  / √ 4  which is √ 5  / 2
Solve Quadratic Equation using the Quadratic Formula
4.4   Solving   -x2-x+1 = 0 by the Quadratic Formula .
According to the Quadratic Formula,  x  , the solution for  Ax2+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                  Â
      - B  ±  √ B2-4AC
 x =  ————————
           2A
 In our case,  A  =   -1
           B  =   -1
           C  =  1
Accordingly, Â B2 Â - Â 4AC Â =
          1 - (-4) =
          5
Applying the quadratic formula :
       1 ± √ 5
 x  =   ————
         -2
 √ 5  , rounded to 4 decimal digits, is  2.2361
So now we are looking at:
     x  =  ( 1 ±  2.236 ) / -2
Two real solutions:
x =(1+√5)/-2=-1.618
or:
x =(1-√5)/-2= 0.618
Solving a Single Variable Equation :
4.5    Solve  :   x+1 = 0 Â
Subtract  1  from both sides of the equation : Â
           x = -1
Hope this helps.