3 + -1x + -3x^2 + x^3 = 0

Question

solve for x 

Answer 1

 

 X^3 - 3X^2 - X + 3 = 0

                                           the rational roots can be;  ± 1, ± 3

 

Try x = 1

 

( 1) ^ 3 - 3 ( 1)^2 - ( 1) + 3 =

 

  1 - 3 - 1 + 3 = 0 , it works.

 

 Rule: Anytime coefficients of any polynomial adds up to 0, like here  1 -3 - 1 + 3 =0, then 

   X = 1 is a root of the polynomial, and Polynomial is divisible by ( X - 1 ).

 

           _X^2 - 2X - 3______________    

 X - 1   l      X^3 -3 X^2 - X +3

-                 X^3 - X ^2

                        +                  

                        - 2X^2 - X

                        - 2X^2 +2X

                                   -        

                                 -3X + 3

                                 - 3X +3

                                  +    -     

                                     0

 X^3 - 3 X^2 - 2X -3 = ( X - 1) ( X^2 -2X - 3 ) =( X-1) ( X -3) ( X+1) =0

 

 

    X = 1   ,  X = -1   , X = 3

 

   

                                   

Answer 2

Hi Floli;

3 + -1x + -3x^2 + x^3 = 0

First, let's note that -1x is -x.

Second, let's note that +-x is -x.

Third, let's note that +-3x² is -3x².

Adding a negative number is the same as subtracting a positive number.

This is now...

3-x-3x²+x³=0

Let's rewrite in descending order...

x³-3x²-x+3=0

For the FOIL...

FIRST must be (x²)(x)=x³

OUTER or INNER must be -3x²

OUTER or INNER must be -x

LAST must be (3)(1) or (1)(3) and both numbers must be negative to result in +3 as well as resulting in a negative OUTER and a negative INNER.

(x²-1)(x-3)=0

Let's FOIL...

FIRST(x²)(x)=x³

OUTER...(x²)(-3)=-3x²

INNER...(-1)(x)=-1x=-x

LAST...(-1)(-3)=3

x³-3x²-x+3

(x²-1)(x-3)=0

Either or both parenthetical equations must equal zero.

x²-1=0

x=1, x=-1

x-3=0

x=3