Solve 2x3 + x2 - 15x completely by factoring.

a.
x = 0,3, 5/2

b.
x= -3, 5

c.
x = 0, -3, 5

d.
x = 0, -3, 5/2

 [D]:  " x = 0, -3, 5/2 "  .

{Assuming:  " 2x³ + x²  − 15x = 0 ".}.

Explanation:

Given:

     2x³ + x² − 15x ;
 
        →  (2x³ + x²) − 15x ;
  
        →   2x³ + x² − 15x  =   x * (2x² + x − 15) ;
         
        →  Factor the expression:  "(2x² + x − 15)" ;
                                     
             → (2x² + x − 15) =  
            
            →   2x² − 5x + 6x − 15 ;

           →  Add the "first TWO (2) terms", and pull out the "like factors" ;                                       →  x*(2x − 5) ;

           →  Add the "last TWO (2) terms, pulling out "common factors";
                            →  3*(2x − 5) ;

           →  Now, add up the previous FOUR (4) terms; to get:

                            → (x + 3)(2x − 5) ;  

           → Now, we have factored the:

                    "(2x² + x − 15)" of:  " x*(2x² + x − 15) " ; 

           →  So we add the "x"; and write the entire factored expression:

                      →  x*(x + 3)(2x − 5) ;           

Now, assume the question is asking to solve for "x" by factoring;
   when the expression is "equal to zero" ;

That is, when:  

                →  x*(x + 3)(2x − 5) = 0 ;

Since we have THREE (3) multiplicands;  and anything times "0" equals "0" ;
    this equation holds true when:

  1) x = 0 ;

  2) (x + 3) = 0 ;
  
Subtract "3" from each side of the equation;

   x + 3 − 3 = 0 − 3 ; 
  
         x = -3 ; 

3)  2x − 5 = 0 ; 

Add "5" to each side of the equation;

2x − 5 + 5 = 0 + 5 ;

2x = 5 ; Now, divide EACH side of the equation by "2" ; to isolate "x" on one side of the equation; and to solve for "x" ; 

2x/2 = 5/2 ;

x = 5/2; or, write as 2.5; or write as 2½ .

Put simply, the equation is true when:

    x = 0, -3, 5/2 ;  which is: Answer choice: [D].

Yes, two vertical lines can be complementary