Find the area of a triangle whose base is x^2 + 2x + 4 and whose height is 2x^2 + 2x + 6

  The area is:  "(x² + x + 3) (x² + 2x + 4)" square units ;
                   or,  write as:

              " x⁴ + 3x³ + 9x² + 10x + 12" square units.

Explanation:

Area of a triangle = ½ * (base length) * (height) ; or:
 
A = ½ * b * h ;   Plug in values given:

A = ½ * (x² + 2x + 4) * (2x² + 2x + 6) ;

↔  A =  ½ * (2x² + 2x + 6) * (x² + 2x + 4) ;


Note:  ½ * (2x² + 2x + 6) = (2x² + 2x + 6) / 2 =

                                                             x² + x + 3 ;

Rewrite:

A = "(x² + x + 3) (x² + 2x + 4)" square units ; or expand further and solve:

  
x² * x² = x⁽²⁺²⁾ = x⁴ ;

x² * 2x =  2x³ ;

x² * 4 =  4x² ;

x * x² = x¹ * x² = x⁽¹⁺²⁾ = x³ ;

x * 2x  = 2x² ;

x * 4 = 4x ;

3 * x² = 3x² ;

3 * 2x = 6x ;

3 * 4 = 12

So, we have:  x⁴ + 2x³ + 4x² + x³ + 2x² + 4x + 3x² + 6x + 12 ;

   → Combine the "like terms" ;

 +2x³ + x³ = 3x³
 +4x² + 2x² + 3x² = 9x²
 +4x + 6x = 10x

And we have:  " x⁴ + 3x³ + 9x² + 10x + 12" square units .

Where are the following equations?

Same length is comparing because it’s saying its the same and not different