Question
The base of a triangle in terms of x is: 2^2+4x+2 and the height is x^2+3x-4
Area=1/2(2x^2+4x+2) (x^2+3x-4)= 6x^2+2 (x^2+3x-4)
Trying to find the area and not sure which direction to go
Area=1/2(2x^2+4x+2) (x^2+3x-4)= 6x^2+2 (x^2+3x-4)
Trying to find the area and not sure which direction to go
Answers
Steve
If you're trying to find the expression for the area, you have for some reason combined the x^2 and x terms (a no-no). The area is
1/2(2x^2+4x+2)(x^2+3x-4)
(x+1)^2 (x+4)(x-1)
As you say, not sure which way to go. You have an expression for the area. Was there some other objective?
1/2(2x^2+4x+2)(x^2+3x-4)
(x+1)^2 (x+4)(x-1)
As you say, not sure which way to go. You have an expression for the area. Was there some other objective?
Ashley
re-evaluate your terms. I got x^4 +5x^3 -x^2 -5x-4.
my work:
.5(2x^2 +4x +2)(x^2 +3x -4)=.5(2x^4+6x^3-8x^2+4x^3+4x^2 -16x+2x^2 +6x-8)
my work:
.5(2x^2 +4x +2)(x^2 +3x -4)=.5(2x^4+6x^3-8x^2+4x^3+4x^2 -16x+2x^2 +6x-8)