Answer:
Perimeter of the rectangle=6x+8 square units
Step-by-step explanation:
Given that area of rectangle is [tex]2x^2+7x+3[/tex]
Area of rectangle=lw square units
[tex]2x^2+7x+3=2x^2+x+6x+3[/tex]
[tex]=x(2x+1)+3(2x+1)[/tex]
[tex]=(x+3)(2x+1)[/tex]
[tex]2x^2+7x+3=(x+3)(2x+1)[/tex]
Comparing the above equation with the given area we get
lw=(x+3)(2x+1)
Therefore length=x+3 and width=2x+1
To find the perimeter :
Perimeter of the rectangle=2(l+w) square units
[tex]=2((x+3)+(2x+1))[/tex]
[tex]=2(x+3+2x+1)[/tex]
[tex]=2(3x+4)[/tex]
[tex]=6x+8[/tex]
Therefore perimeter of the rectangle=6x+8 square units
