Answer:[tex]3x^{2}-4x+1[/tex]
Step-by-step explanation:
Let [tex]l[/tex] be the length of the rectangle.
Let [tex]b[/tex] be the breadth of the rectangle.
The area of the rectangle with length [tex]l[/tex] an breadth [tex]b[/tex] is given by [tex]l\times b[/tex]
Given that [tex]b=x+6[/tex]
Given that area is [tex]3x^{3}+14x^{2}-23x+6[/tex]
Option A:
[tex]lb=(x+6)(3x^{2}-4x+1)=3x^{3}-4x^{2}+x+18x^{2}-24x+6=3x^{3}+14x^{2}-23x+6[/tex]
This is the area given.So,option A is correct.
Option B:
[tex]lb=(x+6)(3x^{2}+4x-1)=3x^{3}+4x^{2}-x+18x^{2}+24x-6=3x^{3}+22x^{2}+23x-6[/tex]
But this is not the area given.So,option B is wrong.
Option C:
[tex]lb=(x+6)(3x^{2}-4x-1)=3x^{3}-4x^{2}-x+18x^{2}-24x-6=3x^{3}+14x^{2}-25x-6[/tex]
But this is not the area given.So,option C is wrong.
Option D:
[tex]lb=(x+6)(3x^{2}+4x+1)=3x^{3}+4x^{2}-x+18x^{2}-24x-6=3x^{3}+14x^{2}-25x-6[/tex]
But this is not the area given.So,option D is wrong.
