Given:
The length of the rectangle is:
[tex]l=x^2-x+3[/tex]
The width of the rectangle is:
[tex]w=4x+1[/tex]
To find:
The area of the given rectangle in the simplest form.
Solution:
We know that, the area of a rectangle is:
[tex]A=l\times w[/tex]
Where, l is the length and w is the width of the rectangle.
Using the above formula, the area of the given rectangle is:
[tex]A=(x^2-x+3)\times (4x+1)[/tex]
[tex]A=x^2(4x)+x^2(1)-x(4x)-x(1)+3(4x)+3(1)[/tex]
[tex]A=4x^3+x^2-4x^2-x+12x+3[/tex]
[tex]A=4x^3-3x^2+11x+3[/tex]
Therefore, the area of the given rectangle is [tex](4x^3-3x^2+11x+3)[/tex] sq. units.
