The area of the base is given by
\(B=\frac{V}{h}\)
We are given
\(V=2x^{3}-9x^{2}+7x+6\) and \(h=2x+1\)
\(\rightarrow B=\frac{2x^{3}-9x^{2}+7x+6}{2x+1}\)
which we can solve with long division:
x^2 -5x+6
2x+1 | 2x^3-9x^2+7x+6
2x^3+ x^2
--------------------------
-10x^2+7x
-10x^2-5x
--------------------
12x+6
12x+6
-----------
0
So \(B=x^{2}-5x+6\)
which can be factored as
\((x-2)(x-3)\)
So the length and width are \(x-2\) and \(x-3\).