The expression

$\frac{1}{\sqrt{4x+1}}\left\{{\left(1+\frac{\sqrt{4x+1}}{2}\right)}^7-{\left(1-\frac{\sqrt{4x+1}}{2}\right)}^7\right\}$

is a polynomial in $x$ of degree.

We have,

$\begin{array}{l}\frac{1}{\sqrt{4x+1}}\left\{{\left(1+\frac{\sqrt{4x+1}}{2}\right)}^7-{\left(1-\frac{\sqrt{4x+1}}{2}\right)}^7\right\}\\ =\frac{1}{2^7\sqrt{4x+1}}\left[2\left\{{ }^7{C}_1\sqrt{4x+1}{+}^7{C}_3(\sqrt{4x+1}{)}^3\right.\left.\left.{+}^7{C}_5(\sqrt{4}x+1{)}^5{+}^7{C}_7(\sqrt{4x+1}{)}^7\right\}\right]\right.\\ \end{array}$

$=\frac{1}{2^6}\left\{{ }^7{C}_1{+}^7{C}_3(4x+1){+}^7{C}_5(4x+1{)}^2{+}^7{C}_7(4x+1{)}^3\right\}$

Clearly, it is a polynomial of degree 3.