binomial therem

((1/2)x-3)^4

$$=\left(\frac{x}{2}-3\right)^4=\left(\frac{x-6}{2}\right)^4=\frac{1}{16}(x-6)^4$$

$$=\frac{1}{16} ( \ \textcolor[rgb]{1,0,0}{1} *x^{\textcolor[rgb]{0,1,0}{4}}*6^{\textcolor[rgb]{0,0,1}{0}} -\textcolor[rgb]{1,0,0}{4} *x^{\textcolor[rgb]{0,1,0}{3}}*6^{\textcolor[rgb]{0,0,1}{1}} +\textcolor[rgb]{1,0,0}{6} *x^{\textcolor[rgb]{0,1,0}{2}}*6^{\textcolor[rgb]{0,0,1}{2}} -\textcolor[rgb]{1,0,0}{4} *x^{\textcolor[rgb]{0,1,0}{1}}*6^{\textcolor[rgb]{0,0,1}{3}} +\textcolor[rgb]{1,0,0}{1} *x^{\textcolor[rgb]{0,1,0}{0}}*6^{\textcolor[rgb]{0,0,1}{4}} \ )$$     $$\left[ \small{\text{Pascal 4: 1,4,6,4,1}} \right]$$

$$=\frac{1}{16} (\ x^4-4*x^3*6+6*x^2*6^2-4*x*6^3+6^4 \ )$$

$$=\frac{1}{16} ( \ x^4-24x^3+216x^2-864x+1296 \ )$$

$$= \frac{1}{16}x^4 -\frac{3}{2} x^3 +\frac{27}{2}x^2 -54x +81$$