Answer:
[tex]\left(\frac{1}{4}m-2n\right)^2=\frac{m^2}{16}-mn+4n^2[/tex]
Step-by-step explanation:
Given the polynomial
[tex]\left(\frac{1}{4}m-2n\right)^2[/tex]
Expanding and simplifying
[tex]\left(\frac{1}{4}m-2n\right)^2[/tex]
as 1/4 = 4/m, so
[tex]=\left(\frac{m}{4}-2n\right)^2[/tex]
Apply perfect square formula: [tex]\left(a-b\right)^2=a^2-2ab+b^2[/tex]
[tex]=\left(\frac{m}{4}\right)^2-2\cdot \frac{m}{4}\cdot \:2n+\left(2n\right)^2[/tex]
simplifying
[tex]=\frac{m^2}{4^2}-mn+2^2n^2[/tex]
[tex]=\frac{m^2}{16}-mn+4n^2[/tex]
Therefore,
[tex]\left(\frac{1}{4}m-2n\right)^2=\frac{m^2}{16}-mn+4n^2[/tex]
