Solve {left(1/4z+2right)}^2{left(1/4z-2right)}^2

$${ \left( \frac{ 1 }{ 4 } z+2 \right) }^{ 2 } { \left( \frac{ 1 }{ 4 } z-2 \right) }^{ 2 }$$

Evaluate

$$\frac{\left(z^{2}-64\right)^{2}}{256}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{1}{4}z+2\right)^{2}.

$$\left(\frac{1}{16}z^{2}+z+4\right)\left(\frac{1}{4}z-2\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{4}z-2\right)^{2}.

$$\left(\frac{1}{16}z^{2}+z+4\right)\left(\frac{1}{16}z^{2}-z+4\right)$$

Use the distributive property to multiply \frac{1}{16}z^{2}+z+4 by \frac{1}{16}z^{2}-z+4 and combine like terms.

$$\frac{1}{256}z^{4}-\frac{1}{2}z^{2}+16$$

Expand

$$\frac{z^{4}}{256}-\frac{z^{2}}{2}+16$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{1}{4}z+2\right)^{2}.

$$\left(\frac{1}{16}z^{2}+z+4\right)\left(\frac{1}{4}z-2\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{4}z-2\right)^{2}.

$$\left(\frac{1}{16}z^{2}+z+4\right)\left(\frac{1}{16}z^{2}-z+4\right)$$

Use the distributive property to multiply \frac{1}{16}z^{2}+z+4 by \frac{1}{16}z^{2}-z+4 and combine like terms.

$$\frac{1}{256}z^{4}-\frac{1}{2}z^{2}+16$$