${2}^{2} \times {\left( -2 \right)}^{2}+8 \times \left( -2 \right)+3 \times \left( -2 \right)+6$
A negative base raised to an even power equals a positive${2}^{2} \times {2}^{2}+8 \times \left( -2 \right)+3 \times \left( -2 \right)+6$
Multiply the numbers${2}^{2} \times {2}^{2}-16+3 \times \left( -2 \right)+6$
Multiply the numbers${2}^{2} \times {2}^{2}-16-6+6$
Calculate the product${2}^{4}-16-6+6$
Since two opposites add up to $0$, remove them from the expression${2}^{4}-16$
Evaluate the power$16-16$
The sum of two opposites equals $0$$0$