Solve 16a^2-54^2

$$16 a ^ { 2 } - 54 ^ { 2 }$$

Factor

$$4\left(2a-27\right)\left(2a+27\right)$$

Solution Steps

Factor out 4.

$$4\left(4a^{2}-729\right)$$

Consider 4a^{2}-729. Rewrite 4a^{2}-729 as \left(2a\right)^{2}-27^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(2a-27\right)\left(2a+27\right)$$

Rewrite the complete factored expression.

$$4\left(2a-27\right)\left(2a+27\right)$$

Evaluate

$$16a^{2}-2916$$

Solution Steps

Calculate 54 to the power of 2 and get 2916.

$$16a^{2}-2916$$