Solve 16n^4-81.(text{Marks}:1)

$$16 n ^ { 4 } - 81 . ( \text { Marks } : 1 )$$

Factor

$$\left(2n-3\right)\left(2n+3\right)\left(4n^{2}+9\right)$$

Solution Steps

Rewrite 16n^{4}-81 as \left(4n^{2}\right)^{2}-9^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(4n^{2}-9\right)\left(4n^{2}+9\right)$$

Consider 4n^{2}-9. Rewrite 4n^{2}-9 as \left(2n\right)^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(2n-3\right)\left(2n+3\right)$$

Rewrite the complete factored expression. Polynomial 4n^{2}+9 is not factored since it does not have any rational roots.

$$\left(2n-3\right)\left(2n+3\right)\left(4n^{2}+9\right)$$

Evaluate

$$16n^{4}-81$$