Solve 6n^3-54n+7n^2-63

$$6 n ^ { 3 } - 54 n + 7 n ^ { 2 } - 63$$

Factor

$$\left(n-3\right)\left(n+3\right)\left(6n+7\right)$$

Solution Steps

Do the grouping 6n^{3}-54n+7n^{2}-63=\left(6n^{3}-54n\right)+\left(7n^{2}-63\right), and factor out 6n in the first and 7 in the second group.

$$6n\left(n^{2}-9\right)+7\left(n^{2}-9\right)$$

Factor out common term n^{2}-9 by using distributive property.

$$\left(n^{2}-9\right)\left(6n+7\right)$$

Consider n^{2}-9. Rewrite n^{2}-9 as n^{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).

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

Rewrite the complete factored expression.

$$\left(n-3\right)\left(n+3\right)\left(6n+7\right)$$

Evaluate

$$\left(6n+7\right)\left(n^{2}-9\right)$$