Solve 3k^3-2k^2-27k+18

$$3 k ^ { 3 } - 2 k ^ { 2 } - 27 k + 18$$

Factor

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

Solution Steps

Do the grouping 3k^{3}-2k^{2}-27k+18=\left(3k^{3}-2k^{2}\right)+\left(-27k+18\right), and factor out k^{2} in the first and -9 in the second group.

$$k^{2}\left(3k-2\right)-9\left(3k-2\right)$$

Factor out common term 3k-2 by using distributive property.

$$\left(3k-2\right)\left(k^{2}-9\right)$$

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

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

Rewrite the complete factored expression.

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

Evaluate

$$\left(3k-2\right)\left(k^{2}-9\right)$$