Solve 6b^3-24b+6b^2-24

$$6 b ^ { 3 } - 24 b + 6 b ^ { 2 } - 24$$

Factor

$$6\left(b-2\right)\left(b+1\right)\left(b+2\right)$$

Solution Steps

Factor out 6.

$$6\left(b^{3}-4b+b^{2}-4\right)$$

Consider b^{3}-4b+b^{2}-4. Do the grouping b^{3}-4b+b^{2}-4=\left(b^{3}-4b\right)+\left(b^{2}-4\right), and factor out b in b^{3}-4b.

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

Factor out common term b^{2}-4 by using distributive property.

$$\left(b^{2}-4\right)\left(b+1\right)$$

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

$$\left(b-2\right)\left(b+2\right)$$

Rewrite the complete factored expression.

$$6\left(b-2\right)\left(b+2\right)\left(b+1\right)$$

Evaluate

$$6\left(b+1\right)\left(b^{2}-4\right)$$