Solve 3bx^3+4bx^2-3bx-4b

$$3 b x ^ { 3 } + 4 b x ^ { 2 } - 3 b x - 4 b$$

Factor

$$b\left(x-1\right)\left(x+1\right)\left(3x+4\right)$$

Solution Steps

Factor out b.

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

Consider 3x^{3}+4x^{2}-3x-4. Do the grouping 3x^{3}+4x^{2}-3x-4=\left(3x^{3}+4x^{2}\right)+\left(-3x-4\right), and factor out x^{2} in the first and -1 in the second group.

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

Factor out common term 3x+4 by using distributive property.

$$\left(3x+4\right)\left(x^{2}-1\right)$$

Consider x^{2}-1. Rewrite x^{2}-1 as x^{2}-1^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(x-1\right)\left(x+1\right)$$

Rewrite the complete factored expression.

$$b\left(3x+4\right)\left(x-1\right)\left(x+1\right)$$

Evaluate

$$b\left(3x+4\right)\left(x^{2}-1\right)$$