Solve 6a^{4}-24a^2b^2+3a^3-12ab^2

$$6 a ^ { 4 } - 24 a ^ { 2 } b ^ { 2 } + 3 a ^ { 3 } - 12 a b ^ { 2 }$$

Factor

$$3a\left(a-2b\right)\left(2a+1\right)\left(a+2b\right)$$

Solution Steps

Factor out 3.

$$3\left(2a^{4}-8a^{2}b^{2}+a^{3}-4ab^{2}\right)$$

Consider 2a^{4}-8a^{2}b^{2}+a^{3}-4ab^{2}. Factor out a.

$$a\left(2a^{3}-8ab^{2}+a^{2}-4b^{2}\right)$$

Consider 2a^{3}-8ab^{2}+a^{2}-4b^{2}. Do the grouping 2a^{3}-8ab^{2}+a^{2}-4b^{2}=\left(2a^{3}-8ab^{2}\right)+\left(a^{2}-4b^{2}\right), and factor out 2a in 2a^{3}-8ab^{2}.

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

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

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

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

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

Rewrite the complete factored expression.

$$3a\left(a-2b\right)\left(a+2b\right)\left(2a+1\right)$$

Evaluate

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