Solve 48a^{4}bc-16b^3a^2c+120a^2bc-40b^3c

$$48 a ^ { 4 } b c - 16 b ^ { 3 } a ^ { 2 } c + 120 a ^ { 2 } b c - 40 b ^ { 3 } c$$

Factor

$$8bc\left(2a^{2}+5\right)\left(3a^{2}-b^{2}\right)$$

Solution Steps

Factor out 8.

$$8\left(6a^{4}bc-2b^{3}a^{2}c+15a^{2}bc-5b^{3}c\right)$$

Consider 6a^{4}bc-2b^{3}a^{2}c+15a^{2}bc-5b^{3}c. Factor out bc.

$$bc\left(6a^{4}-2b^{2}a^{2}+15a^{2}-5b^{2}\right)$$

Consider 6a^{4}-2b^{2}a^{2}+15a^{2}-5b^{2}. Do the grouping 6a^{4}-2b^{2}a^{2}+15a^{2}-5b^{2}=\left(6a^{4}-2b^{2}a^{2}\right)+\left(15a^{2}-5b^{2}\right), and factor out 2a^{2} in the first and 5 in the second group.

$$2a^{2}\left(3a^{2}-b^{2}\right)+5\left(3a^{2}-b^{2}\right)$$

Factor out common term 3a^{2}-b^{2} by using distributive property.

$$\left(3a^{2}-b^{2}\right)\left(2a^{2}+5\right)$$

Rewrite the complete factored expression. Polynomial 2a^{2}+5 is not factored since it does not have any rational roots.

$$8bc\left(3a^{2}-b^{2}\right)\left(2a^{2}+5\right)$$

Evaluate

$$8bc\left(2a^{2}+5\right)\left(3a^{2}-b^{2}\right)$$