Solve 168a^3z-150a^2bc-120a^3c+210a^2bz

$$168 a ^ { 3 } z - 150 a ^ { 2 } b c - 120 a ^ { 3 } c + 210 a ^ { 2 } b z$$

Factor

$$6\left(7z-5c\right)\left(4a+5b\right)a^{2}$$

Solution Steps

Factor out 6.

$$6\left(28a^{3}z-25a^{2}bc-20a^{3}c+35a^{2}bz\right)$$

Consider 28a^{3}z-25a^{2}bc-20a^{3}c+35a^{2}bz. Factor out a^{2}.

$$a^{2}\left(28az-25bc-20ac+35bz\right)$$

Consider 28az-25bc-20ac+35bz. Do the grouping 28az-25bc-20ac+35bz=\left(28az-20ac\right)+\left(35bz-25bc\right), and factor out 4a in the first and 5b in the second group.

$$4a\left(7z-5c\right)+5b\left(7z-5c\right)$$

Factor out common term 7z-5c by using distributive property.

$$\left(7z-5c\right)\left(4a+5b\right)$$

Rewrite the complete factored expression.

$$6a^{2}\left(4a+5b\right)\left(7z-5c\right)$$

Evaluate

$$6\left(7z-5c\right)\left(4a+5b\right)a^{2}$$