Solve 360w^3-780w^2+400w

$$360 w ^ { 3 } - 780 w ^ { 2 } + 400 w$$

Factor

$$20w\left(3w-4\right)\left(6w-5\right)$$

Solution Steps

Factor out 20.

$$20\left(18w^{3}-39w^{2}+20w\right)$$

Consider 18w^{3}-39w^{2}+20w. Factor out w.

$$w\left(18w^{2}-39w+20\right)$$

Consider 18w^{2}-39w+20. Factor the expression by grouping. First, the expression needs to be rewritten as 18w^{2}+aw+bw+20. To find a and b, set up a system to be solved.

$$a+b=-39 ab=18\times 20=360$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 360.

$$-1,-360 -2,-180 -3,-120 -4,-90 -5,-72 -6,-60 -8,-45 -9,-40 -10,-36 -12,-30 -15,-24 -18,-20$$

Calculate the sum for each pair.

$$-1-360=-361 -2-180=-182 -3-120=-123 -4-90=-94 -5-72=-77 -6-60=-66 -8-45=-53 -9-40=-49 -10-36=-46 -12-30=-42 -15-24=-39 -18-20=-38$$

The solution is the pair that gives sum -39.

$$a=-24 b=-15$$

Rewrite 18w^{2}-39w+20 as \left(18w^{2}-24w\right)+\left(-15w+20\right).

$$\left(18w^{2}-24w\right)+\left(-15w+20\right)$$

Factor out 6w in the first and -5 in the second group.

$$6w\left(3w-4\right)-5\left(3w-4\right)$$

Factor out common term 3w-4 by using distributive property.

$$\left(3w-4\right)\left(6w-5\right)$$

Rewrite the complete factored expression.

$$20w\left(3w-4\right)\left(6w-5\right)$$

Evaluate

$$20w\left(3w-4\right)\left(6w-5\right)$$