Solve 8x^{3}y^2-76x^2y^2+140xy^2

$$8 x ^ { 3 } y ^ { 2 } - 76 x ^ { 2 } y ^ { 2 } + 140 x y ^ { 2 }$$

Factor

$$4x\left(x-7\right)\left(2x-5\right)y^{2}$$

Solution Steps

Factor out 4.

$$4\left(2x^{3}y^{2}-19x^{2}y^{2}+35xy^{2}\right)$$

Consider 2x^{3}y^{2}-19x^{2}y^{2}+35xy^{2}. Factor out xy^{2}.

$$xy^{2}\left(2x^{2}-19x+35\right)$$

Consider 2x^{2}-19x+35. Factor the expression by grouping. First, the expression needs to be rewritten as 2x^{2}+ax+bx+35. To find a and b, set up a system to be solved.

$$a+b=-19 ab=2\times 35=70$$

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 70.

$$-1,-70 -2,-35 -5,-14 -7,-10$$

Calculate the sum for each pair.

$$-1-70=-71 -2-35=-37 -5-14=-19 -7-10=-17$$

The solution is the pair that gives sum -19.

$$a=-14 b=-5$$

Rewrite 2x^{2}-19x+35 as \left(2x^{2}-14x\right)+\left(-5x+35\right).

$$\left(2x^{2}-14x\right)+\left(-5x+35\right)$$

Factor out 2x in the first and -5 in the second group.

$$2x\left(x-7\right)-5\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(2x-5\right)$$

Rewrite the complete factored expression.

$$4xy^{2}\left(x-7\right)\left(2x-5\right)$$

Evaluate

$$4x\left(x-7\right)\left(2x-5\right)y^{2}$$