Solve 2xy^2x^3+32xy^2x^2+128x{y}^2x-8{x}^3-128{x}^2-512x

$$2x { y }^{ 2 } { x }^{ 3 } +32x { y }^{ 2 } { x }^{ 2 } +128x { y }^{ 2 } x-8 { x }^{ 3 } -128 { x }^{ 2 } -512x$$

Factor

$$2x\left(x+8\right)^{2}\left(xy^{2}-4\right)$$

Solution Steps

Factor out 2.

$$2\left(xy^{2}x^{3}+16xy^{2}x^{2}+64xy^{2}x-4x^{3}-64x^{2}-256x\right)$$

Consider y^{2}x^{4}+16y^{2}x^{3}+64y^{2}x^{2}-4x^{3}-64x^{2}-256x. Factor out x.

$$x\left(y^{2}x^{3}+16y^{2}x^{2}+64y^{2}x-4x^{2}-64x-256\right)$$

Consider y^{2}x^{3}+16y^{2}x^{2}+64y^{2}x-4x^{2}-64x-256. Do the grouping y^{2}x^{3}+16y^{2}x^{2}+64y^{2}x-4x^{2}-64x-256=\left(y^{2}x^{3}+16y^{2}x^{2}+64y^{2}x\right)+\left(-4x^{2}-64x-256\right), and factor out xy^{2} in the first and -4 in the second group.

$$xy^{2}\left(x^{2}+16x+64\right)-4\left(x^{2}+16x+64\right)$$

Factor out common term x^{2}+16x+64 by using distributive property.

$$\left(x^{2}+16x+64\right)\left(xy^{2}-4\right)$$

Consider x^{2}+16x+64. Use the perfect square formula, a^{2}+2ab+b^{2}=\left(a+b\right)^{2}, where a=x and b=8.

$$\left(x+8\right)^{2}$$

Rewrite the complete factored expression.

$$2x\left(x+8\right)^{2}\left(xy^{2}-4\right)$$

Evaluate

$$2x\left(x+8\right)^{2}\left(xy^{2}-4\right)$$