Question

(x+2) (x+3)+(x-3) (x-2)-2x (x+1)= 0

Answer 1

Simplifying (x + 2)(x + -3) + (x + -3)(x + -2) + -2x(x + 1) = 0

Reorder the terms: (2 + x)(x + -3) + (x + -3)(x + -2) + -2x(x + 1) = 0

Reorder the terms: (2 + x)(-3 + x) + (x + -3)(x + -2) + -2x(x + 1) = 0

Multiply (2 + x) * (-3 + x) (2(-3 + x) + x(-3 + x)) + (x + -3)(x + -2) + -2x(x + 1) = 0 ((-3 * 2 + x * 2) + x(-3 + x)) + (x + -3)(x + -2) + -2x(x + 1) = 0 ((-6 + 2x) + x(-3 + x)) + (x + -3)(x + -2) + -2x(x + 1) = 0 (-6 + 2x + (-3 * x + x * x)) + (x + -3)(x + -2) + -2x(x + 1) = 0 (-6 + 2x + (-3x + x2)) + (x + -3)(x + -2) + -2x(x + 1) = 0

Combine like terms: 2x + -3x = -1x (-6 + -1x + x2) + (x + -3)(x + -2) + -2x(x + 1) = 0

Reorder the terms: -6 + -1x + x2 + (-3 + x)(x + -2) + -2x(x + 1) = 0

Reorder the terms: -6 + -1x + x2 + (-3 + x)(-2 + x) + -2x(x + 1) = 0

Multiply (-3 + x) * (-2 + x) -6 + -1x + x2 + (-3(-2 + x) + x(-2 + x)) + -2x(x + 1) = 0 -6 + -1x + x2 + ((-2 * -3 + x * -3) + x(-2 + x)) + -2x(x + 1) = 0 -6 + -1x + x2 + ((6 + -3x) + x(-2 + x)) + -2x(x + 1) = 0 -6 + -1x + x2 + (6 + -3x + (-2 * x + x * x)) + -2x(x + 1) = 0 -6 + -1x + x2 + (6 + -3x + (-2x + x2)) + -2x(x + 1) = 0

Combine like terms: -3x + -2x = -5x -6 + -1x + x2 + (6 + -5x + x2) + -2x(x + 1) = 0

Reorder the terms: -6 + -1x + x2 + 6 + -5x + x2 + -2x(1 + x) = 0 -6 + -1x + x2 + 6 + -5x + x2 + (1 * -2x + x * -2x) = 0 -6 + -1x + x2 + 6 + -5x + x2 + (-2x + -2x2) = 0

Reorder the terms: -6 + 6 + -1x + -5x + -2x + x2 + x2 + -2x2 = 0

Combine like terms: -6 + 6 = 0 0 + -1x + -5x + -2x + x2 + x2 + -2x2 = 0 -1x + -5x + -2x + x2 + x2 + -2x2 = 0

Combine like terms: -1x + -5x = -6x -6x + -2x + x2 + x2 + -2x2 = 0

Combine like terms: -6x + -2x = -8x -8x + x2 + x2 + -2x2 = 0

Combine like terms: x2 + x2 = 2x2 -8x + 2x2 + -2x2 = 0

Combine like terms: 2x2 + -2x2 = 0 -8x + 0 = 0 -8x = 0

Solving -8x = 0

Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Divide each side by '-8'. x = 0 Simplifying x = 0

Answer 2

The answer is in attachment with explaination and verification..
Hope it helps....