Solve [-2(x-2)(x+2)+3(x-2)^2-(x+2)^2]-8(3x+2)

$$[ - 2 ( x - 2 ) ( x + 2 ) + 3 ( x - 2 ) ^ { 2 } - ( x + 2 ) ^ { 2 } ] - 8 ( 3 x + 2 )$$

Evaluate

$$-40x$$

Solution Steps

Use the distributive property to multiply -2 by x-2.

$$\left(-2x+4\right)\left(x+2\right)+3\left(x-2\right)^{2}-\left(x+2\right)^{2}-8\left(3x+2\right)$$

Use the distributive property to multiply -2x+4 by x+2 and combine like terms.

$$-2x^{2}+8+3\left(x-2\right)^{2}-\left(x+2\right)^{2}-8\left(3x+2\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$-2x^{2}+8+3\left(x^{2}-4x+4\right)-\left(x+2\right)^{2}-8\left(3x+2\right)$$

Use the distributive property to multiply 3 by x^{2}-4x+4.

$$-2x^{2}+8+3x^{2}-12x+12-\left(x+2\right)^{2}-8\left(3x+2\right)$$

Combine -2x^{2} and 3x^{2} to get x^{2}.

$$x^{2}+8-12x+12-\left(x+2\right)^{2}-8\left(3x+2\right)$$

Add 8 and 12 to get 20.

$$x^{2}+20-12x-\left(x+2\right)^{2}-8\left(3x+2\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

$$x^{2}+20-12x-\left(x^{2}+4x+4\right)-8\left(3x+2\right)$$

To find the opposite of x^{2}+4x+4, find the opposite of each term.

$$x^{2}+20-12x-x^{2}-4x-4-8\left(3x+2\right)$$

Combine x^{2} and -x^{2} to get 0.

$$20-12x-4x-4-8\left(3x+2\right)$$

Combine -12x and -4x to get -16x.

$$20-16x-4-8\left(3x+2\right)$$

Subtract 4 from 20 to get 16.

$$16-16x-8\left(3x+2\right)$$

Use the distributive property to multiply -8 by 3x+2.

$$16-16x-24x-16$$

Combine -16x and -24x to get -40x.

$$16-40x-16$$

Subtract 16 from 16 to get 0.

$$-40x$$

Expand

$$-40x$$

Solution Steps

Use the distributive property to multiply -2 by x-2.

$$\left(-2x+4\right)\left(x+2\right)+3\left(x-2\right)^{2}-\left(x+2\right)^{2}-8\left(3x+2\right)$$

Use the distributive property to multiply -2x+4 by x+2 and combine like terms.

$$-2x^{2}+8+3\left(x-2\right)^{2}-\left(x+2\right)^{2}-8\left(3x+2\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$-2x^{2}+8+3\left(x^{2}-4x+4\right)-\left(x+2\right)^{2}-8\left(3x+2\right)$$

Use the distributive property to multiply 3 by x^{2}-4x+4.

$$-2x^{2}+8+3x^{2}-12x+12-\left(x+2\right)^{2}-8\left(3x+2\right)$$

Combine -2x^{2} and 3x^{2} to get x^{2}.

$$x^{2}+8-12x+12-\left(x+2\right)^{2}-8\left(3x+2\right)$$

Add 8 and 12 to get 20.

$$x^{2}+20-12x-\left(x+2\right)^{2}-8\left(3x+2\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

$$x^{2}+20-12x-\left(x^{2}+4x+4\right)-8\left(3x+2\right)$$

To find the opposite of x^{2}+4x+4, find the opposite of each term.

$$x^{2}+20-12x-x^{2}-4x-4-8\left(3x+2\right)$$

Combine x^{2} and -x^{2} to get 0.

$$20-12x-4x-4-8\left(3x+2\right)$$

Combine -12x and -4x to get -16x.

$$20-16x-4-8\left(3x+2\right)$$

Subtract 4 from 20 to get 16.

$$16-16x-8\left(3x+2\right)$$

Use the distributive property to multiply -8 by 3x+2.

$$16-16x-24x-16$$

Combine -16x and -24x to get -40x.

$$16-40x-16$$

Subtract 16 from 16 to get 0.

$$-40x$$