Solve (x-2)^2-3(x+2)^2-(x+2)

$$( x - 2 ) ^ { 2 } - 3 ( x + 2 ) ^ { 2 } - ( x + 2 )$$

Evaluate

$$-2x^{2}-17x-10$$

Solution Steps

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

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

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

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

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

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

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

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

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

$$-2x^{2}-16x+4-12-\left(x+2\right)$$

Subtract 12 from 4 to get -8.

$$-2x^{2}-16x-8-\left(x+2\right)$$

To find the opposite of x+2, find the opposite of each term.

$$-2x^{2}-16x-8-x-2$$

Combine -16x and -x to get -17x.

$$-2x^{2}-17x-8-2$$

Subtract 2 from -8 to get -10.

$$-2x^{2}-17x-10$$

Expand

$$-2x^{2}-17x-10$$

Solution Steps

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

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

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

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

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

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

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

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

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

$$-2x^{2}-16x+4-12-\left(x+2\right)$$

Subtract 12 from 4 to get -8.

$$-2x^{2}-16x-8-\left(x+2\right)$$

To find the opposite of x+2, find the opposite of each term.

$$-2x^{2}-16x-8-x-2$$

Combine -16x and -x to get -17x.

$$-2x^{2}-17x-8-2$$

Subtract 2 from -8 to get -10.

$$-2x^{2}-17x-10$$