Solve (x-1)(x-2)(x+2)-(x-1)(x+1)(x-2)geq(x-1)^2

$$( x - 1 ) ( x - 2 ) ( x + 2 ) - ( x - 1 ) ( x + 1 ) ( x - 2 ) \geq ( x - 1 ) ^ { 2 }$$

Solve for x

$$x\leq 1$$

Solution Steps

Use the distributive property to multiply x-1 by x-2 and combine like terms.

$$\left(x^{2}-3x+2\right)\left(x+2\right)-\left(x-1\right)\left(x+1\right)\left(x-2\right)\geq \left(x-1\right)^{2}$$

Use the distributive property to multiply x^{2}-3x+2 by x+2 and combine like terms.

$$x^{3}-x^{2}-4x+4-\left(x-1\right)\left(x+1\right)\left(x-2\right)\geq \left(x-1\right)^{2}$$

Use the distributive property to multiply x-1 by x+1 and combine like terms.

$$x^{3}-x^{2}-4x+4-\left(x^{2}-1\right)\left(x-2\right)\geq \left(x-1\right)^{2}$$

Use the distributive property to multiply x^{2}-1 by x-2.

$$x^{3}-x^{2}-4x+4-\left(x^{3}-2x^{2}-x+2\right)\geq \left(x-1\right)^{2}$$

To find the opposite of x^{3}-2x^{2}-x+2, find the opposite of each term.

$$x^{3}-x^{2}-4x+4-x^{3}+2x^{2}+x-2\geq \left(x-1\right)^{2}$$

Combine x^{3} and -x^{3} to get 0.

$$-x^{2}-4x+4+2x^{2}+x-2\geq \left(x-1\right)^{2}$$

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

$$x^{2}-4x+4+x-2\geq \left(x-1\right)^{2}$$

Combine -4x and x to get -3x.

$$x^{2}-3x+4-2\geq \left(x-1\right)^{2}$$

Subtract 2 from 4 to get 2.

$$x^{2}-3x+2\geq \left(x-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x^{2}-3x+2\geq x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$x^{2}-3x+2-x^{2}\geq -2x+1$$

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

$$-3x+2\geq -2x+1$$

Add 2x to both sides.

$$-3x+2+2x\geq 1$$

Combine -3x and 2x to get -x.

$$-x+2\geq 1$$

Subtract 2 from both sides.

$$-x\geq 1-2$$

Subtract 2 from 1 to get -1.

$$-x\geq -1$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x\leq \frac{-1}{-1}$$

Divide -1 by -1 to get 1.

$$x\leq 1$$