Solve (x-2)+2(x-2)^2geq2x(x-3)+2

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

Solve for x

$$x\leq 4$$

Solution Steps

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

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

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

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

Combine x and -8x to get -7x.

$$-7x-2+2x^{2}+8\geq 2x\left(x-3\right)+2$$

Add -2 and 8 to get 6.

$$-7x+6+2x^{2}\geq 2x\left(x-3\right)+2$$

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

$$-7x+6+2x^{2}\geq 2x^{2}-6x+2$$

Subtract 2x^{2} from both sides.

$$-7x+6+2x^{2}-2x^{2}\geq -6x+2$$

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

$$-7x+6\geq -6x+2$$

Add 6x to both sides.

$$-7x+6+6x\geq 2$$

Combine -7x and 6x to get -x.

$$-x+6\geq 2$$

Subtract 6 from both sides.

$$-x\geq 2-6$$

Subtract 6 from 2 to get -4.

$$-x\geq -4$$

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

$$x\leq \frac{-4}{-1}$$

Fraction \frac{-4}{-1} can be simplified to 4 by removing the negative sign from both the numerator and the denominator.

$$x\leq 4$$