Solve (x+1)^2>[6-(1-x)]x-2

$$( x + 1 ) ^ { 2 } > [ 6 - ( 1 - x ) ] x - 2$$

Solve for x

$$x<1$$

Solution Steps

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

$$x^{2}+2x+1>\left(6-\left(1-x\right)\right)x-2$$

To find the opposite of 1-x, find the opposite of each term.

$$x^{2}+2x+1>\left(6-1+x\right)x-2$$

Subtract 1 from 6 to get 5.

$$x^{2}+2x+1>\left(5+x\right)x-2$$

Use the distributive property to multiply 5+x by x.

$$x^{2}+2x+1>5x+x^{2}-2$$

Subtract 5x from both sides.

$$x^{2}+2x+1-5x>x^{2}-2$$

Combine 2x and -5x to get -3x.

$$x^{2}-3x+1>x^{2}-2$$

Subtract x^{2} from both sides.

$$x^{2}-3x+1-x^{2}>-2$$

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

$$-3x+1>-2$$

Subtract 1 from both sides.

$$-3x>-2-1$$

Subtract 1 from -2 to get -3.

$$-3x>-3$$

Divide both sides by -3. Since -3 is negative, the inequality direction is changed.

$$x<\frac{-3}{-3}$$

Divide -3 by -3 to get 1.

$$x<1$$