Solve (2x+1)^2-(3x-1)^2+(x-1)(x+2)=-2(2x^2+1)+11x

$$( 2 x + 1 ) ^ { 2 } - ( 3 x - 1 ) ^ { 2 } + ( x - 1 ) ( x + 2 ) = - 2 ( 2 x ^ { 2 } + 1 ) + 11 x$$

Solve for x (complex solution)

$$x\in \mathrm{C}$$

Solution Steps

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

$$4x^{2}+4x+1-\left(3x-1\right)^{2}+\left(x-1\right)\left(x+2\right)=-2\left(2x^{2}+1\right)+11x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$4x^{2}+4x+1-\left(9x^{2}-6x+1\right)+\left(x-1\right)\left(x+2\right)=-2\left(2x^{2}+1\right)+11x$$

To find the opposite of 9x^{2}-6x+1, find the opposite of each term.

$$4x^{2}+4x+1-9x^{2}+6x-1+\left(x-1\right)\left(x+2\right)=-2\left(2x^{2}+1\right)+11x$$

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

$$-5x^{2}+4x+1+6x-1+\left(x-1\right)\left(x+2\right)=-2\left(2x^{2}+1\right)+11x$$

Combine 4x and 6x to get 10x.

$$-5x^{2}+10x+1-1+\left(x-1\right)\left(x+2\right)=-2\left(2x^{2}+1\right)+11x$$

Subtract 1 from 1 to get 0.

$$-5x^{2}+10x+\left(x-1\right)\left(x+2\right)=-2\left(2x^{2}+1\right)+11x$$

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

$$-5x^{2}+10x+x^{2}+x-2=-2\left(2x^{2}+1\right)+11x$$

Combine -5x^{2} and x^{2} to get -4x^{2}.

$$-4x^{2}+10x+x-2=-2\left(2x^{2}+1\right)+11x$$

Combine 10x and x to get 11x.

$$-4x^{2}+11x-2=-2\left(2x^{2}+1\right)+11x$$

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

$$-4x^{2}+11x-2=-4x^{2}-2+11x$$

Add 4x^{2} to both sides.

$$-4x^{2}+11x-2+4x^{2}=-2+11x$$

Combine -4x^{2} and 4x^{2} to get 0.

$$11x-2=-2+11x$$

Subtract -2 from both sides.

$$11x-2-\left(-2\right)=11x$$

The opposite of -2 is 2.

$$11x-2+2=11x$$

Subtract 11x from both sides.

$$11x-2+2-11x=0$$

Add -2 and 2 to get 0.

$$11x-11x=0$$

Combine 11x and -11x to get 0.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{C}$$

Solve for x

$$x\in \mathrm{R}$$

Solution Steps

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

$$4x^{2}+4x+1-\left(3x-1\right)^{2}+\left(x-1\right)\left(x+2\right)=-2\left(2x^{2}+1\right)+11x$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.

$$4x^{2}+4x+1-\left(9x^{2}-6x+1\right)+\left(x-1\right)\left(x+2\right)=-2\left(2x^{2}+1\right)+11x$$

To find the opposite of 9x^{2}-6x+1, find the opposite of each term.

$$4x^{2}+4x+1-9x^{2}+6x-1+\left(x-1\right)\left(x+2\right)=-2\left(2x^{2}+1\right)+11x$$

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

$$-5x^{2}+4x+1+6x-1+\left(x-1\right)\left(x+2\right)=-2\left(2x^{2}+1\right)+11x$$

Combine 4x and 6x to get 10x.

$$-5x^{2}+10x+1-1+\left(x-1\right)\left(x+2\right)=-2\left(2x^{2}+1\right)+11x$$

Subtract 1 from 1 to get 0.

$$-5x^{2}+10x+\left(x-1\right)\left(x+2\right)=-2\left(2x^{2}+1\right)+11x$$

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

$$-5x^{2}+10x+x^{2}+x-2=-2\left(2x^{2}+1\right)+11x$$

Combine -5x^{2} and x^{2} to get -4x^{2}.

$$-4x^{2}+10x+x-2=-2\left(2x^{2}+1\right)+11x$$

Combine 10x and x to get 11x.

$$-4x^{2}+11x-2=-2\left(2x^{2}+1\right)+11x$$

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

$$-4x^{2}+11x-2=-4x^{2}-2+11x$$

Add 4x^{2} to both sides.

$$-4x^{2}+11x-2+4x^{2}=-2+11x$$

Combine -4x^{2} and 4x^{2} to get 0.

$$11x-2=-2+11x$$

Subtract -2 from both sides.

$$11x-2-\left(-2\right)=11x$$

The opposite of -2 is 2.

$$11x-2+2=11x$$

Subtract 11x from both sides.

$$11x-2+2-11x=0$$

Add -2 and 2 to get 0.

$$11x-11x=0$$

Combine 11x and -11x to get 0.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{R}$$