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

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

Solve for x

$$x=-1$$

Steps Using Factoring By Grouping

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

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

Add -1 and 3 to get 2.

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

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

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

Use the distributive property to multiply 2 by 4x^{2}+4x+1.

$$4x^{2}+2+8x^{2}+8x+2=2\times \left(2x\right)^{2}$$

Combine 4x^{2} and 8x^{2} to get 12x^{2}.

$$12x^{2}+2+8x+2=2\times \left(2x\right)^{2}$$

Add 2 and 2 to get 4.

$$12x^{2}+4+8x=2\times \left(2x\right)^{2}$$

Expand \left(2x\right)^{2}.

$$12x^{2}+4+8x=2\times 2^{2}x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$12x^{2}+4+8x=2\times 4x^{2}$$

Multiply 2 and 4 to get 8.

$$12x^{2}+4+8x=8x^{2}$$

Subtract 8x^{2} from both sides.

$$12x^{2}+4+8x-8x^{2}=0$$

Combine 12x^{2} and -8x^{2} to get 4x^{2}.

$$4x^{2}+4+8x=0$$

Divide both sides by 4.

$$x^{2}+1+2x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+2x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\times 1=1$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=1$$

Rewrite x^{2}+2x+1 as \left(x^{2}+x\right)+\left(x+1\right).

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

Factor out x in x^{2}+x.

$$x\left(x+1\right)+x+1$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(x+1\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve x+1=0.

$$x=-1$$

Steps Using the Quadratic Formula

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

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

Add -1 and 3 to get 2.

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

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

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

Use the distributive property to multiply 2 by 4x^{2}+4x+1.

$$4x^{2}+2+8x^{2}+8x+2=2\times \left(2x\right)^{2}$$

Combine 4x^{2} and 8x^{2} to get 12x^{2}.

$$12x^{2}+2+8x+2=2\times \left(2x\right)^{2}$$

Add 2 and 2 to get 4.

$$12x^{2}+4+8x=2\times \left(2x\right)^{2}$$

Expand \left(2x\right)^{2}.

$$12x^{2}+4+8x=2\times 2^{2}x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$12x^{2}+4+8x=2\times 4x^{2}$$

Multiply 2 and 4 to get 8.

$$12x^{2}+4+8x=8x^{2}$$

Subtract 8x^{2} from both sides.

$$12x^{2}+4+8x-8x^{2}=0$$

Combine 12x^{2} and -8x^{2} to get 4x^{2}.

$$4x^{2}+4+8x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$4x^{2}+8x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 8 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 4\times 4}}{2\times 4}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 4\times 4}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-8±\sqrt{64-16\times 4}}{2\times 4}$$

Multiply -16 times 4.

$$x=\frac{-8±\sqrt{64-64}}{2\times 4}$$

Add 64 to -64.

$$x=\frac{-8±\sqrt{0}}{2\times 4}$$

Take the square root of 0.

$$x=-\frac{8}{2\times 4}$$

Multiply 2 times 4.

$$x=-\frac{8}{8}$$

Divide -8 by 8.

$$x=-1$$

Steps for Completing the Square

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

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

Add -1 and 3 to get 2.

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

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

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

Use the distributive property to multiply 2 by 4x^{2}+4x+1.

$$4x^{2}+2+8x^{2}+8x+2=2\times \left(2x\right)^{2}$$

Combine 4x^{2} and 8x^{2} to get 12x^{2}.

$$12x^{2}+2+8x+2=2\times \left(2x\right)^{2}$$

Add 2 and 2 to get 4.

$$12x^{2}+4+8x=2\times \left(2x\right)^{2}$$

Expand \left(2x\right)^{2}.

$$12x^{2}+4+8x=2\times 2^{2}x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$12x^{2}+4+8x=2\times 4x^{2}$$

Multiply 2 and 4 to get 8.

$$12x^{2}+4+8x=8x^{2}$$

Subtract 8x^{2} from both sides.

$$12x^{2}+4+8x-8x^{2}=0$$

Combine 12x^{2} and -8x^{2} to get 4x^{2}.

$$4x^{2}+4+8x=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}+8x=-4$$

Divide both sides by 4.

$$\frac{4x^{2}+8x}{4}=-\frac{4}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{8}{4}x=-\frac{4}{4}$$

Divide 8 by 4.

$$x^{2}+2x=-\frac{4}{4}$$

Divide -4 by 4.

$$x^{2}+2x=-1$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=-1+1^{2}$$

Square 1.

$$x^{2}+2x+1=-1+1$$

Add -1 to 1.

$$x^{2}+2x+1=0$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+1=0 x+1=0$$

Subtract 1 from both sides of the equation.

$$x=-1 x=-1$$

The equation is now solved. Solutions are the same.

$$x=-1$$