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

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

Solve for x

$$x=-1$$

$$x=4$$

Steps Using Factoring By Grouping

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

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

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

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

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

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

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

Combine 3x^{2} and -x^{2} to get 2x^{2}.

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

Combine 4x and 2x to get 6x.

$$2x^{2}+6x+1-1=\left(2x+3\right)\left(2x-3\right)+1$$

Subtract 1 from 1 to get 0.

$$2x^{2}+6x=\left(2x+3\right)\left(2x-3\right)+1$$

Consider \left(2x+3\right)\left(2x-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

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

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

$$2x^{2}+6x=2^{2}x^{2}-9+1$$

Calculate 2 to the power of 2 and get 4.

$$2x^{2}+6x=4x^{2}-9+1$$

Add -9 and 1 to get -8.

$$2x^{2}+6x=4x^{2}-8$$

Subtract 4x^{2} from both sides.

$$2x^{2}+6x-4x^{2}=-8$$

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

$$-2x^{2}+6x=-8$$

Add 8 to both sides.

$$-2x^{2}+6x+8=0$$

Divide both sides by 2.

$$-x^{2}+3x+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=3 ab=-4=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -4.

$$-1,4 -2,2$$

Calculate the sum for each pair.

$$-1+4=3 -2+2=0$$

The solution is the pair that gives sum 3.

$$a=4 b=-1$$

Rewrite -x^{2}+3x+4 as \left(-x^{2}+4x\right)+\left(-x+4\right).

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

Factor out -x in the first and -1 in the second group.

$$-x\left(x-4\right)-\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(-x-1\right)$$

To find equation solutions, solve x-4=0 and -x-1=0.

$$x=4 x=-1$$

Steps Using the Quadratic Formula

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

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

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

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

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

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

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

Combine 3x^{2} and -x^{2} to get 2x^{2}.

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

Combine 4x and 2x to get 6x.

$$2x^{2}+6x+1-1=\left(2x+3\right)\left(2x-3\right)+1$$

Subtract 1 from 1 to get 0.

$$2x^{2}+6x=\left(2x+3\right)\left(2x-3\right)+1$$

Consider \left(2x+3\right)\left(2x-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

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

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

$$2x^{2}+6x=2^{2}x^{2}-9+1$$

Calculate 2 to the power of 2 and get 4.

$$2x^{2}+6x=4x^{2}-9+1$$

Add -9 and 1 to get -8.

$$2x^{2}+6x=4x^{2}-8$$

Subtract 4x^{2} from both sides.

$$2x^{2}+6x-4x^{2}=-8$$

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

$$-2x^{2}+6x=-8$$

Add 8 to both sides.

$$-2x^{2}+6x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 6 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-2\right)\times 8}}{2\left(-2\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-2\right)\times 8}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-6±\sqrt{36+8\times 8}}{2\left(-2\right)}$$

Multiply 8 times 8.

$$x=\frac{-6±\sqrt{36+64}}{2\left(-2\right)}$$

Add 36 to 64.

$$x=\frac{-6±\sqrt{100}}{2\left(-2\right)}$$

Take the square root of 100.

$$x=\frac{-6±10}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-6±10}{-4}$$

Now solve the equation x=\frac{-6±10}{-4} when ± is plus. Add -6 to 10.

$$x=\frac{4}{-4}$$

Divide 4 by -4.

$$x=-1$$

Now solve the equation x=\frac{-6±10}{-4} when ± is minus. Subtract 10 from -6.

$$x=-\frac{16}{-4}$$

Divide -16 by -4.

$$x=4$$

The equation is now solved.

$$x=-1 x=4$$

Steps for Completing the Square

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

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

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

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

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

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

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

Combine 3x^{2} and -x^{2} to get 2x^{2}.

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

Combine 4x and 2x to get 6x.

$$2x^{2}+6x+1-1=\left(2x+3\right)\left(2x-3\right)+1$$

Subtract 1 from 1 to get 0.

$$2x^{2}+6x=\left(2x+3\right)\left(2x-3\right)+1$$

Consider \left(2x+3\right)\left(2x-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

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

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

$$2x^{2}+6x=2^{2}x^{2}-9+1$$

Calculate 2 to the power of 2 and get 4.

$$2x^{2}+6x=4x^{2}-9+1$$

Add -9 and 1 to get -8.

$$2x^{2}+6x=4x^{2}-8$$

Subtract 4x^{2} from both sides.

$$2x^{2}+6x-4x^{2}=-8$$

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

$$-2x^{2}+6x=-8$$

Divide both sides by -2.

$$\frac{-2x^{2}+6x}{-2}=-\frac{8}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{6}{-2}x=-\frac{8}{-2}$$

Divide 6 by -2.

$$x^{2}-3x=-\frac{8}{-2}$$

Divide -8 by -2.

$$x^{2}-3x=4$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=4+\frac{9}{4}$$

Add 4 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{25}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{5}{2} x-\frac{3}{2}=-\frac{5}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=4 x=-1$$