Solve (2-x)^2-(x-1)(x+1)=(x+3)^2-6x

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

Solve for x

$$x=-2$$

Steps Using Factoring By Grouping

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

$$4-4x+x^{2}-\left(x-1\right)\left(x+1\right)=\left(x+3\right)^{2}-6x$$

Consider \left(x-1\right)\left(x+1\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 1.

$$4-4x+x^{2}-\left(x^{2}-1\right)=\left(x+3\right)^{2}-6x$$

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

$$4-4x+x^{2}-x^{2}+1=\left(x+3\right)^{2}-6x$$

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

$$4-4x+1=\left(x+3\right)^{2}-6x$$

Add 4 and 1 to get 5.

$$5-4x=\left(x+3\right)^{2}-6x$$

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

$$5-4x=x^{2}+6x+9-6x$$

Combine 6x and -6x to get 0.

$$5-4x=x^{2}+9$$

Subtract x^{2} from both sides.

$$5-4x-x^{2}=9$$

Subtract 9 from both sides.

$$5-4x-x^{2}-9=0$$

Subtract 9 from 5 to get -4.

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

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

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

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.

$$a+b=-4 ab=-\left(-4\right)=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

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

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -4.

$$a=-2 b=-2$$

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

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

Factor out x in the first and 2 in the second group.

$$x\left(-x-2\right)+2\left(-x-2\right)$$

Factor out common term -x-2 by using distributive property.

$$\left(-x-2\right)\left(x+2\right)$$

To find equation solutions, solve -x-2=0 and x+2=0.

$$x=-2 x=-2$$

Steps Using the Quadratic Formula

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

$$4-4x+x^{2}-\left(x-1\right)\left(x+1\right)=\left(x+3\right)^{2}-6x$$

Consider \left(x-1\right)\left(x+1\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 1.

$$4-4x+x^{2}-\left(x^{2}-1\right)=\left(x+3\right)^{2}-6x$$

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

$$4-4x+x^{2}-x^{2}+1=\left(x+3\right)^{2}-6x$$

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

$$4-4x+1=\left(x+3\right)^{2}-6x$$

Add 4 and 1 to get 5.

$$5-4x=\left(x+3\right)^{2}-6x$$

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

$$5-4x=x^{2}+6x+9-6x$$

Combine 6x and -6x to get 0.

$$5-4x=x^{2}+9$$

Subtract x^{2} from both sides.

$$5-4x-x^{2}=9$$

Subtract 9 from both sides.

$$5-4x-x^{2}-9=0$$

Subtract 9 from 5 to get -4.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -4 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\left(-4\right)}}{2\left(-1\right)}$$

Multiply 4 times -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-16}}{2\left(-1\right)}$$

Add 16 to -16.

$$x=\frac{-\left(-4\right)±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

$$x=-\frac{-4}{2\left(-1\right)}$$

The opposite of -4 is 4.

$$x=\frac{4}{2\left(-1\right)}$$

Multiply 2 times -1.

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

Divide 4 by -2.

$$x=-2$$

Steps for Completing the Square

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

$$4-4x+x^{2}-\left(x-1\right)\left(x+1\right)=\left(x+3\right)^{2}-6x$$

Consider \left(x-1\right)\left(x+1\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 1.

$$4-4x+x^{2}-\left(x^{2}-1\right)=\left(x+3\right)^{2}-6x$$

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

$$4-4x+x^{2}-x^{2}+1=\left(x+3\right)^{2}-6x$$

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

$$4-4x+1=\left(x+3\right)^{2}-6x$$

Add 4 and 1 to get 5.

$$5-4x=\left(x+3\right)^{2}-6x$$

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

$$5-4x=x^{2}+6x+9-6x$$

Combine 6x and -6x to get 0.

$$5-4x=x^{2}+9$$

Subtract x^{2} from both sides.

$$5-4x-x^{2}=9$$

Subtract 5 from both sides.

$$-4x-x^{2}=9-5$$

Subtract 5 from 9 to get 4.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -1.

$$\frac{-x^{2}-4x}{-1}=\frac{4}{-1}$$

Dividing by -1 undoes the multiplication by -1.

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

Divide -4 by -1.

$$x^{2}+4x=\frac{4}{-1}$$

Divide 4 by -1.

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

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square 2.

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

Add -4 to 4.

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

Factor x^{2}+4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+2=0 x+2=0$$

Subtract 2 from both sides of the equation.

$$x=-2 x=-2$$

The equation is now solved. Solutions are the same.

$$x=-2$$