Solve 2(x+2)(1-2x)/(x^2+1)^2=0

$$\frac { 2 ( x + 2 ) ( 1 - 2 x ) } { ( x ^ { 2 } + 1 ) ^ { 2 } } = 0$$

Solve for x

$$x=\frac{1}{2}=0.5$$

$$x=-2$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by \left(x^{2}+1\right)^{2}.

$$2\left(x+2\right)\left(1-2x\right)=0$$

Use the distributive property to multiply 2 by x+2.

$$\left(2x+4\right)\left(1-2x\right)=0$$

Use the distributive property to multiply 2x+4 by 1-2x and combine like terms.

$$-6x-4x^{2}+4=0$$

Divide both sides by 2.

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

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

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

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

$$a+b=-3 ab=-2\times 2=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. 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.

$$a=1 b=-4$$

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

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

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

$$-x\left(2x-1\right)-2\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(-x-2\right)$$

To find equation solutions, solve 2x-1=0 and -x-2=0.

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

Steps Using the Quadratic Formula

Multiply both sides of the equation by \left(x^{2}+1\right)^{2}.

$$2\left(x+2\right)\left(1-2x\right)=0$$

Use the distributive property to multiply 2 by x+2.

$$\left(2x+4\right)\left(1-2x\right)=0$$

Use the distributive property to multiply 2x+4 by 1-2x and combine like terms.

$$-6x-4x^{2}+4=0$$

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.

$$-4x^{2}-6x+4=0$$

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

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

Square -6.

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

Multiply -4 times -4.

$$x=\frac{-\left(-6\right)±\sqrt{36+16\times 4}}{2\left(-4\right)}$$

Multiply 16 times 4.

$$x=\frac{-\left(-6\right)±\sqrt{36+64}}{2\left(-4\right)}$$

Add 36 to 64.

$$x=\frac{-\left(-6\right)±\sqrt{100}}{2\left(-4\right)}$$

Take the square root of 100.

$$x=\frac{-\left(-6\right)±10}{2\left(-4\right)}$$

The opposite of -6 is 6.

$$x=\frac{6±10}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{6±10}{-8}$$

Now solve the equation x=\frac{6±10}{-8} when ± is plus. Add 6 to 10.

$$x=\frac{16}{-8}$$

Divide 16 by -8.

$$x=-2$$

Now solve the equation x=\frac{6±10}{-8} when ± is minus. Subtract 10 from 6.

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

Reduce the fraction \frac{-4}{-8} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{2}$$

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by \left(x^{2}+1\right)^{2}.

$$2\left(x+2\right)\left(1-2x\right)=0$$

Use the distributive property to multiply 2 by x+2.

$$\left(2x+4\right)\left(1-2x\right)=0$$

Use the distributive property to multiply 2x+4 by 1-2x and combine like terms.

$$-6x-4x^{2}+4=0$$

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

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

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.

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

Divide both sides by -4.

$$\frac{-4x^{2}-6x}{-4}=-\frac{4}{-4}$$

Dividing by -4 undoes the multiplication by -4.

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

Reduce the fraction \frac{-6}{-4} to lowest terms by extracting and canceling out 2.

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

Divide -4 by -4.

$$x^{2}+\frac{3}{2}x=1$$

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=1+\frac{9}{16}$$

Add 1 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{25}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{5}{4} x+\frac{3}{4}=-\frac{5}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

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