Solve x^2-2/x^2+x-2+3x+2/x^2-4=1-x+2/x^2-3x+2

$$\frac { x ^ { 2 } - 2 } { x ^ { 2 } + x - 2 } + \frac { 3 x + 2 } { x ^ { 2 } - 4 } = 1 - \frac { x + 2 } { x ^ { 2 } - 3 x + 2 }$$

Solve for x

$$x=-\frac{2}{3}\approx -0.666666667$$

$$x=-1$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -2,1,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x-1\right)\left(x+2\right), the least common multiple of x^{2}+x-2,x^{2}-4,x^{2}-3x+2.

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

Multiply x+2 and x+2 to get \left(x+2\right)^{2}.

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

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

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

Use the distributive property to multiply x-1 by 3x+2 and combine like terms.

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

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

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

Combine -2x and -x to get -3x.

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

Subtract 2 from 4 to get 2.

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

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

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

Use the distributive property to multiply x^{2}-3x+2 by x+2 and combine like terms.

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

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

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

To find the opposite of x^{2}+4x+4, find the opposite of each term.

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

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

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

Combine -4x and -4x to get -8x.

$$x^{3}-3x+x^{2}+2=x^{3}-2x^{2}-8x+4-4$$

Subtract 4 from 4 to get 0.

$$x^{3}-3x+x^{2}+2=x^{3}-2x^{2}-8x$$

Subtract x^{3} from both sides.

$$x^{3}-3x+x^{2}+2-x^{3}=-2x^{2}-8x$$

Combine x^{3} and -x^{3} to get 0.

$$-3x+x^{2}+2=-2x^{2}-8x$$

Add 2x^{2} to both sides.

$$-3x+x^{2}+2+2x^{2}=-8x$$

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

$$-3x+3x^{2}+2=-8x$$

Add 8x to both sides.

$$-3x+3x^{2}+2+8x=0$$

Combine -3x and 8x to get 5x.

$$5x+3x^{2}+2=0$$

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

$$3x^{2}+5x+2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=5 ab=3\times 2=6$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 5.

$$a=2 b=3$$

Rewrite 3x^{2}+5x+2 as \left(3x^{2}+2x\right)+\left(3x+2\right).

$$\left(3x^{2}+2x\right)+\left(3x+2\right)$$

Factor out x in 3x^{2}+2x.

$$x\left(3x+2\right)+3x+2$$

Factor out common term 3x+2 by using distributive property.

$$\left(3x+2\right)\left(x+1\right)$$

To find equation solutions, solve 3x+2=0 and x+1=0.

$$x=-\frac{2}{3} x=-1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,1,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x-1\right)\left(x+2\right), the least common multiple of x^{2}+x-2,x^{2}-4,x^{2}-3x+2.

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

Multiply x+2 and x+2 to get \left(x+2\right)^{2}.

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

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

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

Use the distributive property to multiply x-1 by 3x+2 and combine like terms.

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

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

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

Combine -2x and -x to get -3x.

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

Subtract 2 from 4 to get 2.

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

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

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

Use the distributive property to multiply x^{2}-3x+2 by x+2 and combine like terms.

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

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

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

To find the opposite of x^{2}+4x+4, find the opposite of each term.

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

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

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

Combine -4x and -4x to get -8x.

$$x^{3}-3x+x^{2}+2=x^{3}-2x^{2}-8x+4-4$$

Subtract 4 from 4 to get 0.

$$x^{3}-3x+x^{2}+2=x^{3}-2x^{2}-8x$$

Subtract x^{3} from both sides.

$$x^{3}-3x+x^{2}+2-x^{3}=-2x^{2}-8x$$

Combine x^{3} and -x^{3} to get 0.

$$-3x+x^{2}+2=-2x^{2}-8x$$

Add 2x^{2} to both sides.

$$-3x+x^{2}+2+2x^{2}=-8x$$

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

$$-3x+3x^{2}+2=-8x$$

Add 8x to both sides.

$$-3x+3x^{2}+2+8x=0$$

Combine -3x and 8x to get 5x.

$$5x+3x^{2}+2=0$$

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.

$$3x^{2}+5x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 5 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 3\times 2}}{2\times 3}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 3\times 2}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-5±\sqrt{25-12\times 2}}{2\times 3}$$

Multiply -12 times 2.

$$x=\frac{-5±\sqrt{25-24}}{2\times 3}$$

Add 25 to -24.

$$x=\frac{-5±\sqrt{1}}{2\times 3}$$

Take the square root of 1.

$$x=\frac{-5±1}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-5±1}{6}$$

Now solve the equation x=\frac{-5±1}{6} when ± is plus. Add -5 to 1.

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

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

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

Now solve the equation x=\frac{-5±1}{6} when ± is minus. Subtract 1 from -5.

$$x=-\frac{6}{6}$$

Divide -6 by 6.

$$x=-1$$

The equation is now solved.

$$x=-\frac{2}{3} x=-1$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,1,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x-1\right)\left(x+2\right), the least common multiple of x^{2}+x-2,x^{2}-4,x^{2}-3x+2.

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

Multiply x+2 and x+2 to get \left(x+2\right)^{2}.

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

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

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

Use the distributive property to multiply x-1 by 3x+2 and combine like terms.

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

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

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

Combine -2x and -x to get -3x.

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

Subtract 2 from 4 to get 2.

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

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

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

Use the distributive property to multiply x^{2}-3x+2 by x+2 and combine like terms.

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

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

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

To find the opposite of x^{2}+4x+4, find the opposite of each term.

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

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

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

Combine -4x and -4x to get -8x.

$$x^{3}-3x+x^{2}+2=x^{3}-2x^{2}-8x+4-4$$

Subtract 4 from 4 to get 0.

$$x^{3}-3x+x^{2}+2=x^{3}-2x^{2}-8x$$

Subtract x^{3} from both sides.

$$x^{3}-3x+x^{2}+2-x^{3}=-2x^{2}-8x$$

Combine x^{3} and -x^{3} to get 0.

$$-3x+x^{2}+2=-2x^{2}-8x$$

Add 2x^{2} to both sides.

$$-3x+x^{2}+2+2x^{2}=-8x$$

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

$$-3x+3x^{2}+2=-8x$$

Add 8x to both sides.

$$-3x+3x^{2}+2+8x=0$$

Combine -3x and 8x to get 5x.

$$5x+3x^{2}+2=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$5x+3x^{2}=-2$$

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.

$$3x^{2}+5x=-2$$

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide \frac{5}{3}, the coefficient of the x term, by 2 to get \frac{5}{6}. Then add the square of \frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{5}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=-\frac{2}{3}+\frac{25}{36}$$

Add -\frac{2}{3} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{3}x+\frac{25}{36}=\frac{1}{36}$$

Factor x^{2}+\frac{5}{3}x+\frac{25}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x+\frac{5}{6}=\frac{1}{6} x+\frac{5}{6}=-\frac{1}{6}$$

Subtract \frac{5}{6} from both sides of the equation.

$$x=-\frac{2}{3} x=-1$$