Solve 10/x^2-2x-8+5/x+2+1=0

$$\frac{ 10 }{ { x }^{ 2 } -2x-8 } + \frac{ 5 }{ x+2 } +1=0$$

Solve for x

$$x=-6$$

$$x=3$$

Steps Using Factoring

Variable x cannot be equal to any of the values -2,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x+2\right), the least common multiple of x^{2}-2x-8,x+2.

$$10+\left(x-4\right)\times 5+\left(x-4\right)\left(x+2\right)=0$$

Use the distributive property to multiply x-4 by 5.

$$10+5x-20+\left(x-4\right)\left(x+2\right)=0$$

Subtract 20 from 10 to get -10.

$$-10+5x+\left(x-4\right)\left(x+2\right)=0$$

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

$$-10+5x+x^{2}-2x-8=0$$

Combine 5x and -2x to get 3x.

$$-10+3x+x^{2}-8=0$$

Subtract 8 from -10 to get -18.

$$-18+3x+x^{2}=0$$

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

$$x^{2}+3x-18=0$$

To solve the equation, factor x^{2}+3x-18 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=3 ab=-18$$

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 -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 3.

$$a=-3 b=6$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-3\right)\left(x+6\right)$$

To find equation solutions, solve x-3=0 and x+6=0.

$$x=3 x=-6$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -2,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x+2\right), the least common multiple of x^{2}-2x-8,x+2.

$$10+\left(x-4\right)\times 5+\left(x-4\right)\left(x+2\right)=0$$

Use the distributive property to multiply x-4 by 5.

$$10+5x-20+\left(x-4\right)\left(x+2\right)=0$$

Subtract 20 from 10 to get -10.

$$-10+5x+\left(x-4\right)\left(x+2\right)=0$$

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

$$-10+5x+x^{2}-2x-8=0$$

Combine 5x and -2x to get 3x.

$$-10+3x+x^{2}-8=0$$

Subtract 8 from -10 to get -18.

$$-18+3x+x^{2}=0$$

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

$$x^{2}+3x-18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-18. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-18\right)=-18$$

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 -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 3.

$$a=-3 b=6$$

Rewrite x^{2}+3x-18 as \left(x^{2}-3x\right)+\left(6x-18\right).

$$\left(x^{2}-3x\right)+\left(6x-18\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(x-3\right)+6\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(x+6\right)$$

To find equation solutions, solve x-3=0 and x+6=0.

$$x=3 x=-6$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x+2\right), the least common multiple of x^{2}-2x-8,x+2.

$$10+\left(x-4\right)\times 5+\left(x-4\right)\left(x+2\right)=0$$

Use the distributive property to multiply x-4 by 5.

$$10+5x-20+\left(x-4\right)\left(x+2\right)=0$$

Subtract 20 from 10 to get -10.

$$-10+5x+\left(x-4\right)\left(x+2\right)=0$$

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

$$-10+5x+x^{2}-2x-8=0$$

Combine 5x and -2x to get 3x.

$$-10+3x+x^{2}-8=0$$

Subtract 8 from -10 to get -18.

$$-18+3x+x^{2}=0$$

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.

$$x^{2}+3x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-18\right)}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-18\right)}}{2}$$

Multiply -4 times -18.

$$x=\frac{-3±\sqrt{9+72}}{2}$$

Add 9 to 72.

$$x=\frac{-3±\sqrt{81}}{2}$$

Take the square root of 81.

$$x=\frac{-3±9}{2}$$

Now solve the equation x=\frac{-3±9}{2} when ± is plus. Add -3 to 9.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{-3±9}{2} when ± is minus. Subtract 9 from -3.

$$x=-\frac{12}{2}$$

Divide -12 by 2.

$$x=-6$$

The equation is now solved.

$$x=3 x=-6$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x+2\right), the least common multiple of x^{2}-2x-8,x+2.

$$10+\left(x-4\right)\times 5+\left(x-4\right)\left(x+2\right)=0$$

Use the distributive property to multiply x-4 by 5.

$$10+5x-20+\left(x-4\right)\left(x+2\right)=0$$

Subtract 20 from 10 to get -10.

$$-10+5x+\left(x-4\right)\left(x+2\right)=0$$

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

$$-10+5x+x^{2}-2x-8=0$$

Combine 5x and -2x to get 3x.

$$-10+3x+x^{2}-8=0$$

Subtract 8 from -10 to get -18.

$$-18+3x+x^{2}=0$$

Add 18 to both sides. Anything plus zero gives itself.

$$3x+x^{2}=18$$

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.

$$x^{2}+3x=18$$

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.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=18+\left(\frac{3}{2}\right)^{2}$$

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

$$x^{2}+3x+\frac{9}{4}=18+\frac{9}{4}$$

Add 18 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{81}{4}$$

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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

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

Subtract \frac{3}{2} from both sides of the equation.

$$x=3 x=-6$$