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

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

Solve for x

$$x=2$$

Solution Steps

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by x+1.

$$\left(x+1\right)x^{2}-2x\left(x+1\right)+\left(x+1\right)\left(-8\right)+x^{2}-x-2+\left(x+1\right)\times 8=0$$

Use the distributive property to multiply x+1 by x^{2}.

$$x^{3}+x^{2}-2x\left(x+1\right)+\left(x+1\right)\left(-8\right)+x^{2}-x-2+\left(x+1\right)\times 8=0$$

Use the distributive property to multiply -2x by x+1.

$$x^{3}+x^{2}-2x^{2}-2x+\left(x+1\right)\left(-8\right)+x^{2}-x-2+\left(x+1\right)\times 8=0$$

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

$$x^{3}-x^{2}-2x+\left(x+1\right)\left(-8\right)+x^{2}-x-2+\left(x+1\right)\times 8=0$$

Use the distributive property to multiply x+1 by -8.

$$x^{3}-x^{2}-2x-8x-8+x^{2}-x-2+\left(x+1\right)\times 8=0$$

Combine -2x and -8x to get -10x.

$$x^{3}-x^{2}-10x-8+x^{2}-x-2+\left(x+1\right)\times 8=0$$

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

$$x^{3}-10x-8-x-2+\left(x+1\right)\times 8=0$$

Combine -10x and -x to get -11x.

$$x^{3}-11x-8-2+\left(x+1\right)\times 8=0$$

Subtract 2 from -8 to get -10.

$$x^{3}-11x-10+\left(x+1\right)\times 8=0$$

Use the distributive property to multiply x+1 by 8.

$$x^{3}-11x-10+8x+8=0$$

Combine -11x and 8x to get -3x.

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

Add -10 and 8 to get -2.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -2 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-3x-2 by x+1 to get x^{2}-x-2. Solve the equation where the result equals to 0.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -1 for b, and -2 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{1±3}{2}$$

Solve the equation x^{2}-x-2=0 when ± is plus and when ± is minus.

$$x=-1 x=2$$

Remove the values that the variable cannot be equal to.

$$x=2$$

List all found solutions.

$$x=-1 x=2$$

Variable x cannot be equal to -1.

$$x=2$$