Solve (x-2)(x+4)(x+2)/(x^2-4)=0

$$\frac { ( x - 2 ) ( x + 4 ) ( x + 2 ) } { ( x ^ { 2 } - 4 ) } = 0$$

Solve for x

$$x=-4$$

Solution Steps

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+2\right).

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

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

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

Use the distributive property to multiply x^{2}+2x-8 by x+2 and combine like terms.

$$x^{3}+4x^{2}-4x-16=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -16 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±16,±8,±4,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+4x^{2}-4x-16 by x-2 to get x^{2}+6x+8. Solve the equation where the result equals to 0.

$$x^{2}+6x+8=0$$

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, 6 for b, and 8 for c in the quadratic formula.

$$x=\frac{-6±\sqrt{6^{2}-4\times 1\times 8}}{2}$$

Do the calculations.

$$x=\frac{-6±2}{2}$$

Solve the equation x^{2}+6x+8=0 when ± is plus and when ± is minus.

$$x=-4 x=-2$$

Remove the values that the variable cannot be equal to.

$$x=-4$$

List all found solutions.

$$x=2 x=-4 x=-2$$

Variable x cannot be equal to any of the values 2,-2.

$$x=-4$$