Find the Roots (Zeros) f(x)=(x^2-2x-8)/(x^2-4x)

$f (x) = \frac{x^{2} - 2 x - 8}{x^{2} - 4 x}$

Step 1

Set $\frac{x^{2} - 2 x - 8}{x^{2} - 4 x}$ equal to $0$ .

$\frac{x^{2} - 2 x - 8}{x^{2} - 4 x} = 0$

Step 2

Solve for $x$ .

Set the numerator equal to zero.

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

Solve the equation for $x$ .

Factor $x^{2} - 2 x - 8$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 8$ and whose sum is $- 2$ .

$- 4, 2$

Write the factored form using these integers.

$(x - 4) (x + 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 4 = 0$

$x + 2 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $(x - 4) (x + 2) = 0$ true.

$x = 4, - 2$

Exclude the solutions that do not make $\frac{x^{2} - 2 x - 8}{x^{2} - 4 x} = 0$ true.

$x = - 2$

Step 3

Find the Roots (Zeros) f(x)=(x^2-2x-8)/(x^2-4x)

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