Find the Excluded Values (x+5)/(x^3-7x^2+12x)

$\frac{x + 5}{x^{3} - 7 x^{2} + 12 x}$

Set the denominator in $\frac{x + 5}{x^{3} - 7 x^{2} + 12 x}$ equal to $0$ to find where the expression is undefined.

$x^{3} - 7 x^{2} + 12 x = 0$

Solve for $x$ .

Factor the left side of the equation.

Factor $x$ out of $x^{3} - 7 x^{2} + 12 x$ .

Factor $x$ out of $x^{3}$ .

$x \cdot x^{2} - 7 x^{2} + 12 x = 0$

Factor $x$ out of $- 7 x^{2}$ .

$x \cdot x^{2} + x (- 7 x) + 12 x = 0$

Factor $x$ out of $12 x$ .

$x \cdot x^{2} + x (- 7 x) + x \cdot 12 = 0$

Factor $x$ out of $x \cdot x^{2} + x (- 7 x)$ .

$x (x^{2} - 7 x) + x \cdot 12 = 0$

Factor $x$ out of $x (x^{2} - 7 x) + x \cdot 12$ .

$x (x^{2} - 7 x + 12) = 0$

Factor.

Factor $x^{2} - 7 x + 12$ 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 $12$ and whose sum is $- 7$ .

$- 4, - 3$

Write the factored form using these integers.

$x ((x - 4) (x - 3)) = 0$

Remove unnecessary parentheses.

$x (x - 4) (x - 3) = 0$

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

$x = 0$

$x - 4 = 0$

$x - 3 = 0$

Set $x$ equal to $0$ .

$x = 0$

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 $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $x (x - 4) (x - 3) = 0$ true.

$x = 0, 4, 3$

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = 0, x = 3, x = 4$

Find the Excluded Values (x+5)/(x^3-7x^2+12x)

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