Find the Domain x^2y-4y=x^2+4

$x^{2} y - 4 y = x^{2} + 4$

Step 1

Factor $y$ out of $x^{2} y - 4 y$ .

Factor $y$ out of $x^{2} y$ .

$y x^{2} - 4 y = x^{2} + 4$

Factor $y$ out of $- 4 y$ .

$y x^{2} + y \cdot - 4 = x^{2} + 4$

Factor $y$ out of $y x^{2} + y \cdot - 4$ .

$y (x^{2} - 4) = x^{2} + 4$

Step 2

Rewrite $4$ as $2^{2}$ .

$y (x^{2} - 2^{2}) = x^{2} + 4$

Step 3

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

$y ((x + 2) (x - 2)) = x^{2} + 4$

Remove unnecessary parentheses.

$y (x + 2) (x - 2) = x^{2} + 4$

Step 4

Divide each term in $y (x + 2) (x - 2) = x^{2} + 4$ by $(x + 2) (x - 2)$ and simplify.

Divide each term in $y (x + 2) (x - 2) = x^{2} + 4$ by $(x + 2) (x - 2)$ .

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

Simplify the left side.

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Step 5

Set the denominator in $\frac{x^{2}}{(x + 2) (x - 2)}$ equal to $0$ to find where the expression is undefined.

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

Step 6

Solve for $x$ .

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

$x + 2 = 0$

$x - 2 = 0$

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.

$x = - 2$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

$x = - 2, 2$

Step 7

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, - 2) \cup (- 2, 2) \cup (2, \infty)$

Set-Builder Notation:

${x | x \neq 2, - 2}$

Step 8

Find the Domain x^2y-4y=x^2+4

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