Find Where Undefined/Discontinuous (4x^2+23x+15)/(16x^2-9)*(4x^2+9x-9)/(x^2+2x-15)

$\frac{4 x^{2} + 23 x + 15}{16 x^{2} - 9} \cdot \frac{4 x^{2} + 9 x - 9}{x^{2} + 2 x - 15}$

Step 1

Set the denominator in $\frac{4 x^{2} + 23 x + 15}{16 x^{2} - 9}$ equal to $0$ to find where the expression is undefined.

$16 x^{2} - 9 = 0$

Step 2

Solve for $x$ .

Add $9$ to both sides of the equation.

$16 x^{2} = 9$

Divide each term in $16 x^{2} = 9$ by $16$ and simplify.

Divide each term in $16 x^{2} = 9$ by $16$ .

$\frac{16 x^{2}}{16} = \frac{9}{16}$

Simplify the left side.

Cancel the common factor of $16$ .

Cancel the common factor.

$\frac{16 x^{2}}{16} = \frac{9}{16}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{9}{16}$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{\frac{9}{16}}$

Simplify $\pm \sqrt{\frac{9}{16}}$ .

Rewrite $\sqrt{\frac{9}{16}}$ as $\frac{\sqrt{9}}{\sqrt{16}}$ .

$x = \pm \frac{\sqrt{9}}{\sqrt{16}}$

Simplify the numerator.

Rewrite $9$ as $3^{2}$ .

$x = \pm \frac{\sqrt{3^{2}}}{\sqrt{16}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm \frac{3}{\sqrt{16}}$

Simplify the denominator.

Rewrite $16$ as $4^{2}$ .

$x = \pm \frac{3}{\sqrt{4^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm \frac{3}{4}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = \frac{3}{4}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \frac{3}{4}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \frac{3}{4}, - \frac{3}{4}$

Step 3

Set the denominator in $\frac{4 x^{2} + 9 x - 9}{x^{2} + 2 x - 15}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 2 x - 15 = 0$

Step 4

Solve for $x$ .

Factor $x^{2} + 2 x - 15$ 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 $- 15$ and whose sum is $2$ .

$- 3, 5$

Write the factored form using these integers.

$(x - 3) (x + 5) = 0$

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

$x - 3 = 0$

$x + 5 = 0$

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.

$x = 3$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $(x - 3) (x + 5) = 0$ true.

$x = 3, - 5$

Step 5

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 $0$ .

$x = - 5, x = - \frac{3}{4}, x = \frac{3}{4}, x = 3$

Step 6

Find Where Undefined/Discontinuous (4x^2+23x+15)/(16x^2-9)*(4x^2+9x-9)/(x^2+2x-15)

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