Find Where Undefined/Discontinuous (3x^2+16x+5)/(9x^2-1)*(3x^2+2x-1)/(x^2+4x-5)

$\frac{3 x^{2} + 16 x + 5}{9 x^{2} - 1} \cdot \frac{3 x^{2} + 2 x - 1}{x^{2} + 4 x - 5}$

Step 1

Set the denominator in $\frac{3 x^{2} + 16 x + 5}{9 x^{2} - 1}$ equal to $0$ to find where the expression is undefined.

$9 x^{2} - 1 = 0$

Step 2

Solve for $x$ .

Add $1$ to both sides of the equation.

$9 x^{2} = 1$

Divide each term in $9 x^{2} = 1$ by $9$ and simplify.

Divide each term in $9 x^{2} = 1$ by $9$ .

$\frac{9 x^{2}}{9} = \frac{1}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 x^{2}}{9} = \frac{1}{9}$

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

$x^{2} = \frac{1}{9}$

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

$x = \pm \sqrt{\frac{1}{9}}$

Simplify $\pm \sqrt{\frac{1}{9}}$ .

Rewrite $\sqrt{\frac{1}{9}}$ as $\frac{\sqrt{1}}{\sqrt{9}}$ .

$x = \pm \frac{\sqrt{1}}{\sqrt{9}}$

Any root of $1$ is $1$ .

$x = \pm \frac{1}{\sqrt{9}}$

Simplify the denominator.

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

$x = \pm \frac{1}{\sqrt{3^{2}}}$

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

$x = \pm \frac{1}{3}$

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.

$x = \frac{1}{3}$

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

$x = - \frac{1}{3}$

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

$x = \frac{1}{3}, - \frac{1}{3}$

Step 3

Set the denominator in $\frac{3 x^{2} + 2 x - 1}{x^{2} + 4 x - 5}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 4 x - 5 = 0$

Step 4

Solve for $x$ .

Factor $x^{2} + 4 x - 5$ 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 $- 5$ and whose sum is $4$ .

$- 1, 5$

Write the factored form using these integers.

$(x - 1) (x + 5) = 0$

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

$x - 1 = 0$

$x + 5 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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 $(x - 1) (x + 5) = 0$ true.

$x = 1, - 5$

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 $0$ .

$x = - 5, x = - \frac{1}{3}, x = \frac{1}{3}, x = 1$

Step 6

Find Where Undefined/Discontinuous (3x^2+16x+5)/(9x^2-1)*(3x^2+2x-1)/(x^2+4x-5)

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