Precalculus Examples

$\frac{(2 x^{2} - 10 x)}{(x - 5) (x^{2} - 1)} \cdot (3 x^{2} + 4 x + 1)$

Step 1

Set the denominator in $\frac{2 x^{2} - 10 x}{(x - 5) (x^{2} - 1)}$ equal to $0$ to find where the expression is undefined.

$(x - 5) (x^{2} - 1) = 0$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

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

$x - 5 = 0$

$x^{2} - 1 = 0$

Step 2.2

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

Tap for more steps...

Step 2.2.1

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

$x - 5 = 0$

Step 2.2.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.3

Set $x^{2} - 1$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.3.1

Set $x^{2} - 1$ equal to $0$ .

$x^{2} - 1 = 0$

Step 2.3.2

Solve $x^{2} - 1 = 0$ for $x$ .

Tap for more steps...

Step 2.3.2.1

Add $1$ to both sides of the equation.

$x^{2} = 1$

Step 2.3.2.2

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

$x = \pm \sqrt{1}$

Step 2.3.2.3

Any root of $1$ is $1$ .

$x = \pm 1$

Step 2.3.2.4

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

Tap for more steps...

Step 2.3.2.4.1

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

$x = 1$

Step 2.3.2.4.2

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

$x = - 1$

Step 2.3.2.4.3

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

$x = 1, - 1$

Step 2.4

The final solution is all the values that make $(x - 5) (x^{2} - 1) = 0$ true.

$x = 5, 1, - 1$

Step 3

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

Interval Notation:

$(- \infty, - 1) \cup (- 1, 1) \cup (1, 5) \cup (5, \infty)$

Set-Builder Notation:

${x | x \neq 1, - 1, 5}$

Step 4