Linear Algebra Examples

$\frac{2 x^{2} + 16 x + 30}{5 x^{2} + 13 x - 6}$

Step 1

Set the denominator in $\frac{2 x^{2} + 16 x + 30}{5 x^{2} + 13 x - 6}$ equal to $0$ to find where the expression is undefined.

$5 x^{2} + 13 x - 6 = 0$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Factor by grouping.

Tap for more steps...

Step 2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 5 \cdot - 6 = - 30$ and whose sum is $b = 13$ .

Tap for more steps...

Step 2.1.1.1

Factor $13$ out of $13 x$ .

$5 x^{2} + 13 (x) - 6 = 0$

Step 2.1.1.2

Rewrite $13$ as $- 2$ plus $15$

$5 x^{2} + (- 2 + 15) x - 6 = 0$

Step 2.1.1.3

Apply the distributive property.

$5 x^{2} - 2 x + 15 x - 6 = 0$

Step 2.1.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 2.1.2.1

Group the first two terms and the last two terms.

$(5 x^{2} - 2 x) + 15 x - 6 = 0$

Step 2.1.2.2

Factor out the greatest common factor (GCF) from each group.

$x (5 x - 2) + 3 (5 x - 2) = 0$

Step 2.1.3

Factor the polynomial by factoring out the greatest common factor, $5 x - 2$ .

$(5 x - 2) (x + 3) = 0$

Step 2.2

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

$5 x - 2 = 0$

$x + 3 = 0$

Step 2.3

Set $5 x - 2$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.3.1

Set $5 x - 2$ equal to $0$ .

$5 x - 2 = 0$

Step 2.3.2

Solve $5 x - 2 = 0$ for $x$ .

Tap for more steps...

Step 2.3.2.1

Add $2$ to both sides of the equation.

$5 x = 2$

Step 2.3.2.2

Divide each term in $5 x = 2$ by $5$ and simplify.

Tap for more steps...

Step 2.3.2.2.1

Divide each term in $5 x = 2$ by $5$ .

$\frac{5 x}{5} = \frac{2}{5}$

Step 2.3.2.2.2

Simplify the left side.

Tap for more steps...

Step 2.3.2.2.2.1

Cancel the common factor of $5$ .

Tap for more steps...

Step 2.3.2.2.2.1.1

Cancel the common factor.

$\frac{5 x}{5} = \frac{2}{5}$

Step 2.3.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{2}{5}$

Step 2.4

Set $x + 3$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.4.1

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

$x + 3 = 0$

Step 2.4.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.5

The final solution is all the values that make $(5 x - 2) (x + 3) = 0$ true.

$x = \frac{2}{5}, - 3$

Step 3

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

Interval Notation:

$(- \infty, - 3) \cup (- 3, \frac{2}{5}) \cup (\frac{2}{5}, \infty)$

Set-Builder Notation:

${x | x \neq \frac{2}{5}, - 3}$

Step 4