Find Where Undefined/Discontinuous (2x^2-32)/(6x^2+13x-5)

$\frac{2 x^{2} - 32}{6 x^{2} + 13 x - 5}$

Step 1

Set the denominator in $\frac{2 x^{2} - 32}{6 x^{2} + 13 x - 5}$ equal to $0$ to find where the expression is undefined.

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

Step 2

Solve for $x$ .

Factor by grouping.

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 $a \cdot c = 6 \cdot - 5 = - 30$ and whose sum is $b = 13$ .

Factor $13$ out of $13 x$ .

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

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

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$2 x (3 x - 1) + 5 (3 x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

$(3 x - 1) (2 x + 5) = 0$

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

$3 x - 1 = 0$

$2 x + 5 = 0$

Set $3 x - 1$ equal to $0$ and solve for $x$ .

Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Solve $3 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$3 x = 1$

Divide each term in $3 x = 1$ by $3$ and simplify.

Divide each term in $3 x = 1$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

$2 x + 5 = 0$

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

Subtract $5$ from both sides of the equation.

$2 x = - 5$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

The final solution is all the values that make $(3 x - 1) (2 x + 5) = 0$ true.

$x = \frac{1}{3}, - \frac{5}{2}$

Step 3

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 $0$ .

$x = - \frac{5}{2}, x = \frac{1}{3}$

Step 4

Find Where Undefined/Discontinuous (2x^2-32)/(6x^2+13x-5)

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