Find Where Undefined/Discontinuous (-8x^2+8x-2)/(32x^2-4x-6)

$\frac{- 8 x^{2} + 8 x - 2}{32 x^{2} - 4 x - 6}$

Step 1

Set the denominator in $\frac{- 8 x^{2} + 8 x - 2}{32 x^{2} - 4 x - 6}$ equal to $0$ to find where the expression is undefined.

$32 x^{2} - 4 x - 6 = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $2$ out of $32 x^{2} - 4 x - 6$ .

Factor $2$ out of $32 x^{2}$ .

$2 (16 x^{2}) - 4 x - 6 = 0$

Factor $2$ out of $- 4 x$ .

$2 (16 x^{2}) + 2 (- 2 x) - 6 = 0$

Factor $2$ out of $- 6$ .

$2 (16 x^{2}) + 2 (- 2 x) + 2 (- 3) = 0$

Factor $2$ out of $2 (16 x^{2}) + 2 (- 2 x)$ .

$2 (16 x^{2} - 2 x) + 2 (- 3) = 0$

Factor $2$ out of $2 (16 x^{2} - 2 x) + 2 (- 3)$ .

$2 (16 x^{2} - 2 x - 3) = 0$

Factor.

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 $a \cdot c = 16 \cdot - 3 = - 48$ and whose sum is $b = - 2$ .

Factor $- 2$ out of $- 2 x$ .

$2 (16 x^{2} - 2 x - 3) = 0$

Rewrite $- 2$ as $6$ plus $- 8$

$2 (16 x^{2} + (6 - 8) x - 3) = 0$

Apply the distributive property.

$2 (16 x^{2} + 6 x - 8 x - 3) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$2 ((16 x^{2} + 6 x) - 8 x - 3) = 0$

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

$2 (2 x (8 x + 3) - (8 x + 3)) = 0$

Factor the polynomial by factoring out the greatest common factor, $8 x + 3$ .

$2 ((8 x + 3) (2 x - 1)) = 0$

Remove unnecessary parentheses.

$2 (8 x + 3) (2 x - 1) = 0$

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

$8 x + 3 = 0$

$2 x - 1 = 0$

Set $8 x + 3$ equal to $0$ and solve for $x$ .

Set $8 x + 3$ equal to $0$ .

$8 x + 3 = 0$

Solve $8 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$8 x = - 3$

Divide each term in $8 x = - 3$ by $8$ and simplify.

Divide each term in $8 x = - 3$ by $8$ .

$\frac{8 x}{8} = \frac{- 3}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 x}{8} = \frac{- 3}{8}$

Divide $x$ by $1$ .

$x = \frac{- 3}{8}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{8}$

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

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

The final solution is all the values that make $2 (8 x + 3) (2 x - 1) = 0$ true.

$x = - \frac{3}{8}, \frac{1}{2}$

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 $0$ .

$x = - \frac{3}{8}, x = \frac{1}{2}$

Step 4

Find Where Undefined/Discontinuous (-8x^2+8x-2)/(32x^2-4x-6)

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