Find Where Undefined/Discontinuous f(x)=(-11x+1)/(-30x^2-25x+5)

$f (x) = \frac{- 11 x + 1}{- 30 x^{2} - 25 x + 5}$

Set the denominator in $\frac{- 11 x + 1}{- 30 x^{2} - 25 x + 5}$ equal to $0$ to find where the expression is undefined.

$- 30 x^{2} - 25 x + 5 = 0$

Solve for $x$ .

Factor the left side of the equation.

Factor $- 5$ out of $- 30 x^{2} - 25 x + 5$ .

Factor $- 5$ out of $- 30 x^{2}$ .

$- 5 (6 x^{2}) - 25 x + 5 = 0$

Factor $- 5$ out of $- 25 x$ .

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

Factor $- 5$ out of $5$ .

$- 5 (6 x^{2}) - 5 (5 x) - 5 \cdot - 1 = 0$

Factor $- 5$ out of $- 5 (6 x^{2}) - 5 (5 x)$ .

$- 5 (6 x^{2} + 5 x) - 5 \cdot - 1 = 0$

Factor $- 5$ out of $- 5 (6 x^{2} + 5 x) - 5 (- 1)$ .

$- 5 (6 x^{2} + 5 x - 1) = 0$

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 $a \cdot c = 6 \cdot - 1 = - 6$ and whose sum is $b = 5$ .

Factor $5$ out of $5 x$ .

$- 5 (6 x^{2} + 5 (x) - 1) = 0$

Rewrite $5$ as $- 1$ plus $6$

$- 5 (6 x^{2} + (- 1 + 6) x - 1) = 0$

Apply the distributive property.

$- 5 (6 x^{2} - 1 x + 6 x - 1) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- 5 ((6 x^{2} - 1 x) + 6 x - 1) = 0$

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

$- 5 (x (6 x - 1) + 1 (6 x - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $6 x - 1$ .

$- 5 ((6 x - 1) (x + 1)) = 0$

Remove unnecessary parentheses.

$- 5 (6 x - 1) (x + 1) = 0$

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

$6 x - 1 = 0$

$x + 1 = 0$

Set $6 x - 1$ equal to $0$ and solve for $x$ .

Set $6 x - 1$ equal to $0$ .

$6 x - 1 = 0$

Solve $6 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$6 x = 1$

Divide each term in $6 x = 1$ by $6$ and simplify.

Divide each term in $6 x = 1$ by $6$ .

$\frac{6 x}{6} = \frac{1}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} = \frac{1}{6}$

Divide $x$ by $1$ .

$x = \frac{1}{6}$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $- 5 (6 x - 1) (x + 1) = 0$ true.

$x = \frac{1}{6}, - 1$

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 $0$ .

$x = - 1, x = \frac{1}{6}$

Find Where Undefined/Discontinuous f(x)=(-11x+1)/(-30x^2-25x+5)

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