Find Where Undefined/Discontinuous (5k^2-26k+5)/(25k^3-k)

$\frac{5 k^{2} - 26 k + 5}{25 k^{3} - k}$

Step 1

Set the denominator in $\frac{5 k^{2} - 26 k + 5}{25 k^{3} - k}$ equal to $0$ to find where the expression is undefined.

$25 k^{3} - k = 0$

Step 2

Solve for $k$ .

Factor the left side of the equation.

Factor $k$ out of $25 k^{3} - k$ .

Factor $k$ out of $25 k^{3}$ .

$k (25 k^{2}) - k = 0$

Factor $k$ out of $- k$ .

$k (25 k^{2}) + k \cdot - 1 = 0$

Factor $k$ out of $k (25 k^{2}) + k \cdot - 1$ .

$k (25 k^{2} - 1) = 0$

Rewrite $25 k^{2}$ as ${(5 k)}^{2}$ .

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

Rewrite $1$ as $1^{2}$ .

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 5 k$ and $b = 1$ .

$k ((5 k + 1) (5 k - 1)) = 0$

Remove unnecessary parentheses.

$k (5 k + 1) (5 k - 1) = 0$

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

$k = 0$

$5 k + 1 = 0$

$5 k - 1 = 0$

Set $k$ equal to $0$ .

$k = 0$

Set $5 k + 1$ equal to $0$ and solve for $k$ .

Set $5 k + 1$ equal to $0$ .

$5 k + 1 = 0$

Solve $5 k + 1 = 0$ for $k$ .

Subtract $1$ from both sides of the equation.

$5 k = - 1$

Divide each term in $5 k = - 1$ by $5$ and simplify.

Divide each term in $5 k = - 1$ by $5$ .

$\frac{5 k}{5} = \frac{- 1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 k}{5} = \frac{- 1}{5}$

Divide $k$ by $1$ .

$k = \frac{- 1}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$k = - \frac{1}{5}$

Set $5 k - 1$ equal to $0$ and solve for $k$ .

Set $5 k - 1$ equal to $0$ .

$5 k - 1 = 0$

Solve $5 k - 1 = 0$ for $k$ .

Add $1$ to both sides of the equation.

$5 k = 1$

Divide each term in $5 k = 1$ by $5$ and simplify.

Divide each term in $5 k = 1$ by $5$ .

$\frac{5 k}{5} = \frac{1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 k}{5} = \frac{1}{5}$

Divide $k$ by $1$ .

$k = \frac{1}{5}$

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

$k = 0, - \frac{1}{5}, \frac{1}{5}$

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 $0$ .

$k = - \frac{1}{5}, k = 0, k = \frac{1}{5}$

Step 4

Find Where Undefined/Discontinuous (5k^2-26k+5)/(25k^3-k)

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