Find Where Undefined/Discontinuous (4x^4+6x^3)/(6x^2+9x)

$\frac{4 x^{4} + 6 x^{3}}{6 x^{2} + 9 x}$

Step 1

Set the denominator in $\frac{4 x^{4} + 6 x^{3}}{6 x^{2} + 9 x}$ equal to $0$ to find where the expression is undefined.

$6 x^{2} + 9 x = 0$

Step 2

Solve for $x$ .

Factor $3 x$ out of $6 x^{2} + 9 x$ .

Factor $3 x$ out of $6 x^{2}$ .

$3 x (2 x) + 9 x = 0$

Factor $3 x$ out of $9 x$ .

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

Factor $3 x$ out of $3 x (2 x) + 3 x (3)$ .

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

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

$x = 0$

$2 x + 3 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $2 x + 3$ equal to $0$ and solve for $x$ .

Set $2 x + 3$ equal to $0$ .

$2 x + 3 = 0$

Solve $2 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Divide each term in $2 x = - 3$ by $2$ and simplify.

Divide each term in $2 x = - 3$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

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

$x = 0, - \frac{3}{2}$

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 $0$ .

$x = - \frac{3}{2}, x = 0$

Step 4

Find Where Undefined/Discontinuous (4x^4+6x^3)/(6x^2+9x)

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