Find the Domain (16x^3-16x^2+4x)/(12x^3+12x^2+3x)

$\frac{16 x^{3} - 16 x^{2} + 4 x}{12 x^{3} + 12 x^{2} + 3 x}$

Step 1

Set the denominator in $\frac{16 x^{3} - 16 x^{2} + 4 x}{12 x^{3} + 12 x^{2} + 3 x}$ equal to $0$ to find where the expression is undefined.

$12 x^{3} + 12 x^{2} + 3 x = 0$

Step 2

Solve for $x$ .

Factor the left side of the equation.

Factor $3 x$ out of $12 x^{3} + 12 x^{2} + 3 x$ .

Factor $3 x$ out of $12 x^{3}$ .

$3 x (4 x^{2}) + 12 x^{2} + 3 x = 0$

Factor $3 x$ out of $12 x^{2}$ .

$3 x (4 x^{2}) + 3 x (4 x) + 3 x = 0$

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

$3 x (4 x^{2}) + 3 x (4 x) + 3 x (1) = 0$

Factor $3 x$ out of $3 x (4 x^{2}) + 3 x (4 x)$ .

$3 x (4 x^{2} + 4 x) + 3 x (1) = 0$

Factor $3 x$ out of $3 x (4 x^{2} + 4 x) + 3 x (1)$ .

$3 x (4 x^{2} + 4 x + 1) = 0$

Factor using the perfect square rule.

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$3 x ({(2 x)}^{2} + 4 x + 1) = 0$

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

$3 x ({(2 x)}^{2} + 4 x + 1^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot (2 x) \cdot 1$

Rewrite the polynomial.

$3 x ({(2 x)}^{2} + 2 \cdot (2 x) \cdot 1 + 1^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 2 x$ and $b = 1$ .

$3 x {(2 x + 1)}^{2} = 0$

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

$x = 0$

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

Set $x$ equal to $0$ .

$x = 0$

Set ${(2 x + 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(2 x + 1)}^{2}$ equal to $0$ .

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

Solve ${(2 x + 1)}^{2} = 0$ for $x$ .

Set the $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Solve for $x$ .

Subtract $1$ from 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 $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

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

$x = 0, - \frac{1}{2}$

Step 3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, - \frac{1}{2}) \cup (- \frac{1}{2}, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0, - \frac{1}{2}}$

Step 4

Find the Domain (16x^3-16x^2+4x)/(12x^3+12x^2+3x)

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