Find Where Undefined/Discontinuous ((6n+42)/(n^3-18n^2+81n))÷((n^2+14n+49)/(9n^4))

$\frac{6 n + 42}{n^{3} - 18 n^{2} + 81 n} \div \frac{n^{2} + 14 n + 49}{9 n^{4}}$

Step 1

Set the denominator in $\frac{6 n + 42}{n^{3} - 18 n^{2} + 81 n}$ equal to $0$ to find where the expression is undefined.

$n^{3} - 18 n^{2} + 81 n = 0$

Step 2

Solve for $n$ .

Factor the left side of the equation.

Factor $n$ out of $n^{3} - 18 n^{2} + 81 n$ .

Factor $n$ out of $n^{3}$ .

$n \cdot n^{2} - 18 n^{2} + 81 n = 0$

Factor $n$ out of $- 18 n^{2}$ .

$n \cdot n^{2} + n (- 18 n) + 81 n = 0$

Factor $n$ out of $81 n$ .

$n \cdot n^{2} + n (- 18 n) + n \cdot 81 = 0$

Factor $n$ out of $n \cdot n^{2} + n (- 18 n)$ .

$n (n^{2} - 18 n) + n \cdot 81 = 0$

Factor $n$ out of $n (n^{2} - 18 n) + n \cdot 81$ .

$n (n^{2} - 18 n + 81) = 0$

Factor using the perfect square rule.

Rewrite $81$ as $9^{2}$ .

$n (n^{2} - 18 n + 9^{2}) = 0$

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

$18 n = 2 \cdot n \cdot 9$

Rewrite the polynomial.

$n (n^{2} - 2 \cdot n \cdot 9 + 9^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = n$ and $b = 9$ .

$n {(n - 9)}^{2} = 0$

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

$n = 0$

${(n - 9)}^{2} = 0$

Set $n$ equal to $0$ .

$n = 0$

Set ${(n - 9)}^{2}$ equal to $0$ and solve for $n$ .

Set ${(n - 9)}^{2}$ equal to $0$ .

${(n - 9)}^{2} = 0$

Solve ${(n - 9)}^{2} = 0$ for $n$ .

Set the $n - 9$ equal to $0$ .

$n - 9 = 0$

Add $9$ to both sides of the equation.

$n = 9$

The final solution is all the values that make $n {(n - 9)}^{2} = 0$ true.

$n = 0, 9$

Step 3

Set the denominator in $\frac{n^{2} + 14 n + 49}{9 n^{4}}$ equal to $0$ to find where the expression is undefined.

$9 n^{4} = 0$

Step 4

Solve for $n$ .

Divide each term in $9 n^{4} = 0$ by $9$ and simplify.

Divide each term in $9 n^{4} = 0$ by $9$ .

$\frac{9 n^{4}}{9} = \frac{0}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 n^{4}}{9} = \frac{0}{9}$

Divide $n^{4}$ by $1$ .

$n^{4} = \frac{0}{9}$

Simplify the right side.

Divide $0$ by $9$ .

$n^{4} = 0$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$n = \pm \sqrt[4]{0}$

Simplify $\pm \sqrt[4]{0}$ .

Rewrite $0$ as $0^{4}$ .

$n = \pm \sqrt[4]{0^{4}}$

Pull terms out from under the radical, assuming positive real numbers.

$n = \pm 0$

Plus or minus $0$ is $0$ .

$n = 0$

Step 5

Set the denominator in $\frac{\frac{6 n + 42}{n^{3} - 18 n^{2} + 81 n}}{\frac{n^{2} + 14 n + 49}{9 n^{4}}}$ equal to $0$ to find where the expression is undefined.

$\frac{n^{2} + 14 n + 49}{9 n^{4}} = 0$

Step 6

Solve for $n$ .

Set the numerator equal to zero.

$n^{2} + 14 n + 49 = 0$

Solve the equation for $n$ .

Factor using the perfect square rule.

Rewrite $49$ as $7^{2}$ .

$n^{2} + 14 n + 7^{2} = 0$

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

$14 n = 2 \cdot n \cdot 7$

Rewrite the polynomial.

$n^{2} + 2 \cdot n \cdot 7 + 7^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = n$ and $b = 7$ .

${(n + 7)}^{2} = 0$

Set the $n + 7$ equal to $0$ .

$n + 7 = 0$

Subtract $7$ from both sides of the equation.

$n = - 7$

Step 7

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 $0$ .

$n = - 7, n = 0, n = 9$

Step 8

Find Where Undefined/Discontinuous ((6n+42)/(n^3-18n^2+81n))÷((n^2+14n+49)/(9n^4))

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