Find Where Undefined/Discontinuous ((3x-6)/(x^2+21x+54))÷((40x^2)/(x^2+6x+9))

$\frac{3 x - 6}{x^{2} + 21 x + 54} \div \frac{40 x^{2}}{x^{2} + 6 x + 9}$

Step 1

Set the denominator in $\frac{3 x - 6}{x^{2} + 21 x + 54}$ equal to $0$ to find where the expression is undefined.

$x^{2} + 21 x + 54 = 0$

Step 2

Solve for $x$ .

Factor $x^{2} + 21 x + 54$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $54$ and whose sum is $21$ .

$3, 18$

Write the factored form using these integers.

$(x + 3) (x + 18) = 0$

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

$x + 3 = 0$

$x + 18 = 0$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Set $x + 18$ equal to $0$ and solve for $x$ .

Set $x + 18$ equal to $0$ .

$x + 18 = 0$

Subtract $18$ from both sides of the equation.

$x = - 18$

The final solution is all the values that make $(x + 3) (x + 18) = 0$ true.

$x = - 3, - 18$

Step 3

Set the denominator in $\frac{40 x^{2}}{x^{2} + 6 x + 9}$ equal to $0$ to find where the expression is undefined.

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

Step 4

Solve for $x$ .

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

$x^{2} + 6 x + 3^{2} = 0$

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

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 3$ .

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 5

Set the denominator in $\frac{\frac{3 x - 6}{x^{2} + 21 x + 54}}{\frac{40 x^{2}}{x^{2} + 6 x + 9}}$ equal to $0$ to find where the expression is undefined.

$\frac{40 x^{2}}{x^{2} + 6 x + 9} = 0$

Step 6

Solve for $x$ .

Set the numerator equal to zero.

$40 x^{2} = 0$

Solve the equation for $x$ .

Divide each term in $40 x^{2} = 0$ by $40$ and simplify.

Divide each term in $40 x^{2} = 0$ by $40$ .

$\frac{40 x^{2}}{40} = \frac{0}{40}$

Simplify the left side.

Cancel the common factor of $40$ .

Cancel the common factor.

$\frac{40 x^{2}}{40} = \frac{0}{40}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{0}{40}$

Simplify the right side.

Divide $0$ by $40$ .

$x^{2} = 0$

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

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 $0$ .

$x = - 18, x = - 3, x = 0$

Step 8

Find Where Undefined/Discontinuous ((3x-6)/(x^2+21x+54))÷((40x^2)/(x^2+6x+9))

Related Questions

Support LCMGCF.com by disabling your adblocker.