Find the LCD x/(x^2-9)(3x)/(x^2-5x+6)

$\frac{x}{x^{2} - 9} \cdot \frac{3 x}{x^{2} - 5 x + 6}$

Step 1

Simplify the denominator.

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

$\frac{x}{x^{2} - 3^{2}} \cdot \frac{3 x}{x^{2} - 5 x + 6}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$\frac{x}{(x + 3) (x - 3)} \cdot \frac{3 x}{x^{2} - 5 x + 6}$

Step 2

Factor $x^{2} - 5 x + 6$ 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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

$\frac{x}{(x + 3) (x - 3)} \cdot \frac{3 x}{(x - 3) (x - 2)}$

Step 3

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$(x + 3) (x - 3), (x - 3) (x - 2)$

Step 4

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 5

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 6

The LCM of $1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

Step 7

The factor for $x + 3$ is $x + 3$ itself.

$(x + 3) = x + 3$

$(x + 3)$ occurs $1$ time.

Step 8

The factor for $x - 3$ is $x - 3$ itself.

$(x - 3) = x - 3$

$(x - 3)$ occurs $1$ time.

Step 9

The factor for $x - 2$ is $x - 2$ itself.

$(x - 2) = x - 2$

$(x - 2)$ occurs $1$ time.

Step 10

The LCM of $x + 3, x - 3, x - 3, x - 2$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(x + 3) (x - 3) (x - 2)$

Find the LCD x/(x^2-9)(3x)/(x^2-5x+6)

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