Find the LCD (2h^2)/(2h+6)-(4h-1)/(h^2-9)

$\frac{2 h^{2}}{2 h + 6} - \frac{4 h - 1}{h^{2} - 9}$

Step 1

Cancel the common factor of $2$ and $2 h + 6$ .

Factor $2$ out of $2 h^{2}$ .

$\frac{2 (h^{2})}{2 h + 6} - \frac{4 h - 1}{h^{2} - 9}$

Cancel the common factors.

Factor $2$ out of $2 h$ .

$\frac{2 (h^{2})}{2 (h) + 6} - \frac{4 h - 1}{h^{2} - 9}$

Factor $2$ out of $6$ .

$\frac{2 (h^{2})}{2 h + 2 \cdot 3} - \frac{4 h - 1}{h^{2} - 9}$

Factor $2$ out of $2 h + 2 \cdot 3$ .

$\frac{2 (h^{2})}{2 (h + 3)} - \frac{4 h - 1}{h^{2} - 9}$

Cancel the common factor.

$\frac{2 h^{2}}{2 (h + 3)} - \frac{4 h - 1}{h^{2} - 9}$

Rewrite the expression.

$\frac{h^{2}}{h + 3} - \frac{4 h - 1}{h^{2} - 9}$

Step 2

Simplify the denominator.

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

$\frac{h^{2}}{h + 3} - \frac{4 h - 1}{h^{2} - 3^{2}}$

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

$\frac{h^{2}}{h + 3} - \frac{4 h - 1}{(h + 3) (h - 3)}$

Step 3

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

$h + 3, (h + 3) (h - 3)$

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 $h + 3$ is $h + 3$ itself.

$(h + 3) = h + 3$

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

Step 8

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

$(h - 3) = h - 3$

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

Step 9

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

$(h + 3) (h - 3)$

Find the LCD (2h^2)/(2h+6)-(4h-1)/(h^2-9)

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