Find the LCD (9x)/(2x^2-7x+5) and 5/(2x^2-9x+10)

$\frac{9 x}{2 x^{2} - 7 x + 5}$ and $\frac{5}{2 x^{2} - 9 x + 10}$

Step 1

Simplify each polynomial.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 5 = 10$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

$\frac{9 x}{2 x^{2} - 7 x + 5}, \frac{5}{2 x^{2} - 9 x + 10}$

Rewrite $- 7$ as $- 2$ plus $- 5$

$\frac{9 x}{2 x^{2} + (- 2 - 5) x + 5}, \frac{5}{2 x^{2} - 9 x + 10}$

Apply the distributive property.

$\frac{9 x}{2 x^{2} - 2 x - 5 x + 5}, \frac{5}{2 x^{2} - 9 x + 10}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{9 x}{(2 x^{2} - 2 x) - 5 x + 5}, \frac{5}{2 x^{2} - 9 x + 10}$

Factor out the greatest common factor (GCF) from each group.

$\frac{9 x}{2 x (x - 1) - 5 (x - 1)}, \frac{5}{2 x^{2} - 9 x + 10}$

Factor the polynomial by factoring out the greatest common factor, $x - 1$ .

$\frac{9 x}{(x - 1) (2 x - 5)}, \frac{5}{2 x^{2} - 9 x + 10}$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 10 = 20$ and whose sum is $b = - 9$ .

Factor $- 9$ out of $- 9 x$ .

$\frac{9 x}{(x - 1) (2 x - 5)}, \frac{5}{2 x^{2} - 9 x + 10}$

Rewrite $- 9$ as $- 4$ plus $- 5$

$\frac{9 x}{(x - 1) (2 x - 5)}, \frac{5}{2 x^{2} + (- 4 - 5) x + 10}$

Apply the distributive property.

$\frac{9 x}{(x - 1) (2 x - 5)}, \frac{5}{2 x^{2} - 4 x - 5 x + 10}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{9 x}{(x - 1) (2 x - 5)}, \frac{5}{(2 x^{2} - 4 x) - 5 x + 10}$

Factor out the greatest common factor (GCF) from each group.

$\frac{9 x}{(x - 1) (2 x - 5)}, \frac{5}{2 x (x - 2) - 5 (x - 2)}$

Factor the polynomial by factoring out the greatest common factor, $x - 2$ .

$\frac{9 x}{(x - 1) (2 x - 5)}, \frac{5}{(x - 2) (2 x - 5)}$

Step 2

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

$(x - 1) (2 x - 5), (x - 2) (2 x - 5)$

Step 3

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 4

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

Not prime

Step 5

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 6

The factor for $x - 1$ is $x - 1$ itself.

$(x - 1) = x - 1$

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

Step 7

The factor for $2 x - 5$ is $2 x - 5$ itself.

$(2 x - 5) = 2 x - 5$

$(2 x - 5)$ occurs $1$ time.

Step 8

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

$(x - 2) = x - 2$

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

Step 9

The factor for $2 x - 5$ is $2 x - 5$ itself.

$(2 x - 5) = 2 x - 5$

$(2 x - 5)$ occurs $1$ time.

Step 10

The LCM of $x - 1, 2 x - 5, x - 2, 2 x - 5$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(x - 1) (2 x - 5) (x - 2)$

Find the LCD (9x)/(2x^2-7x+5) and 5/(2x^2-9x+10)

Related Questions

Support LCMGCF.com by disabling your adblocker.