Simplify ((2x-2)/(x+5))÷((x^2-4x+3)/(2x^2+13x+15))

$\frac{2 x - 2}{x + 5} \div \frac{x^{2} - 4 x + 3}{2 x^{2} + 13 x + 15}$

To divide by a fraction, multiply by its reciprocal.

$\frac{2 x - 2}{x + 5} \cdot \frac{2 x^{2} + 13 x + 15}{x^{2} - 4 x + 3}$

Factor $2$ out of $2 x - 2$ .

Factor $2$ out of $2 x$ .

$\frac{2 (x) - 2}{x + 5} \cdot \frac{2 x^{2} + 13 x + 15}{x^{2} - 4 x + 3}$

Factor $2$ out of $- 2$ .

$\frac{2 (x) + 2 (- 1)}{x + 5} \cdot \frac{2 x^{2} + 13 x + 15}{x^{2} - 4 x + 3}$

Factor $2$ out of $2 (x) + 2 (- 1)$ .

$\frac{2 (x - 1)}{x + 5} \cdot \frac{2 x^{2} + 13 x + 15}{x^{2} - 4 x + 3}$

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 $a \cdot c = 2 \cdot 15 = 30$ and whose sum is $b = 13$ .

Factor $13$ out of $13 x$ .

$\frac{2 (x - 1)}{x + 5} \cdot \frac{2 x^{2} + 13 (x) + 15}{x^{2} - 4 x + 3}$

Rewrite $13$ as $3$ plus $10$

$\frac{2 (x - 1)}{x + 5} \cdot \frac{2 x^{2} + (3 + 10) x + 15}{x^{2} - 4 x + 3}$

Apply the distributive property.

$\frac{2 (x - 1)}{x + 5} \cdot \frac{2 x^{2} + 3 x + 10 x + 15}{x^{2} - 4 x + 3}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{2 (x - 1)}{x + 5} \cdot \frac{(2 x^{2} + 3 x) + 10 x + 15}{x^{2} - 4 x + 3}$

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

$\frac{2 (x - 1)}{x + 5} \cdot \frac{x (2 x + 3) + 5 (2 x + 3)}{x^{2} - 4 x + 3}$

Factor the polynomial by factoring out the greatest common factor, $2 x + 3$ .

$\frac{2 (x - 1)}{x + 5} \cdot \frac{(2 x + 3) (x + 5)}{x^{2} - 4 x + 3}$

Factor $x^{2} - 4 x + 3$ 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 $3$ and whose sum is $- 4$ .

$- 3, - 1$

Write the factored form using these integers.

$\frac{2 (x - 1)}{x + 5} \cdot \frac{(2 x + 3) (x + 5)}{(x - 3) (x - 1)}$

Simplify terms.

Cancel the common factor of $x - 1$ .

Factor $x - 1$ out of $2 (x - 1)$ .

$\frac{(x - 1) \cdot 2}{x + 5} \cdot \frac{(2 x + 3) (x + 5)}{(x - 3) (x - 1)}$

Factor $x - 1$ out of $(x - 3) (x - 1)$ .

$\frac{(x - 1) \cdot 2}{x + 5} \cdot \frac{(2 x + 3) (x + 5)}{(x - 1) (x - 3)}$

Cancel the common factor.

$\frac{(x - 1) \cdot 2}{x + 5} \cdot \frac{(2 x + 3) (x + 5)}{(x - 1) (x - 3)}$

Rewrite the expression.

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

Cancel the common factor of $x + 5$ .

Factor $x + 5$ out of $(2 x + 3) (x + 5)$ .

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

Cancel the common factor.

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

Rewrite the expression.

$2 \frac{2 x + 3}{x - 3}$

Combine $2$ and $\frac{2 x + 3}{x - 3}$ .

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

Simplify ((2x-2)/(x+5))÷((x^2-4x+3)/(2x^2+13x+15))

Related Questions

Support LCMGCF.com by disabling your adblocker.