Divide ((x^2-3x-10)/(2x^2-11x+5))÷((x^2-5x+6)/(2x^2-7x+3))

$\frac{x^{2} - 3 x - 10}{2 x^{2} - 11 x + 5} \div \frac{x^{2} - 5 x + 6}{2 x^{2} - 7 x + 3}$

To divide by a fraction, multiply by its reciprocal.

$\frac{x^{2} - 3 x - 10}{2 x^{2} - 11 x + 5} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} - 5 x + 6}$

Factor $x^{2} - 3 x - 10$ 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 $- 10$ and whose sum is $- 3$ .

$- 5, 2$

Write the factored form using these integers.

$\frac{(x - 5) (x + 2)}{2 x^{2} - 11 x + 5} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} - 5 x + 6}$

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 $b = - 11$ .

Factor $- 11$ out of $- 11 x$ .

$\frac{(x - 5) (x + 2)}{2 x^{2} - 11 (x) + 5} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} - 5 x + 6}$

Rewrite $- 11$ as $- 1$ plus $- 10$

$\frac{(x - 5) (x + 2)}{2 x^{2} + (- 1 - 10) x + 5} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} - 5 x + 6}$

Apply the distributive property.

$\frac{(x - 5) (x + 2)}{2 x^{2} - 1 x - 10 x + 5} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} - 5 x + 6}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(x - 5) (x + 2)}{(2 x^{2} - 1 x) - 10 x + 5} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} - 5 x + 6}$

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

$\frac{(x - 5) (x + 2)}{x (2 x - 1) - 5 (2 x - 1)} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} - 5 x + 6}$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$\frac{(x - 5) (x + 2)}{(2 x - 1) (x - 5)} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} - 5 x + 6}$

Cancel the common factor of $x - 5$ .

Cancel the common factor.

$\frac{(x - 5) (x + 2)}{(2 x - 1) (x - 5)} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} - 5 x + 6}$

Rewrite the expression.

$\frac{x + 2}{2 x - 1} \cdot \frac{2 x^{2} - 7 x + 3}{x^{2} - 5 x + 6}$

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 $a \cdot c = 2 \cdot 3 = 6$ and whose sum is $b = - 7$ .

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

$\frac{x + 2}{2 x - 1} \cdot \frac{2 x^{2} - 7 (x) + 3}{x^{2} - 5 x + 6}$

Rewrite $- 7$ as $- 1$ plus $- 6$

$\frac{x + 2}{2 x - 1} \cdot \frac{2 x^{2} + (- 1 - 6) x + 3}{x^{2} - 5 x + 6}$

Apply the distributive property.

$\frac{x + 2}{2 x - 1} \cdot \frac{2 x^{2} - 1 x - 6 x + 3}{x^{2} - 5 x + 6}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{x + 2}{2 x - 1} \cdot \frac{(2 x^{2} - 1 x) - 6 x + 3}{x^{2} - 5 x + 6}$

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

$\frac{x + 2}{2 x - 1} \cdot \frac{x (2 x - 1) - 3 (2 x - 1)}{x^{2} - 5 x + 6}$

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$\frac{x + 2}{2 x - 1} \cdot \frac{(2 x - 1) (x - 3)}{x^{2} - 5 x + 6}$

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.

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

Cancel the common factor of $2 x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x - 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $x + 2$ by $\frac{1}{x - 2}$ .

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

Divide ((x^2-3x-10)/(2x^2-11x+5))÷((x^2-5x+6)/(2x^2-7x+3))

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