Simplify ((10x^2-7x-12)/(10x+8))÷((2x^2-5x-12)/(x^2-4x))

$\frac{10 x^{2} - 7 x - 12}{10 x + 8} \div \frac{2 x^{2} - 5 x - 12}{x^{2} - 4 x}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{10 x^{2} - 7 x - 12}{10 x + 8} \cdot \frac{x^{2} - 4 x}{2 x^{2} - 5 x - 12}$

Step 2

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 $a \cdot c = 10 \cdot - 12 = - 120$ and whose sum is $b = - 7$ .

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

$\frac{10 x^{2} - 7 (x) - 12}{10 x + 8} \cdot \frac{x^{2} - 4 x}{2 x^{2} - 5 x - 12}$

Rewrite $- 7$ as $8$ plus $- 15$

$\frac{10 x^{2} + (8 - 15) x - 12}{10 x + 8} \cdot \frac{x^{2} - 4 x}{2 x^{2} - 5 x - 12}$

Apply the distributive property.

$\frac{10 x^{2} + 8 x - 15 x - 12}{10 x + 8} \cdot \frac{x^{2} - 4 x}{2 x^{2} - 5 x - 12}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(10 x^{2} + 8 x) - 15 x - 12}{10 x + 8} \cdot \frac{x^{2} - 4 x}{2 x^{2} - 5 x - 12}$

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

$\frac{2 x (5 x + 4) - 3 (5 x + 4)}{10 x + 8} \cdot \frac{x^{2} - 4 x}{2 x^{2} - 5 x - 12}$

Factor the polynomial by factoring out the greatest common factor, $5 x + 4$ .

$\frac{(5 x + 4) (2 x - 3)}{10 x + 8} \cdot \frac{x^{2} - 4 x}{2 x^{2} - 5 x - 12}$

Step 3

Factor $2$ out of $10 x + 8$ .

Factor $2$ out of $10 x$ .

$\frac{(5 x + 4) (2 x - 3)}{2 (5 x) + 8} \cdot \frac{x^{2} - 4 x}{2 x^{2} - 5 x - 12}$

Factor $2$ out of $8$ .

$\frac{(5 x + 4) (2 x - 3)}{2 (5 x) + 2 (4)} \cdot \frac{x^{2} - 4 x}{2 x^{2} - 5 x - 12}$

Factor $2$ out of $2 (5 x) + 2 (4)$ .

$\frac{(5 x + 4) (2 x - 3)}{2 (5 x + 4)} \cdot \frac{x^{2} - 4 x}{2 x^{2} - 5 x - 12}$

Step 4

Cancel the common factor of $5 x + 4$ .

Cancel the common factor.

$\frac{(5 x + 4) (2 x - 3)}{2 (5 x + 4)} \cdot \frac{x^{2} - 4 x}{2 x^{2} - 5 x - 12}$

Rewrite the expression.

$\frac{2 x - 3}{2} \cdot \frac{x^{2} - 4 x}{2 x^{2} - 5 x - 12}$

Step 5

Factor $x$ out of $x^{2} - 4 x$ .

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

$\frac{2 x - 3}{2} \cdot \frac{x \cdot x - 4 x}{2 x^{2} - 5 x - 12}$

Factor $x$ out of $- 4 x$ .

$\frac{2 x - 3}{2} \cdot \frac{x \cdot x + x \cdot - 4}{2 x^{2} - 5 x - 12}$

Factor $x$ out of $x \cdot x + x \cdot - 4$ .

$\frac{2 x - 3}{2} \cdot \frac{x (x - 4)}{2 x^{2} - 5 x - 12}$

Step 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 $a \cdot c = 2 \cdot - 12 = - 24$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 x$ .

$\frac{2 x - 3}{2} \cdot \frac{x (x - 4)}{2 x^{2} - 5 (x) - 12}$

Rewrite $- 5$ as $3$ plus $- 8$

$\frac{2 x - 3}{2} \cdot \frac{x (x - 4)}{2 x^{2} + (3 - 8) x - 12}$

Apply the distributive property.

$\frac{2 x - 3}{2} \cdot \frac{x (x - 4)}{2 x^{2} + 3 x - 8 x - 12}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{2 x - 3}{2} \cdot \frac{x (x - 4)}{(2 x^{2} + 3 x) - 8 x - 12}$

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

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

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

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

Step 7

Cancel the common factor of $x - 4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 8

Multiply $\frac{2 x - 3}{2}$ by $\frac{x}{2 x + 3}$ .

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

Step 9

Reorder factors in $\frac{(2 x - 3) x}{2 (2 x + 3)}$ .

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

Simplify ((10x^2-7x-12)/(10x+8))÷((2x^2-5x-12)/(x^2-4x))

Related Questions

Support LCMGCF.com by disabling your adblocker.