Divide (4x^2-10x-24)÷(2x+3)

$(4 x^{2} - 10 x - 24) \div (2 x + 3)$

Step 1

Rewrite the division as a fraction.

$\frac{4 x^{2} - 10 x - 24}{2 x + 3}$

Step 2

Simplify the numerator.

Factor $2$ out of $4 x^{2} - 10 x - 24$ .

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

$\frac{2 (2 x^{2}) - 10 x - 24}{2 x + 3}$

Factor $2$ out of $- 10 x$ .

$\frac{2 (2 x^{2}) + 2 (- 5 x) - 24}{2 x + 3}$

Factor $2$ out of $- 24$ .

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

Factor $2$ out of $2 (2 x^{2}) + 2 (- 5 x)$ .

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

Factor $2$ out of $2 (2 x^{2} - 5 x) + 2 (- 12)$ .

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

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

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

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

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

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

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

Step 3

Cancel the common factor of $2 x + 3$ .

Cancel the common factor.

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

Divide $2 (x - 4)$ by $1$ .

$2 (x - 4)$

Step 4

Apply the distributive property.

$2 x + 2 \cdot - 4$

Step 5

Multiply $2$ by $- 4$ .

$2 x - 8$

Divide (4x^2-10x-24)÷(2x+3)

Related Questions

Support LCMGCF.com by disabling your adblocker.