Divide (2v^2+10v-48)/(8v+64)

$\frac{2 v^{2} + 10 v - 48}{8 v + 64}$

Step 1

Cancel the common factor of $2 v^{2} + 10 v - 48$ and $8 v + 64$ .

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

$\frac{2 (v^{2}) + 10 v - 48}{8 v + 64}$

Factor $2$ out of $10 v$ .

$\frac{2 (v^{2}) + 2 (5 v) - 48}{8 v + 64}$

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

$\frac{2 (v^{2} + 5 v) - 48}{8 v + 64}$

Factor $2$ out of $- 48$ .

$\frac{2 (v^{2} + 5 v) + 2 \cdot - 24}{8 v + 64}$

Factor $2$ out of $2 (v^{2} + 5 v) + 2 \cdot - 24$ .

$\frac{2 (v^{2} + 5 v - 24)}{8 v + 64}$

Cancel the common factors.

Factor $2$ out of $8 v$ .

$\frac{2 (v^{2} + 5 v - 24)}{2 (4 v) + 64}$

Factor $2$ out of $64$ .

$\frac{2 (v^{2} + 5 v - 24)}{2 (4 v) + 2 (32)}$

Factor $2$ out of $2 (4 v) + 2 (32)$ .

$\frac{2 (v^{2} + 5 v - 24)}{2 (4 v + 32)}$

Cancel the common factor.

$\frac{2 (v^{2} + 5 v - 24)}{2 (4 v + 32)}$

Rewrite the expression.

$\frac{v^{2} + 5 v - 24}{4 v + 32}$

Step 2

Factor $v^{2} + 5 v - 24$ 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 $- 24$ and whose sum is $5$ .

$- 3, 8$

Write the factored form using these integers.

$\frac{(v - 3) (v + 8)}{4 v + 32}$

Step 3

Factor $4$ out of $4 v + 32$ .

Factor $4$ out of $4 v$ .

$\frac{(v - 3) (v + 8)}{4 (v) + 32}$

Factor $4$ out of $32$ .

$\frac{(v - 3) (v + 8)}{4 v + 4 \cdot 8}$

Factor $4$ out of $4 v + 4 \cdot 8$ .

$\frac{(v - 3) (v + 8)}{4 (v + 8)}$

Step 4

Cancel the common factor of $v + 8$ .

Cancel the common factor.

$\frac{(v - 3) (v + 8)}{4 (v + 8)}$

Rewrite the expression.

$\frac{v - 3}{4}$

Step 5

Split the fraction $\frac{v - 3}{4}$ into two fractions.

$\frac{v}{4} + \frac{- 3}{4}$

Step 6

Move the negative in front of the fraction.

$\frac{v}{4} - \frac{3}{4}$

Divide (2v^2+10v-48)/(8v+64)

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