Pre-Algebra Examples

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

Step 1

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

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Step 1.1

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

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

Step 1.2

Factor $2$ out of $10 x$ .

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

Step 1.3

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

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

Step 1.4

Factor $2$ out of $- 48$ .

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

Step 1.5

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

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

Step 1.6

Cancel the common factors.

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Step 1.6.1

Factor $2$ out of $8 x$ .

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

Step 1.6.2

Factor $2$ out of $64$ .

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

Step 1.6.3

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

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

Step 1.6.4

Cancel the common factor.

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

Step 1.6.5

Rewrite the expression.

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

Step 2

Factor $x^{2} + 5 x - 24$ using the AC method.

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Step 2.1

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 $5$ .

$- 3, 8$

Step 2.2

Write the factored form using these integers.

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

Step 3

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

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Step 3.1

Factor $4$ out of $4 x$ .

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

Step 3.2

Factor $4$ out of $32$ .

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

Step 3.3

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

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

Step 4

Cancel the common factor of $x + 8$ .

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Step 4.1

Cancel the common factor.

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

Step 4.2

Rewrite the expression.

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

Step 5

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

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

Step 6

Move the negative in front of the fraction.

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