Simplify (x-3)/(x^2+5x-24)*(x^2-64)/(7x-49)

$\frac{x - 3}{x^{2} + 5 x - 24} \cdot \frac{x^{2} - 64}{7 x - 49}$

Step 1

Factor $x^{2} + 5 x - 24$ 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.

$\frac{x - 3}{(x - 3) (x + 8)} \cdot \frac{x^{2} - 64}{7 x - 49}$

Step 2

Simplify the numerator.

Rewrite $64$ as $8^{2}$ .

$\frac{x - 3}{(x - 3) (x + 8)} \cdot \frac{x^{2} - 8^{2}}{7 x - 49}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 8$ .

$\frac{x - 3}{(x - 3) (x + 8)} \cdot \frac{(x + 8) (x - 8)}{7 x - 49}$

Step 3

Simplify terms.

Factor $7$ out of $7 x - 49$ .

Factor $7$ out of $7 x$ .

$\frac{x - 3}{(x - 3) (x + 8)} \cdot \frac{(x + 8) (x - 8)}{7 (x) - 49}$

Factor $7$ out of $- 49$ .

$\frac{x - 3}{(x - 3) (x + 8)} \cdot \frac{(x + 8) (x - 8)}{7 x + 7 \cdot - 7}$

Factor $7$ out of $7 x + 7 \cdot - 7$ .

$\frac{x - 3}{(x - 3) (x + 8)} \cdot \frac{(x + 8) (x - 8)}{7 (x - 7)}$

Cancel the common factor of $x + 8$ .

Factor $x + 8$ out of $(x - 3) (x + 8)$ .

$\frac{x - 3}{(x + 8) (x - 3)} \cdot \frac{(x + 8) (x - 8)}{7 (x - 7)}$

Cancel the common factor.

$\frac{x - 3}{(x + 8) (x - 3)} \cdot \frac{(x + 8) (x - 8)}{7 (x - 7)}$

Rewrite the expression.

$\frac{x - 3}{x - 3} \cdot \frac{x - 8}{7 (x - 7)}$

Multiply $\frac{x - 3}{x - 3}$ by $\frac{x - 8}{7 (x - 7)}$ .

$\frac{(x - 3) (x - 8)}{(x - 3) (7 (x - 7))}$

Cancel the common factor of $x - 3$ .

Cancel the common factor.

$\frac{(x - 3) (x - 8)}{(x - 3) (7 (x - 7))}$

Rewrite the expression.

$\frac{x - 8}{7 (x - 7)}$

Simplify (x-3)/(x^2+5x-24)*(x^2-64)/(7x-49)

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