Simplify (6x^2-384)/(2x^2-36x+160)*(x-10)/(x^2+12x+32)

$\frac{6 x^{2} - 384}{2 x^{2} - 36 x + 160} \cdot \frac{x - 10}{x^{2} + 12 x + 32}$

Step 1

Simplify the numerator.

Factor $6$ out of $6 x^{2} - 384$ .

Factor $6$ out of $6 x^{2}$ .

$\frac{6 (x^{2}) - 384}{2 x^{2} - 36 x + 160} \cdot \frac{x - 10}{x^{2} + 12 x + 32}$

Factor $6$ out of $- 384$ .

$\frac{6 x^{2} + 6 \cdot - 64}{2 x^{2} - 36 x + 160} \cdot \frac{x - 10}{x^{2} + 12 x + 32}$

Factor $6$ out of $6 x^{2} + 6 \cdot - 64$ .

$\frac{6 (x^{2} - 64)}{2 x^{2} - 36 x + 160} \cdot \frac{x - 10}{x^{2} + 12 x + 32}$

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

$\frac{6 (x^{2} - 8^{2})}{2 x^{2} - 36 x + 160} \cdot \frac{x - 10}{x^{2} + 12 x + 32}$

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 $b = 8$ .

$\frac{6 (x + 8) (x - 8)}{2 x^{2} - 36 x + 160} \cdot \frac{x - 10}{x^{2} + 12 x + 32}$

Step 2

Simplify the denominator.

Factor $2$ out of $2 x^{2} - 36 x + 160$ .

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

$\frac{6 (x + 8) (x - 8)}{2 (x^{2}) - 36 x + 160} \cdot \frac{x - 10}{x^{2} + 12 x + 32}$

Factor $2$ out of $- 36 x$ .

$\frac{6 (x + 8) (x - 8)}{2 (x^{2}) + 2 (- 18 x) + 160} \cdot \frac{x - 10}{x^{2} + 12 x + 32}$

Factor $2$ out of $160$ .

$\frac{6 (x + 8) (x - 8)}{2 x^{2} + 2 (- 18 x) + 2 \cdot 80} \cdot \frac{x - 10}{x^{2} + 12 x + 32}$

Factor $2$ out of $2 x^{2} + 2 (- 18 x)$ .

$\frac{6 (x + 8) (x - 8)}{2 (x^{2} - 18 x) + 2 \cdot 80} \cdot \frac{x - 10}{x^{2} + 12 x + 32}$

Factor $2$ out of $2 (x^{2} - 18 x) + 2 \cdot 80$ .

$\frac{6 (x + 8) (x - 8)}{2 (x^{2} - 18 x + 80)} \cdot \frac{x - 10}{x^{2} + 12 x + 32}$

Factor $x^{2} - 18 x + 80$ 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 $80$ and whose sum is $- 18$ .

$- 10, - 8$

Write the factored form using these integers.

$\frac{6 (x + 8) (x - 8)}{2 ((x - 10) (x - 8))} \cdot \frac{x - 10}{x^{2} + 12 x + 32}$

$\frac{6 (x + 8) (x - 8)}{2 (x - 10) (x - 8)} \cdot \frac{x - 10}{x^{2} + 12 x + 32}$

Step 3

Factor $x^{2} + 12 x + 32$ 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 $32$ and whose sum is $12$ .

$4, 8$

Write the factored form using these integers.

$\frac{6 (x + 8) (x - 8)}{2 (x - 10) (x - 8)} \cdot \frac{x - 10}{(x + 4) (x + 8)}$

Step 4

Simplify terms.

Cancel the common factor of $x + 8$ .

Factor $x + 8$ out of $6 (x + 8) (x - 8)$ .

$\frac{(x + 8) (6 (x - 8))}{2 (x - 10) (x - 8)} \cdot \frac{x - 10}{(x + 4) (x + 8)}$

Factor $x + 8$ out of $(x + 4) (x + 8)$ .

$\frac{(x + 8) (6 (x - 8))}{2 (x - 10) (x - 8)} \cdot \frac{x - 10}{(x + 8) (x + 4)}$

Cancel the common factor.

$\frac{(x + 8) (6 (x - 8))}{2 (x - 10) (x - 8)} \cdot \frac{x - 10}{(x + 8) (x + 4)}$

Rewrite the expression.

$\frac{6 (x - 8)}{2 (x - 10) (x - 8)} \cdot \frac{x - 10}{x + 4}$

Cancel the common factor of $x - 10$ .

Factor $x - 10$ out of $2 (x - 10) (x - 8)$ .

$\frac{6 (x - 8)}{(x - 10) (2 (x - 8))} \cdot \frac{x - 10}{x + 4}$

Cancel the common factor.

$\frac{6 (x - 8)}{(x - 10) (2 (x - 8))} \cdot \frac{x - 10}{x + 4}$

Rewrite the expression.

$\frac{6 (x - 8)}{2 (x - 8)} \cdot \frac{1}{x + 4}$

Multiply $\frac{6 (x - 8)}{2 (x - 8)}$ by $\frac{1}{x + 4}$ .

$\frac{6 (x - 8)}{2 (x - 8) (x + 4)}$

Cancel the common factor of $6$ and $2$ .

Factor $2$ out of $6 (x - 8)$ .

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

Cancel the common factors.

Factor $2$ out of $2 (x - 8) (x + 4)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x - 8$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify (6x^2-384)/(2x^2-36x+160)*(x-10)/(x^2+12x+32)

Related Questions

Support LCMGCF.com by disabling your adblocker.