Simplify (x^2+4x-5)/(5x^2-8x+3)*(20x-12)/(x^2-6x-55)

$\frac{x^{2} + 4 x - 5}{5 x^{2} - 8 x + 3} \cdot \frac{20 x - 12}{x^{2} - 6 x - 55}$

Step 1

Factor $x^{2} + 4 x - 5$ 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 $- 5$ and whose sum is $4$ .

$- 1, 5$

Write the factored form using these integers.

$\frac{(x - 1) (x + 5)}{5 x^{2} - 8 x + 3} \cdot \frac{20 x - 12}{x^{2} - 6 x - 55}$

Step 2

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 $a \cdot c = 5 \cdot 3 = 15$ and whose sum is $b = - 8$ .

Factor $- 8$ out of $- 8 x$ .

$\frac{(x - 1) (x + 5)}{5 x^{2} - 8 (x) + 3} \cdot \frac{20 x - 12}{x^{2} - 6 x - 55}$

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

$\frac{(x - 1) (x + 5)}{5 x^{2} + (- 3 - 5) x + 3} \cdot \frac{20 x - 12}{x^{2} - 6 x - 55}$

Apply the distributive property.

$\frac{(x - 1) (x + 5)}{5 x^{2} - 3 x - 5 x + 3} \cdot \frac{20 x - 12}{x^{2} - 6 x - 55}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(x - 1) (x + 5)}{(5 x^{2} - 3 x) - 5 x + 3} \cdot \frac{20 x - 12}{x^{2} - 6 x - 55}$

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

$\frac{(x - 1) (x + 5)}{x (5 x - 3) - (5 x - 3)} \cdot \frac{20 x - 12}{x^{2} - 6 x - 55}$

Factor the polynomial by factoring out the greatest common factor, $5 x - 3$ .

$\frac{(x - 1) (x + 5)}{(5 x - 3) (x - 1)} \cdot \frac{20 x - 12}{x^{2} - 6 x - 55}$

Step 3

Factor $4$ out of $20 x - 12$ .

Factor $4$ out of $20 x$ .

$\frac{(x - 1) (x + 5)}{(5 x - 3) (x - 1)} \cdot \frac{4 (5 x) - 12}{x^{2} - 6 x - 55}$

Factor $4$ out of $- 12$ .

$\frac{(x - 1) (x + 5)}{(5 x - 3) (x - 1)} \cdot \frac{4 (5 x) + 4 (- 3)}{x^{2} - 6 x - 55}$

Factor $4$ out of $4 (5 x) + 4 (- 3)$ .

$\frac{(x - 1) (x + 5)}{(5 x - 3) (x - 1)} \cdot \frac{4 (5 x - 3)}{x^{2} - 6 x - 55}$

Step 4

Factor $x^{2} - 6 x - 55$ 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 $- 55$ and whose sum is $- 6$ .

$- 11, 5$

Write the factored form using these integers.

$\frac{(x - 1) (x + 5)}{(5 x - 3) (x - 1)} \cdot \frac{4 (5 x - 3)}{(x - 11) (x + 5)}$

Step 5

Simplify terms.

Cancel the common factor of $x + 5$ .

Factor $x + 5$ out of $(x - 1) (x + 5)$ .

$\frac{(x + 5) (x - 1)}{(5 x - 3) (x - 1)} \cdot \frac{4 (5 x - 3)}{(x - 11) (x + 5)}$

Factor $x + 5$ out of $(x - 11) (x + 5)$ .

$\frac{(x + 5) (x - 1)}{(5 x - 3) (x - 1)} \cdot \frac{4 (5 x - 3)}{(x + 5) (x - 11)}$

Cancel the common factor.

$\frac{(x + 5) (x - 1)}{(5 x - 3) (x - 1)} \cdot \frac{4 (5 x - 3)}{(x + 5) (x - 11)}$

Rewrite the expression.

$\frac{x - 1}{(5 x - 3) (x - 1)} \cdot \frac{4 (5 x - 3)}{x - 11}$

Cancel the common factor of $5 x - 3$ .

Factor $5 x - 3$ out of $4 (5 x - 3)$ .

$\frac{x - 1}{(5 x - 3) (x - 1)} \cdot \frac{(5 x - 3) \cdot 4}{x - 11}$

Cancel the common factor.

$\frac{x - 1}{(5 x - 3) (x - 1)} \cdot \frac{(5 x - 3) \cdot 4}{x - 11}$

Rewrite the expression.

$\frac{x - 1}{x - 1} \cdot \frac{4}{x - 11}$

Multiply $\frac{x - 1}{x - 1}$ by $\frac{4}{x - 11}$ .

$\frac{(x - 1) \cdot 4}{(x - 1) (x - 11)}$

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$\frac{(x - 1) \cdot 4}{(x - 1) (x - 11)}$

Rewrite the expression.

$\frac{4}{x - 11}$

Simplify (x^2+4x-5)/(5x^2-8x+3)*(20x-12)/(x^2-6x-55)

Related Questions

Support LCMGCF.com by disabling your adblocker.