Simplify (5x-5)/(10x^2-25x+15)

$\frac{5 x - 5}{10 x^{2} - 25 x + 15}$

Step 1

Cancel the common factor of $5 x - 5$ and $10 x^{2} - 25 x + 15$ .

Factor $5$ out of $5 x$ .

$\frac{5 (x) - 5}{10 x^{2} - 25 x + 15}$

Factor $5$ out of $- 5$ .

$\frac{5 (x) + 5 (- 1)}{10 x^{2} - 25 x + 15}$

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

$\frac{5 (x - 1)}{10 x^{2} - 25 x + 15}$

Cancel the common factors.

Factor $5$ out of $10 x^{2}$ .

$\frac{5 (x - 1)}{5 (2 x^{2}) - 25 x + 15}$

Factor $5$ out of $- 25 x$ .

$\frac{5 (x - 1)}{5 (2 x^{2}) + 5 (- 5 x) + 15}$

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

$\frac{5 (x - 1)}{5 (2 x^{2} - 5 x) + 15}$

Factor $5$ out of $15$ .

$\frac{5 (x - 1)}{5 (2 x^{2} - 5 x) + 5 (3)}$

Factor $5$ out of $5 (2 x^{2} - 5 x) + 5 (3)$ .

$\frac{5 (x - 1)}{5 (2 x^{2} - 5 x + 3)}$

Cancel the common factor.

$\frac{5 (x - 1)}{5 (2 x^{2} - 5 x + 3)}$

Rewrite the expression.

$\frac{x - 1}{2 x^{2} - 5 x + 3}$

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 $a \cdot c = 2 \cdot 3 = 6$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 x$ .

$\frac{x - 1}{2 x^{2} - 5 (x) + 3}$

Rewrite $- 5$ as $- 2$ plus $- 3$

$\frac{x - 1}{2 x^{2} + (- 2 - 3) x + 3}$

Apply the distributive property.

$\frac{x - 1}{2 x^{2} - 2 x - 3 x + 3}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{x - 1}{(2 x^{2} - 2 x) - 3 x + 3}$

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

$\frac{x - 1}{2 x (x - 1) - 3 (x - 1)}$

Factor the polynomial by factoring out the greatest common factor, $x - 1$ .

$\frac{x - 1}{(x - 1) (2 x - 3)}$

Step 3

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$\frac{x - 1}{(x - 1) (2 x - 3)}$

Rewrite the expression.

$\frac{1}{2 x - 3}$

Simplify (5x-5)/(10x^2-25x+15)

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