Question

simplify.
\\(\frac{x^2 + 2x - 15}{5x} \cdot \frac{x^2 - 7x}{x^2 - 25} \div \frac{x - 3}{x - 5}\\)

Explanation:

Step1: Factor all polynomials

Factor $x^2 + 2x - 15$: We need two numbers that multiply to $-15$ and add to $2$. Those numbers are $5$ and $-3$. So $x^2 + 2x - 15=(x + 5)(x - 3)$.

Factor $x^2 - 7x$: Take out the common factor $x$, so $x^2 - 7x=x(x - 7)$.

Factor $x^2 - 25$: This is a difference of squares, so $x^2 - 25=(x + 5)(x - 5)$.

Now the expression becomes:
$\frac{(x + 5)(x - 3)}{5x}\cdot\frac{x(x - 7)}{(x + 5)(x - 5)}\div\frac{x - 3}{x - 5}$

Step2: Change division to multiplication

Recall that dividing by a fraction is the same as multiplying by its reciprocal. So we rewrite the division as multiplication by the reciprocal of $\frac{x - 3}{x - 5}$, which is $\frac{x - 5}{x - 3}$.

The expression now is:
$\frac{(x + 5)(x - 3)}{5x}\cdot\frac{x(x - 7)}{(x + 5)(x - 5)}\cdot\frac{x - 5}{x - 3}$

Step3: Cancel out common factors

• Cancel out $(x + 5)$ from the numerator of the first fraction and the denominator of the second fraction.
• Cancel out $x$ from the numerator of the second fraction and the denominator of the first fraction.
• Cancel out $(x - 3)$ from the numerator of the first fraction and the denominator of the third fraction.
• Cancel out $(x - 5)$ from the numerator of the third fraction and the denominator of the second fraction.

After canceling, we are left with:
$\frac{x - 7}{5}$

Answer:

$\frac{x - 7}{5}$