Simplify (9x)/(x^2-25)*(x^2+5x)/(2x-4)*(x^2+3x-10)/(3x^4)

Consider the form ${\displaystyle x^2+bx+c}$. Find a pair of integers whose product is ${\displaystyle c}$ and whose sum is ${\displaystyle b}$. In this case, whose product is ${\displaystyle -10}$ and whose sum is ${\displaystyle 3}$.

Write the factored form using these integers.

Rewrite ${\displaystyle 25}$ as ${\displaystyle 5^2}$.

Since both terms are perfect squares, factor using the difference of squares formula, ${\displaystyle a^2-b^2=(a+b)(a-b)}$ where ${\displaystyle a=x}$ and ${\displaystyle b=5}$.

Factor ${\displaystyle x}$ out of ${\displaystyle x^2+5x}$.

Factor ${\displaystyle 2}$ out of ${\displaystyle 2x-4}$.

Cancel the common factor of ${\displaystyle x+5}$.

Multiply ${\displaystyle \frac{9x}{x-5}}$ by ${\displaystyle \frac{x}{2(x-2)}}$.

Factor ${\displaystyle 3}$ out of ${\displaystyle 9x^2\left((x-2)(x+5)\right)}$.

Cancel the common factors.

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

Cancel the common factors.

Cancel the common factor.

Rewrite the expression.

Move ${\displaystyle 2}$ to the left of ${\displaystyle x-5}$.

Reorder factors in ${\displaystyle \frac{3(x+5)}{2(x-5)x^2}}$.