# How do you simplify (2x^2+x-15)/(2x^2-11x-21)*(6x+9)div(2x-5)/(3x-21)?

Apr 20, 2017

$9 \left(x + 3\right)$

#### Explanation:

Given: $\frac{2 {x}^{2} + x - 15}{2 {x}^{2} - 11 x - 21} \cdot \left(6 x + 9\right) \div \frac{2 x - 5}{3 x - 21}$

Change the division problem to multiplication by reciprocating the last rational function:

$\frac{2 {x}^{2} + x - 15}{2 {x}^{2} - 11 x - 21} \cdot \left(6 x + 9\right) \cdot \frac{3 x - 21}{2 x - 5}$

Make the second polynomial a fraction:

$\frac{2 {x}^{2} + x - 15}{2 {x}^{2} - 11 x - 21} \cdot \frac{6 x + 9}{1} \cdot \frac{3 x - 21}{2 x - 5}$

Factor each polynomial:

$\frac{\left(x + 3\right) \left(2 x - 5\right)}{\left(2 x + 3\right) \left(x - 7\right)} \cdot \frac{3 \left(2 x + 3\right)}{1} \cdot \frac{3 \left(x - 7\right)}{2 x - 5}$

Cancel all factors that occur in both the numerator & denominator:

$\frac{\textcolor{red}{x + 3} \cancel{2 x - 5}}{\cancel{2 x + 3} \cancel{x - 7}} \cdot \frac{\textcolor{red}{3} \cancel{2 x + 3}}{1} \cdot \frac{\textcolor{red}{3} \cancel{x - 7}}{\cancel{2 x - 5}}$

$= 9 \left(x + 3\right)$