# How do you find the coefficient of a of the term ax^4y^5 in the expansion of the binomial (3x-2y)^9?

Jan 28, 2018

$a = \textcolor{b l u e}{- 326592}$

#### Explanation:

Pattern for power 9 as per Pascal's Triangle

1 9 36 84 126 126 84 36 9 1

$a {x}^{4} {y}^{5}$ is the ${6}_{t} h$ term of the expansion.

${6}_{t} h$ term is $126 {\left(3 x\right)}^{4} {\left(- 2 y\right)}^{5}$

$= - 126 \cdot {\left(3\right)}^{4} {x}^{4} \left({2}^{5}\right) {y}^{5}$

$= - \left(126\right) \cdot \left(81\right) \cdot \left(32\right) {x}^{4} {y}^{5}$

$= - \left(326592\right) {x}^{4} {y}^{5}$

Hence $a = \textcolor{b l u e}{- 326592}$