# How do you find the coefficient to the x^6 term in the expansion of (x-3)^9?

Dec 8, 2016

The coefficient is ${\textcolor{w h i t e}{1}}_{9} {C}_{6} {\left(\text{-} 3\right)}^{9 - 6} = - 2268.$

#### Explanation:

This is a classic example of binomial expansion.

A few simple expansion of $\left(a + b\right)$ to different powers:
${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$
${\left(a + b\right)}^{3} = {a}^{3} + 3 {a}^{2} b + 3 a {b}^{2} + {b}^{3}$
${\left(a + b\right)}^{4} = {a}^{4} + 4 {a}^{3} b + 6 {a}^{2} {b}^{2} + 4 a {b}^{3} + {b}^{4}$

You may begin to see a pattern in the coefficients. They come from Pascal's Triangle:

$\text{ 1}$
$\text{ 1 1}$
$\text{ 1 2 1 } \textcolor{g r e y}{{\left(a + b\right)}^{2}}$
$\text{ 1 3 3 1 } \textcolor{g r e y}{{\left(a + b\right)}^{3}}$
$\text{ 1 4 6 4 1 } \textcolor{g r e y}{{\left(a + b\right)}^{4}}$
$\text{1 5 10 10 5 1}$
$\ldots$

Each number is the sum of the two numbers above it. The entries are also equal to ""_nC_k, where $n$ is the row number and $k$ is the (diagonal) column number. (The first row is "row 0", and the first column is "column 0".)

To expand a general binomial like ${\left(a + b\right)}^{n}$, we use the following form:

${\left(a + b\right)}^{n} = {\text{_nC_0a^n+""_nC_1a^(n-1)b^1+""_nC_2a^(n-2)b^2+...+}}_{n} {C}_{n} {b}^{n}$

As the terms go by, the $a$ exponent counts down from $n$, while the $b$ exponent counts up from $0.$ So the term with ${a}^{k}$ in it will be:

${\textcolor{w h i t e}{1}}_{n} {C}_{n - k} {a}^{k} {b}^{n - k}$.

In this question, $n = 9 ,$ $k = 6 ,$ $a = x ,$ and $b = \text{-} 3.$ Thus, the term with ${x}^{6}$ in it is

${\textcolor{w h i t e}{1}}_{9} {C}_{9 - 6} {x}^{6} {\left(\text{-} 3\right)}^{9 - 6}$
=""_9C_3x^6("-"3)^3
$= 84 {x}^{6} \left(\text{-} 27\right)$
$= - 2268 {x}^{6}$

So the coefficient is $- 2268$.

## Bonus:

We can generalize this to find the ${x}^{k}$ term in a general binomial ${\left(a x + b y\right)}^{n} ,$ where $a$ and $b$ are numbers. The formula will be:

$\left({\textcolor{w h i t e}{1}}_{n} {C}_{k} \cdot {a}^{k} \cdot {b}^{n - k}\right) {x}^{k} {y}^{n - k}$

and the product in the brackets will be our coefficient. For example, in the above problem, we have $a = 1 , b = \text{-3}$ (and $y = 1$). When $n = 9$ and $k = 6$, this gives:

${\textcolor{w h i t e}{1}}_{9} {C}_{6} \cdot {\left(1\right)}^{6} \cdot {\left(\text{-3}\right)}^{9 - 6}$

which is the same value as before. (Note: ${\textcolor{w h i t e}{1}}_{n} {C}_{k} = {\textcolor{w h i t e}{1}}_{n} {C}_{n - k} .$)