# How do you find the binomial expansion for (2x+3)^3?

Jul 10, 2015

${\left(2 x + 3\right)}^{3} = 8 {x}^{3} + 36 {x}^{2} + 54 x + 27$

#### Explanation:

With the Pascal's triangle, it's easy to find every binomial expansion :

Each term, of this triangle, is the result of the sum of two terms on the top-line. (example in red)

$1$
$1. 1$
$\textcolor{b l u e}{1. 2. 1}$
$1. \textcolor{red}{3} . \textcolor{red}{3.} 1$
$1. 4. \textcolor{red}{6.} 4. 1$
...

More, each line has the information of one binomial expansion :

The 1st line, for the power $0$
The 2nd, for the power $1$
The 3rd, for the power $2$...

For example : ${\left(a + b\right)}^{2}$ we will use the 3rd line in blue following this expansion :

${\left(a + b\right)}^{2} = \textcolor{b l u e}{1} \cdot {a}^{2} \cdot {b}^{0} + \textcolor{b l u e}{2} \cdot {a}^{1} \cdot {b}^{1} + \textcolor{b l u e}{1} \cdot {a}^{0} \cdot {b}^{2}$

Then : ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

To the power $3$ :

${\left(a + b\right)}^{3} = \textcolor{g r e e n}{1} \cdot {a}^{3} \cdot {b}^{0} + \textcolor{g r e e n}{3} \cdot {a}^{2} \cdot {b}^{1} + \textcolor{g r e e n}{3} \cdot {a}^{1} \cdot {b}^{2} + \textcolor{g r e e n}{1} \cdot {a}^{0} \cdot {b}^{3}$

Then ${\left(a + b\right)}^{3} = {a}^{3} + 3 {a}^{2} b + 3 a {b}^{2} + {b}^{3}$

So here we have $\textcolor{red}{a = 2 x}$ and $\textcolor{b l u e}{b = 3}$ :

And ${\left(2 x + 3\right)}^{3} = {\textcolor{red}{\left(2 x\right)}}^{3} + 3 \cdot {\textcolor{red}{\left(2 x\right)}}^{2} \cdot \textcolor{b l u e}{3} + 3 \cdot \textcolor{red}{\left(2 x\right)} \cdot {\textcolor{b l u e}{3}}^{2} + {\textcolor{b l u e}{3}}^{3}$

Therefore : ${\left(2 x + 3\right)}^{3} = 8 {x}^{3} + 36 {x}^{2} + 54 x + 27$

Jul 10, 2015

${\left(2 x + 3\right)}^{3} = 8 {x}^{3} + 36 {x}^{2} + 54 x + 27$

#### Explanation:

${\left(2 x + 3\right)}^{3}$

Use the cube of a sum method, in which ${\left(a + b\right)}^{3} = {a}^{3} + 3 {a}^{2} b + 3 a {b}^{2} + {b}^{3}$.

a=2x; $b = 3$

${\left(2 x + 3\right)}^{3} = {\left(2 x\right)}^{3} + \left(3 \cdot 2 {x}^{2} \cdot 3\right) + \left(3 \cdot 2 x \cdot {3}^{2}\right) + {3}^{3}$ =

$8 {x}^{3} + \left(3 \cdot 4 {x}^{2} \cdot 3\right) + \left(3 \cdot 2 x \cdot 9\right) + 27$ =

$8 {x}^{3} + \left(9 \cdot 4 {x}^{2}\right) + \left(27 \cdot 2 x\right) + 27$ =

$8 {x}^{3} + 36 {x}^{2} + 54 x + 27$