# How do you expand (x+3)^5?

May 9, 2017

${x}^{5} + 15 {x}^{4} + 90 {x}^{3} + 270 {x}^{2} + 405 x + 243$

#### Explanation:

The simplest way is by Pascals triangle

the coefficients for expansions are found by :

$1$
${\left(x + y\right)}^{2} \rightarrow \text{ } 1 , 2 , 1$
${\left(x + y\right)}^{3} \rightarrow \text{ } 1 , 3 , 3 , 1$
${\left(x + y\right)}^{4} \rightarrow \text{ } 1 , 4 , 6 , 4 , 1$
$\textcolor{b l u e}{{\left(x + y\right)}^{5} \rightarrow \text{ } 1 , 5 , 10 , 10 , 5 , 1}$

etc....

the powers for${\left(x + y\right)}^{5}$

will be$\rightarrow {x}^{5} , {x}^{4} y , {x}^{3} {y}^{2} {x}^{2} {y}^{3} , x {y}^{4} , {y}^{5}$

so putting the two together we have:

${\left(x + 3\right)}^{5}$
$= \textcolor{b l u e}{1} {x}^{5} + \textcolor{b l u e}{5} {x}^{4} \cdot 3 + \textcolor{b l u e}{10} {x}^{3} \cdot {3}^{2} + \textcolor{b l u e}{10} {x}^{2} \cdot {3}^{3} + \textcolor{b l u e}{5} x \cdot {3}^{4} + \textcolor{b l u e}{1} \cdot {3}^{5}$

$= {x}^{5} + 15 {x}^{4} + 90 {x}^{3} + 270 {x}^{2} + 405 x + 243$