# How do you multiply (x+2)^5?

Jul 4, 2017

${x}^{5} + 10 {x}^{4} + 40 {x}^{3} + 80 {x}^{2} + 80 x + 32$

#### Explanation:

Use Pascal's triangle (n=5 so 5th row) giving you 1,5,10,10,5,1. The powers of x go from ${x}^{5}$ down to ${x}^{0}$ i.e. 1 and the 2nd term contributes ${2}^{0}$ i.e. 1 up to ${2}^{5}$.
So 1st term is 1 (from Pascal) multiplied by ${x}^{5}$ multiplied by ${2}^{0}$
2nd term is 5 multiplied by x^4 then 2^1
and so on.
In an exam, you won't have the triangle so you can use the nCr button to find the multipliers.

Jul 4, 2017

See a solution process below:

#### Explanation:

We can use Pascal's triangle to solve this problem.

The triangle values for the exponent 5 are:

$\textcolor{red}{1} \textcolor{w h i t e}{\ldots \ldots \ldots} \textcolor{red}{5} \textcolor{w h i t e}{\ldots \ldots \ldots} \textcolor{red}{10} \textcolor{w h i t e}{\ldots \ldots \ldots} \textcolor{red}{10} \textcolor{w h i t e}{\ldots \ldots \ldots} \textcolor{red}{5} \textcolor{w h i t e}{\ldots \ldots \ldots} \textcolor{red}{1}$

Therefore ${\left(\textcolor{b l u e}{x} + \textcolor{g r e e n}{2}\right)}^{5}$ can be multiplied as:

$\textcolor{red}{1} \left({\textcolor{g r e e n}{2}}^{0} {\textcolor{b l u e}{x}}^{5}\right) + \textcolor{red}{5} \left({\textcolor{g r e e n}{2}}^{1} {\textcolor{b l u e}{x}}^{4}\right) + \textcolor{red}{10} \left({\textcolor{g r e e n}{2}}^{2} {\textcolor{b l u e}{x}}^{3}\right) + \textcolor{red}{10} \left({\textcolor{g r e e n}{2}}^{3} {\textcolor{b l u e}{x}}^{2}\right) + \textcolor{red}{5} \left({\textcolor{g r e e n}{2}}^{4} {\textcolor{b l u e}{x}}^{1}\right) + \textcolor{red}{1} \left({\textcolor{g r e e n}{2}}^{5} {\textcolor{b l u e}{x}}^{0}\right)$

$\textcolor{red}{1} \left(\textcolor{g r e e n}{1} {\textcolor{b l u e}{x}}^{5}\right) + \textcolor{red}{5} \left(\textcolor{g r e e n}{2} {\textcolor{b l u e}{x}}^{4}\right) + \textcolor{red}{10} \left(\textcolor{g r e e n}{4} {\textcolor{b l u e}{x}}^{3}\right) + \textcolor{red}{10} \left(\textcolor{g r e e n}{8} {\textcolor{b l u e}{x}}^{2}\right) + \textcolor{red}{5} \left(\textcolor{g r e e n}{16} {\textcolor{b l u e}{x}}^{1}\right) + \textcolor{red}{1} \left(\textcolor{g r e e n}{32} \cdot \textcolor{b l u e}{1}\right)$

${x}^{5} + 10 {x}^{4} + 40 {x}^{3} + 80 {x}^{2} + 80 x + 32$

Jul 4, 2017

You can use the Binomial Theorem to expand.

#### Explanation:

${\left(x + 2\right)}^{5}$

$= \left(\begin{matrix}5 \\ 0\end{matrix}\right) \cdot {x}^{5} \cdot {2}^{0} + \left(\begin{matrix}5 \\ 1\end{matrix}\right) \cdot {x}^{4} \cdot {2}^{1} + \left(\begin{matrix}5 \\ 2\end{matrix}\right) \cdot {x}^{3} \cdot {2}^{2} + \left(\begin{matrix}5 \\ 3\end{matrix}\right) \cdot {x}^{2} \cdot {2}^{3} + \left(\begin{matrix}5 \\ 4\end{matrix}\right) \cdot {x}^{1} \cdot {2}^{4} + \left(\begin{matrix}5 \\ 5\end{matrix}\right) \cdot {x}^{0} \cdot {2}^{5}$

$= {x}^{5} + 10 {x}^{4} + 40 {x}^{3} + 80 {x}^{2} + 80 x + 32$

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By way of explanation using 1 example

$\left(\begin{matrix}5 \\ 2\end{matrix}\right)$ is another way of writing ${\textcolor{w h i t e}{}}^{5} {C}_{2}$ which is:

(5!)/((5-2)!2!) = (5xx4xx3!)/((3!)2!) = (5xx4)/(2xx1)xx(3!)/(3!)=10

The above is called a Combination which in general terms is:
" "(n!)/((n-r)!r!)

This is different to Permutations which in general terms is:
" "(n!)/((n-r)!)