Francisca Rodden

2021-12-26

How do I find the binomial expansion of ${\left(2x+1\right)}^{3}$ ?

eninsala06

We must use our knowledge of the binomial expansion:
Method 1:
We can use:
${\left(x+1\right)}^{n}=1+nx+\frac{n\left(n-1\right)}{2!}{x}^{2}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{x}^{3}+\dots$
Substituting $n=3$ and x for $2x⇒$
${\left(2x+1\right)}^{3}=1+\left(3\cdot 2x\right)+\frac{3\cdot 2}{2!}\cdot {\left(2x\right)}^{2}+\frac{3\cdot 2\cdot 1}{3!}\cdot {\left(2x\right)}^{3}$
$=1+6x+12{x}^{2}+8{x}^{3}$

ramirezhereva

Method 2:
We can use:
${\left(A+B\right)}^{n}={A}^{n}+\left(\begin{array}{c}n\\ 1\end{array}\right){A}^{n-1}{B}^{1}+\left(\begin{array}{c}n\\ 2\end{array}\right){A}^{n-2}{B}^{2}+\dots$
Letting $A=2x$ and $B=1$ for this circumstance:
${\left(2x+1\right)}^{3}={\left(2x\right)}^{3}+\left(\begin{array}{c}3\\ 1\end{array}\right){\left(2x\right)}^{2}\left(1\right)+\left(\begin{array}{c}3\\ 2\end{array}\right){\left(2x\right)}^{1}{\left(1\right)}^{2}+\left(\begin{array}{c}3\\ 3\end{array}\right){\left(2x\right)}^{0}{\left(1\right)}^{3}$
$=8{x}^{3}+12{x}^{2}+6x+1$

nick1337

Explanation:
we use Pascal's triangle for the coefficients
$\left(a+b{\right)}^{2}\to 1,2,1$
$\left(a+b{\right)}^{3}\to 1,3,3,1$
So we need 1,3,3,1
the powers of the terms will sum to 3
and startingthe first term will be 3 and descend in 1 s thus
$\left(2x+1{\right)}^{3}=1\left(2x{\right)}^{3}+3\left(2x{\right)}^{2}\left(1\right)+3\left(2x\right)\left(1{\right)}^{2}+1\cdot {1}^{3}$
$\left(2x+1{\right)}^{3}=8{x}^{3}+12{x}^{2}+6x+1$

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