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# Algebraic Expansion

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What is the coefficient of x^3  in the expansion of (x + 2*sqrt(3))^5?

Jun 27, 2021

#1
+139
+3

ahh i do not really remember how to work these, but i will do what i can

we are asked to find the coefficent of $x^3$ in $\left(x+2\sqrt{3}\right)^5$

in these types of binomials, your exponent power value is going to be the $n$ and the $s$, or $i$, or $a$, i do not know what you know them as is going to be the coefficient thinge -- the one we are looking for is $i=3$

so we know that $n=5$ and $i=3$

plug it all into $\frac{n!}{i!\left(n-i\right)!}x^{\left(n-i\right)}\left(2\sqrt{3}\right)^i$

as following, we work it out:

$\frac{5!}{3!\left(5-3\right)!}x^2\left(2\sqrt{3}\right)^3$

$\frac{120}{6\cdot \:2}x^2\left(2\sqrt{3}\right)^3$

$10\left(2\sqrt{3}\right)^3x^2$

$10\cdot \: 8\cdot \: 3^{\frac{1}{2}\cdot \:3} x^2$

$10\cdot \: 8\cdot \: 3^{\frac{3}{2}} x^2$

$10\cdot \: 8\cdot \: 3^{\frac{3}{2}} x^2$

$10\cdot \:8\cdot \:3\sqrt{3}\cdot \: x^2$

$\boxed{240\cdot \sqrt{3}\cdot \: x^2 }$

if you want to get into decimals it would be:

$240\cdot 1.73\cdot \: x^2$

$\boxed{ 415.2\cdot \:x^2 }$

:D

Jun 27, 2021
#2
+121000
+3

2sqrt 3  =  =sqrt 12

So

The coefficient  is

C ( 5 ,  2)   ( x)^(5-2)  (sqrt 12)^2 =

10  x^3  * 12    =

120x^3

Jun 27, 2021
#3
+139
+3

wait actually, no, i do not think thats right.

the cube root and the exponential thingie do not just cancel -- it would be $3\sqrt{3}$ no?

edited by UsernameTooShort  Jun 27, 2021
edited by UsernameTooShort  Jun 27, 2021
#4
+121000
+2

No biggie.....I still gave you a point for a  good  effort  !!!!

CPhill  Jun 27, 2021
#5
+121000
+2

See here  :   https://www.wolframalpha.com/input/?i=%28x++%2B+sqrt+12%29%5E5

CPhill  Jun 27, 2021
#6
+139
+2

well, we both are talking about different things -- though yeah it is 120x^3 as its in the power of 3, not two -- good catch 👍:D