Question

The coefficient of the third term in the expansion of the binomial (3x2 + 2y3)4 is

Answer 1

@jim_thompson5910

Answer 2

@iGreen

Answer 3

We have to expand it first..

Answer 4

Is this it?

\(3x^2 + 2y^3)^4\)

Answer 5

\((3x^2 + 2y^3)^4\)*

Answer 6

yess

Answer 7

and idk how to expand this

Answer 8

Okay, expanded form:

\((3x^2 + 2y^3)(3x^2 + 2y^3)(3x^2 + 2y^3)(3x^2 + 2y^3)\)

Then we just keep multiplying..this is gonna get messy..I'll just use a calculator..lol.

\(81x^8 + 216x^6y^3 + 216x^4y^6 + 96x^2y^9 + 16y^{12}\)

Answer 9

omg thats so much! and okay

Answer 10

But I think that's backwards..'cause usually we go from the highest exponent to the lowest..and this is lowest to highest..

\(16y^{12} + 96x^2y^9 + 216x^4y^6 + 216x^6y^3 + 81x^8\)

So yeah, I think you can find the third terms coefficient

Answer 11

so would the answer be 216x^4y^6 ?

Answer 12

Well it says the coefficient..which, I think is just the number..(216).

Answer 13

ohhh okay got it, thanks so much!

Answer 14

No problem

Answer 15

there is a shortcut for \(r^{th}\) term
\(\large \begin{align} \color{black}{\normalsize \text{rth term of } \hspace{.33em}\\~\\
(a+b)^n \hspace{.33em}\\~\\
\normalsize \text{is given as} \hspace{.33em}\\~\\
=\dbinom{n}{r-1} a^{n-(r-1)}b^{r-1}\hspace{.33em}\\~\\
\normalsize \text{so 3rd term for } \hspace{.33em}\\~\\
(3x^2+2y^3)^4\hspace{.33em}\\~\\
=\dbinom{4}{3-1} (3x^2)^{4-(3-1)}(2y^3)^{3-1}\hspace{.33em}\\~\\
}\end{align}\)