College Algebra, Chapter 9, 9.6, Section 9.6, Problem 28

Expand the expression $(2A + B^2)^4$ using the Binomial Theorem
Recall that the Binomial Theorem is defined as

$
(2A + B^2)^4
=
\left(
\begin{array}{c}
4\\
0
\end{array}
\right)
(2A)^4 +
\left(
\begin{array}{c}
4\\
1
\end{array}
\right)
(2A)^3(B^2) +
\left(
\begin{array}{c}
4\\
2
\end{array}
\right)
(2A)^2(B^2)^2 +
\left(
\begin{array}{c}
4\\
3
\end{array}
\right)
(2A)(B^2)^3 +
\left(
\begin{array}{c}
4\\
4
\end{array}
\right)
(B^2)^4
$

From the 4th row of the Pascal's Triangle,

$
\left(
\begin{array}{c}
4\\
0
\end{array}
\right)
= 1,
\quad
\left(
\begin{array}{c}
4\\
1
\end{array}
\right)
= 4.
\quad
\left(
\begin{array}{c}
4\\
2
\end{array}
\right)
= 6,
\quad
\left(
\begin{array}{c}
4\\
3
\end{array}
\right)
= 4,
\quad
\left(
\begin{array}{c}
4\\
4
\end{array}
\right)
=1
$

Thus,

$
\begin{equation}
\begin{aligned}
(2A + B^2)^4 &= (1)(2A)64 + (4)(2A)^3 (B^2) + (6)(2A)^2(B^2)^2 + (4)(2A)(B^2)^3 + (1) (B^2)^4\\
\\
&= 16A^4 + 32 A^3 B^2 + 24 A^2 B^4 + 8AB^6 + B^8
\end{aligned}
\end{equation}
$