Answer
$(2A+B^{2})^{4}=16A^{4}+32A^{3}B^{2}+24A^{2}B^{4}+8AB^{6}+B^{8}$
Work Step by Step
We expand using the Binomial Theorem:
$(2A+B^{2})^{4}=\left(\begin{array}{l} 4\\ 0 \end{array}\right)(2A)^{4}+\left(\begin{array}{l} 4\\ 1 \end{array}\right)(2A)^{3}(B^{2})^{1}+\left(\begin{array}{l} 4\\ 2 \end{array}\right)(2A)^{2}(B^{2})^{2}+\left(\begin{array}{l} 4\\ 3 \end{array}\right)(2A)^{1}(B^{2})^{3}+\left(\begin{array}{l} 4\\ 4 \end{array}\right)(B^{2})^{4}=1*16A^4+4*8A^3B^2+6*4A^2B^4+4*2AB^6+1*B^8=16A^{4}+32A^{3}B^{2}+24A^{2}B^{4}+8AB^{6}+B^{8}$
