We first expand each of the factors of the given product using Binomial Theorem. We have

${(1 + 2a)^4} = {\,^4}{C_0} + {\,^4}{C_1}(2a) + {\,^4}{C_2}{(2a)^2} + {\,^4}{C_3}{(2a)^3} + {\,^4}{C_4}{(2a)^4}$

$=1+4(2 a)+6\left(4 a^{2}\right)+4\left(8 a^{3}\right)+16 a^{4}$

$=1+8 a+24 a^{2}+32 a^{3}+16 a^{4}$

and  ${(2 – a)^5} = {\,^5}{C_0}{(2)^5} – {\,^5}{C_1}{(2)^4}(a) + {\,^5}{C_2}{(2)^3}{(a)^2} – {\,^5}{C_3}{(2)^2}{(a)^3}$

                $ + {\,^5}{C_4}(2){(a)^4} – {\,^5}{C_5}{(a)^5}$

$=32-80 a+80 a^{2}-40 a^{3}+10 a^{4}-a^{5}$

Thus $(1+2 a)^{4}(2-a)^{5}$

$=\left(1+8 a+24 a^{2}+32 a^{3}+16 a^{4}\right)$

$\left(32-80 a+80 a^{2}-40 a^{3}+10 a^{4}-a^{5}\right)$

The complete multiplication of the two brackets need not be carried out. We write only those terms which involve $a^{4}$. This can be done if we note that ${a^r}.{a^{4 – r}} = {a^4}.$ The terms containing $a^{4}$ are

$1\left(10 a^{4}\right)+(8 a)\left(-40 a^{3}\right)+\left(24 a^{2}\right)\left(80 a^{2}\right)+\left(32 a^{3}\right)(-80 a)+\left(16 a^{4}\right)(32)=-438 a^{4}$