Question

Find the coefficient of $a^4$ in the product $(1 + 2a)^4 (2 - a)^5$.

• A. $- 435$
• B. $- 438$
• C. $- 424$
• D. $- 430$

Answer 1

Answer: (B)

$\left(1+2a\right)^{4} = \,{}^{4}C_{0}+ \,{}^{4}C_{1}\left(2a\right)+ \,{}^{4}C_{2}\left(2a\right)^{2}+ \,{}^{4}C_{3}\left(2a\right)^{3}+$
$ \,{}^{4}C_{4}\left(2a\right)^{4} = 1+8a+24a^{2}+32a^{3}+16a^{4}$
and $\left(2-a\right)^{5} = \,{}^{5}C_{0}\left(2\right)^{5}-\,{}^{5}C_{1}\left(2\right)^{4}\left(a\right)+\,{}^{5}C_{4}\left(2\right)^{3}\left(a\right)^{2}$
$- \,{}^{5}C_{3}\left(2\right)^{2}\left(a\right)^{3}+\,{}^{5}C_{4}\left(2\right)\left(a\right)^{4} -\,{}^{5}C_{5}\left(a\right)^{5}$
$= 32 - 80a + 80a^{2} - 40a^{3} + 10a^{4} - a^{5}$
Thus $\left(1 + 2a\right)^{4}\left(2 - a\right)^{5}$
$= \left(1 + 8a + 24a^{2} + 32a^{3} + 16a^{4}\right)\left(32 - 80a + 80a^{2}- 40a^{3} + 10a^{4} - a^{5}\right)$
$\therefore $ The terms containing $a^{4}$ are
$1\left(10a^{4}\right) + \left(8a\right) \left(- 40a^{3}\right) + \left(24a^{2}\right)\left(80a^{2}\right) + \left(32a^{3}\right)\left(-80a\right) +$
$\left(16a^{4}\right)\left(32\right) = -438a^{4}$
Thus, the coefficient of $a^{4}$ in the given product is $-438$.