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 $-438$.