Question

Find the coefficient of $x^4$ in the expansion of the product $(1+2x{)}^4(2-x{)}^5$.

Answer 1

First expand the term ${\left(1+2x\right)}^4$ by binomial expansion.

${\left(1+2x\right)}^4={}^4{C}_0{\left(1\right)}^4{\left(2x\right)}^0+{}^4{C}_1{\left(1\right)}^3{\left(2x\right)}^1+{}^4{C}_2{\left(1\right)}^2{\left(2x\right)}^2+{}^4{C}_3{\left(1\right)}^1{\left(2x\right)}^3+{}^4{C}_4{\left(1\right)}^0{\left(2x\right)}^4$

$=1+8x+24x^2+32x^3+16x^4$ (1)

Now expand the term ${\left(2-x\right)}^5$ by binomial expansion,

${\left(2-x\right)}^5={}^5{C}_0{\left(2\right)}^5{\left(x\right)}^0-{}^5{C}_1{\left(2\right)}^4{\left(x\right)}^1+{}^5{C}_2{\left(2\right)}^3{\left(x\right)}^2-{}^5{C}_3{\left(2\right)}^2{\left(x\right)}^3+{}^5{C}_4{\left(2\right)}^1{\left(x\right)}^4-{}^5{C}_5{\left(2\right)}^0{\left(x\right)}^5$

$=32-80x+80x^2-40x^3+10x^4-x^5$ (2)

Multiply the coefficients of those powers which can give the term $x^4$ and then add from equation (1) and (2).

$=1\times 10+8\left(-40\right)+24\left(80\right)+32\left(-80\right)+16\left(32\right)$

$=-438$

Therefore, the coefficient of $x^4$ is $-438$.