Question

Find the coefficients of $x^2$ and $x$ in $(\frac{x}{2}+1)(\frac{x}{2}+2)(\frac{x}{2}+3)$

Answer 1

The given binomials are $(\frac{x}{2}+1)$, $(\frac{x}{2}+2)$ and $(\frac{x}{2}+3)$.

Firstly multiply the first and second term as follows:

$(\frac{x}{2}+1)(\frac{x}{2}+2)(\frac{x}{2}+3)=[(\frac{x}{2}\times \frac{x}{2})+(\frac{x}{2}\times 2)+(1\times \frac{x}{2})+(1\times 2)](\frac{x}{2}+3)=[\frac{x^2}{4}+x+\frac{x}{2}+2](\frac{x}{2}+3)$

$=[\frac{x^2}{4}+\frac{3x}{2}+2](\frac{x}{2}+3)$

Now multiply the resulting expression:

$[\frac{x^2}{4}+\frac{3x}{2}+2](\frac{x}{2}+3)=(\frac{x^2}{4}\times \frac{x}{2})+(\frac{x^2}{4}\times 3)+(\frac{3x}{2}\times \frac{x}{2})+(\frac{3x}{2}\times 3)+(2\times \frac{x}{2})+(2\times 3)$ $=\frac{x^3}{8}+\frac{3x^2}{4}+\frac{3x^2}{4}+\frac{9x}{2}+x+6=\frac{x^3}{8}+\frac{6x^2}{4}+\frac{11x}{2}+6=\frac{x^3}{8}+\frac{3x^2}{2}+\frac{11x}{2}+6$

Therefore, $(\frac{x}{2}+1)(\frac{x}{2}+2)(\frac{x}{2}+3)=\frac{x^3}{8}+\frac{3x^2}{2}+\frac{11x}{2}+6$.

Hence, the coefficient of $x^2$ is $\frac{3}{2}$ and coefficient of $x$ is $\frac{11}{2}$.