Write down and simplify the first \(4\) terms in the expansion by the binomial theorem of \(\left(1 - \tfrac{1}{2}x\right)^{10}\).
The binomial theorem says that, if \(x\) is a real number and \(n\) is a positive integer, then \[
(1+x)^n = \sum_{i=0}^n \binom{n}{i} x^i
\] where
\[\begin{align*}
\binom{n}{0} &= 1, \\
\binom{n}{i} &= \frac{n(n-1) \dotsm (n-(i-1))}{i(i-1) \dotsm 1} \quad\text{for each integer $1 \le i \le n$.}
\end{align*}\]
By applying this to the expression \(\left(1 - \tfrac{1}{2}x\right)^{10}\), we have that \[
\left(1 - \tfrac{1}{2}x\right)^{10} = \binom{10}{0} \left( -\frac{x}{2} \right)^0 + \binom{10}{1} \left( -\frac{x}{2} \right)^1 + \binom{10}{2} \left( -\frac{x}{2} \right)^2 + \binom{10}{3} \left( -\frac{x}{2} \right)^3 + \dotsb
\] We can simplify this a little.
\[\begin{align*}
\binom{10}{1} &= \frac{10}{1} = 10, \\
\binom{10}{2} &= \frac{10 \times 9}{2 \times 1} = 45, \\
\binom{10}{3} &= \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120,
\end{align*}\]
so that \[ \left(1 - \tfrac{1}{2}x\right)^{10} = 1 - 5x + \frac{45}{4} x^2 - 15x^3 + \dotsb \]
Find the coefficient of \(x^2\) in the expansion of \[ (5 + 4x)\left(1 - \tfrac{1}{2}x\right)^{10}. \]
There are only two contributing terms to the coefficient of \(x^2\).
The \(5\) from the factor \((5 + 4x)\) pairs with the \(\dfrac{45}{4}x^2\) term in the expansion of \(\left(1 - \tfrac{1}{2}x\right)^{10}\), and the \(4x\) term pairs with the \(-5x\) term. Since we have already calculated the coefficients in this expansion this is a straightforward task.
Thus, we have that the coefficient of \(x^2\) in the product is \[\begin{align*} 5 \times \frac{45}{4} + 4 \times (-5) &= \frac{225}{4} - 20 \\ &= \frac{145}{4}. \end{align*}\]
