\[
16(2+t)(22-t)
\]
To make the numbers result in 0 I must make the $t$ in $(2+t)$ be -2 .
and the $t$ in $(22-t)$ must be 22 .
\[
\begin{array}{l}
22-22=0 \\
{[-2,22] \text { - are my values } t} \\
\text { to resultin } 0 \text {. Value. }
\end{array}
\]
to resultin O. Value,
c. Why would the rocket scientists want to know when the expression has a value of 0 ? $t$ seconds
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Step 1 ：The problem is asking for the values of t that make the expression \(16(2+t)(22-t)\) equal to zero. This is a common problem in algebra, where we are often interested in finding the roots of an equation, i.e., the values that make the equation equal to zero. In this case, the expression will be zero when either \((2+t) = 0\) or \((22-t) = 0\). This is because anything multiplied by zero is zero.

Step 2 ：To make the numbers result in 0, I must make the t in \((2+t)\) be -2 and the t in \((22-t)\) must be 22. Therefore, \(22-22=0\) and \([-2,22]\) are my values for t to result in 0.

Step 3 ：The final answer is: The values of t that make the expression \(16(2+t)(22-t)\) equal to zero are \(\boxed{-2}\) and \(\boxed{22}\).