Question

factoring the numerator, we have

$v(2)=\lim_{t\to 2}\frac{(50t - 16t^{2})-36}{t - 2}$

$=\lim_{t\to 2}\frac{(-16t+\square)(t - \square)}{t - 2}$.

Explanation:

Step1: Expand the numerator

First, expand \((-16t + a)(t - b)=-16t^{2}+(16b + a)t - ab\). The original numerator is \(50t-16t^{2}-36=-16t^{2}+50t - 36\).

Step2: Compare coefficients

Compare the coefficients:

• For the coefficient of \(t\): \(16b + a = 50\)
• For the constant term: \(ab = 36\)

Since we know that when factoring, and considering the limit as \(t
ightarrow2\), we can also use polynomial long - division or trial and error.
If we assume \(b = 2\) (because of the denominator \(t - 2\)), then substitute \(b = 2\) into \(16b+a=50\).
We get \(16\times2+a = 50\), so \(a=50 - 32=18\).
Check \(ab\): \(18\times2=36\)

Answer:

The first blank is \(18\) and the second blank is \(2\)