Question

evaluate the limit by factoring the numerator.
\lim_{t \to -3} \frac{3t^2 + 12t + 9}{2t + 6}
simplify any fractions and radicals in your answer.

Explanation:

Step1: Factor the numerator and denominator

First, factor the numerator \(3t^{2}+12t + 9\). We can factor out a 3 first: \(3(t^{2}+4t + 3)\). Then factor the quadratic inside the parentheses: \(t^{2}+4t + 3=(t + 1)(t+ 3)\), so the numerator becomes \(3(t + 1)(t + 3)\).

Next, factor the denominator \(2t+6\). We can factor out a 2: \(2(t + 3)\).

So the expression becomes \(\lim_{t
ightarrow - 3}\frac{3(t + 1)(t + 3)}{2(t + 3)}\)

Step2: Cancel out the common factor

Since \(t
ightarrow - 3\) but \(t
eq - 3\) (we are taking the limit, not evaluating at \(t=-3\)), we can cancel out the common factor \((t + 3)\) from the numerator and the denominator.

After canceling, we get \(\lim_{t
ightarrow - 3}\frac{3(t + 1)}{2}\)

Step3: Substitute \(t=-3\)

Now we can substitute \(t = - 3\) into the simplified expression \(\frac{3(t + 1)}{2}\).

Substitute \(t=-3\): \(\frac{3(-3 + 1)}{2}=\frac{3\times(-2)}{2}=\frac{-6}{2}=-3\)

Answer:

\(-3\)