To solve the given expression:
[tex]\[ \frac{t-3}{t+4} \div \left(t^2 + 7t + 12\right) \][/tex]
we need to follow these steps:
1. Rewrite the division as multiplication by the reciprocal:
[tex]\[ \frac{t - 3}{t + 4} \div \left(t^2 + 7t + 12\right) \quad \text{becomes} \quad \frac{t - 3}{t + 4} \times \frac{1}{t^2 + 7t + 12} \][/tex]
2. Factorize [tex]\( t^2 + 7t + 12 \)[/tex]:
To factorize [tex]\( t^2 + 7t + 12 \)[/tex], we find the roots of the quadratic equation. The factors of 12 that add up to 7 are 3 and 4. Therefore,
[tex]\[ t^2 + 7t + 12 = (t + 3)(t + 4) \][/tex]
3. Substitute the factorized form back into the expression:
[tex]\[
\frac{t - 3}{t + 4} \times \frac{1}{(t + 3)(t + 4)}
\][/tex]
4. Simplify the expression:
[tex]\[
\frac{t - 3}{t + 4} \times \frac{1}{(t + 3)(t+ 4)} = \frac{t - 3}{(t + 4)(t + 3)(t + 4)}
\][/tex]
[tex]\[
= \frac{t - 3}{(t + 3)(t + 4)^2}
\][/tex]
So, the quotient after simplification is:
[tex]\[
\frac{t - 3}{(t + 3)(t + 4)^2}
\][/tex]
This matches our provided answer to the problem.
