we have
[tex]\frac{t+3}{\frac{t+4}{t^{2}+7t+12}}[/tex]
we know that
[tex](t^2+7t+12)=(t+4)(t+3)[/tex]
substitute
[tex]\frac{t+3}{\frac{t+4}{t^{2}+7t+12}}=\frac{t+3}{\frac{t+4}{(t+4)(t+3)}} \\ \\=\frac{t+3}{t+4}*\frac{1}{(t+4)(t+3)}\\ \\= \frac{1}{(t+4)^{2}}\\ \\=\frac{1}{t^{2}+8t+16}[/tex]
therefore
the answer is
[tex]\frac{1}{t^{2}+8t+16}[/tex]
You can make the quadratic polynomial factored into smaller pieces to get to the answer.
The quotient of given expression is [tex]t^2 + 2t + 6[/tex]
You can use the fact that division can be taken as multiplication but with the denominator's multiplicative inverse.
Thus,
[tex]\dfrac{a}{\frac{b}{c}} = a \times \dfrac{1}{\frac{b}{c} } = a \times \dfrac{c}{b} = \dfrac{a \times c}{b}[/tex]
The polynomial [tex]t^2 + 7t + 12[/tex] can be factored as:
[tex]t^2 + 7t + 12 = t^2 + 4t + 3t + 12 = t(t + 4) + 3(t+ 4) = (t+3)(t+4)[/tex]
Using that, we have the denominator as:
[tex]\dfrac{t+4}{t^2 + 7t + 12} = \dfrac{t+4}{(t+4)(t+3)} = \dfrac{1}{(t+3)}[/tex]
Thus, the quotient for given numbers is found as:
[tex]\dfrac{t+3}{\frac{t+4}{t^2 + 7t + 12}} = \dfrac{t+3}{\frac{t+4}{(t+3)(t+4)}} = \dfrac{t+3}{\frac{1}{t+3}} = (t+3)^2 = t^2 + 2t + 6[/tex]
Thus,
The quotient of given expression is [tex]t^2 + 2t + 6[/tex]
Learn more about polynomials here:
https://brainly.com/question/19508384
