## What is the quotient? StartFraction t + 3 Over t + 4 EndFraction divided by (t squared + 7 t + 12)

Question

What is the quotient?

StartFraction t + 3 Over t + 4 EndFraction divided by (t squared + 7 t + 12)

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3 months 2022-02-10T10:26:55+00:00 2 Answers 0 views 0

1. The quotient is:

$$\frac{1}{(t+4)^2} \text{ or } \frac{1}{t^2+8t+16}$$

Solution:

Given expression is:

$$\frac{t+3}{\frac{t+4}{t^{2}+7t+12}}$$

We have to find the quotient

From given,

$$\frac{t+3}{\frac{t+4}{t^{2}+7t+12}} = \frac{t+3}{t+4} \times \frac{1}{(t^2+7t+12)} —— eqn 1$$

$$Let\ us\ factor\ (t^2+7t+12)$$

Split the middle term

$$t^2+7t+12 = t^2 + 3t+4t+12$$

Group the terms

$$(t^2+3t) + (4t+12)$$

Take the common factor out

$$t(t+3)+4(t+3)$$

Again take (t+3) as common factor

$$(t+3)(t+4)$$

Substitute in eqn 1

$$\frac{t+3}{\frac{t+4}{t^{2}+7t+12}} = \frac{t+3}{t+4} \times \frac{1}{(t+3)(t+4)}$$

Cancel the common factors

$$\frac{t+3}{\frac{t+4}{t^{2}+7t+12}} = \frac{1}{(t+4)^2}$$

Therefore, quotient is:

$$\frac{t+3}{\frac{t+4}{t^{2}+7t+12}} = \frac{1}{(t+4)^2} = \frac{1}{t^2+8t+16}$$