How do you simplify $\dfrac{{x + 3}}{{{x^2} + 7x + 12}}$?

Hint: First, simplify the denominator by factoring using middle term splitting method. For this, find the product of the first and last constant term of the given expression. Then, choose the factors of product value in such a way that addition or subtraction of those factors is the middle constant term. Then split the middle constant term or coefficient of $x$ in these factors and take common terms out in first terms and last two terms. Then again take common terms out of terms obtained. Then, we will get the factors of the given denominator. Then, write the fraction given and replace the denominator with its factors and cancel the common factor. We will get the desired result.

Complete step by step solution:
We have to simplify $\dfrac{{x + 3}}{{{x^2} + 7x + 12}}$.
First, we will simplify the denominator.
Factor using the middle term splitting method.
Denominator: ${x^2} + 7x + 12$
We have to factor this trinomial.
To factor this trinomial, first we have to find the product of the first and last constant term of the expression.
Here, the first constant term in ${x^2} + 7x + 12$ is $1$, as it is the coefficient of ${x^2}$ and last constant term is $12$, as it is a constant value.
Now, we have to multiply the coefficient of ${x^2}$ with the constant value in ${x^2} + 7x + 12$, i.e., multiply $1$ with $12$.
Multiplying $1$ and $12$, we get
$1 \times 12 = 12$
Now, we have to find the factors of $12$ in such a way that addition or subtraction of those factors is the middle constant term.
Middle constant term or coefficient of $x$ in ${x^2} + 7x + 12$ is $7$.
So, we have to find two factors of $12$, which on multiplying gives $12$ and in addition gives $7$.
We can do this by determining all factors of $12$.
Factors of $12$ are $ \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$.
Now among these values find two factors of $12$, which on multiplying gives $12$ and in addition gives $7$.
After observing, we can see that
$3 \times 4 = 12$ and $3 + 4 = 7$
So, these factors are suitable for factorising the given trinomial.
Now, the next step is to split the middle constant term or coefficient of $x$ in these factors.
That is, write $7x$ as $3x + 4x$ in ${x^2} + 7x + 12$.
After writing $7x$ as $3x + 4x$ in ${x^2} + 7x + 12$, we get
$ \Rightarrow {x^2} + 3x + 4x + 12$
Now, taking $x$ common in ${x^2} + 3x$ and putting in above equation, we get
$ \Rightarrow x\left( {x + 3} \right) + 4x + 12$
Now, taking $4$ common in $4x + 12$ and putting in above equation, we get
$ \Rightarrow x\left( {x + 3} \right) + 4\left( {x + 3} \right)$
Now, taking \[\left( {x + 3} \right)\] common in \[x\left( {x + 3} \right) + 4\left( {x + 3} \right)\] and putting in above equation, we get
$ \Rightarrow \left( {x + 3} \right)\left( {x + 4} \right)$
Therefore, the trinomial ${x^2} + 7x + 12$ can be factored as $\left( {x + 3} \right)\left( {x + 4} \right)$.
Write the fraction given and replace ${x^2} + 7x + 12$ with $\left( {x + 3} \right)\left( {x + 4} \right)$.
$\dfrac{{x + 3}}{{\left( {x + 3} \right)\left( {x + 4} \right)}}$
Cancel the common factor.
$ \Rightarrow \dfrac{1}{{x + 4}}$

Therefore, $\dfrac{{x + 3}}{{{x^2} + 7x + 12}} = \dfrac{1}{{x + 4}}$.

Note: We can also find the simplified version of $\dfrac{{x + 3}}{{{x^2} + 7x + 12}}$ by dividing ${x^2} + 7x + 12$ by $x + 3$.

$ \Rightarrow \left( {x + 3} \right)\left( {x + 4} \right) = {x^2} + 7x + 12$
$ \Rightarrow \dfrac{{x + 3}}{{{x^2} + 7x + 12}} = \dfrac{1}{{x + 4}}$