Question

How do you combine $\dfrac{3n+8}{{{n}^{2}}+6n+8}-\dfrac{4n+2}{{{n}^{2}}+n-12}$ ?

Answer 1

Hint: Here in this question we have been asked to combine the given expression $\dfrac{3n+8}{{{n}^{2}}+6n+8}-\dfrac{4n+2}{{{n}^{2}}+n-12}$ . For combining the expression we will perform basic arithmetic operations on similar terms in the expression and make it in simplified and reduced form. We will factorise the denominators of both the terms and then take put common terms from the numerator, thereby simplifying the terms separately. Then, we will take LCM and subtract the terms.

Complete step by step solution:
Now considering from the question we have been asked to combine the given expression $\dfrac{3n+8}{{{n}^{2}}+6n+8}-\dfrac{4n+2}{{{n}^{2}}+n-12}$ .
For combining the expression we will perform simple basic arithmetic operations between similar terms in the expression and make it in simplified and reduced form.
For that we will factorize the denominators. We will use the traditional method of factorization that is we will write the coefficient of $n$ as the sum of factors of product of constant and coefficient of ${{n}^{2}}$ .
For ${{n}^{2}}+6n+8$ factorization can be done as follows
$\begin{align}
  & \Rightarrow {{n}^{2}}+4n+2n+8=n\left( n+4 \right)+2\left( n+4 \right) \\
 & \Rightarrow \left( n+4 \right)\left( n+2 \right) \\
\end{align}$ .
For ${{n}^{2}}+n-12$ factorization can be done as follows
$\begin{align}
  & \Rightarrow {{n}^{2}}+4n-3n-12=n\left( n+4 \right)-3\left( n+4 \right) \\
 & \Rightarrow \left( n+4 \right)\left( n-3 \right) \\
\end{align}$ .
By substituting these values in the expression we will have $\dfrac{3n+8}{\left( n+4 \right)\left( n+2 \right)}-\dfrac{4n+2}{\left( n+4 \right)\left( n-3 \right)}$
Now we will make the expression to have a common denominator that we will have
$\begin{align}
  & \dfrac{\left( 3n+8 \right)\left( n-3 \right)}{\left( n+4 \right)\left( n+2 \right)\left( n-3 \right)}-\dfrac{\left( 4n+2 \right)\left( n+2 \right)}{\left( n+4 \right)\left( n-3 \right)\left( n+2 \right)} \\
 & \Rightarrow \dfrac{\left( 3n+8 \right)\left( n-3 \right)-\left( 4n+2 \right)\left( n+2 \right)}{\left( n+4 \right)\left( n+2 \right)\left( n-3 \right)} \\
\end{align}$.
Now we will multiply the expressions in the numerator then we will have $\dfrac{3{{n}^{2}}-9n+8n-24-\left( 4{{n}^{2}}+8n+2n+4 \right)}{\left( n+4 \right)\left( n+2 \right)\left( n-3 \right)}$ .
Now we will further simplify this. Then we will have
$\begin{align}
  & \dfrac{-{{n}^{2}}-11n-28}{\left( n+4 \right)\left( n+2 \right)\left( n-3 \right)}=\dfrac{-\left( {{n}^{2}}+11n+28 \right)}{\left( n+2 \right)\left( n+4 \right)\left( n-3 \right)} \\
 & \Rightarrow \dfrac{-\left( {{n}^{2}}+7n+4n+28 \right)}{\left( n+4 \right)\left( n+2 \right)\left( n-3 \right)}=\dfrac{-\left( n\left( n+7 \right)+4\left( n+7 \right) \right)}{\left( n+4 \right)\left( n+2 \right)\left( n-3 \right)} \\
 & \Rightarrow \dfrac{-\left( n+7 \right)\left( n+4 \right)}{\left( n+4 \right)\left( n+2 \right)\left( n-3 \right)}=\dfrac{-\left( n+7 \right)}{\left( n+2 \right)\left( n-3 \right)} \\
\end{align}$ .
Therefore we can conclude that the simplified form of the given expression $\dfrac{3n+8}{{{n}^{2}}+6n+8}-\dfrac{4n+2}{{{n}^{2}}+n-12}$ is $\dfrac{-\left( n+7 \right)}{\left( n+2 \right)\left( n-3 \right)}$ .

Note: This type of questions require a lot of care while performing the calculations. For factorization of the any quadratic expression we can also use the formula for finding roots of the expression in the form of $a{{x}^{2}}+bx+c$ is given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . For example consider for the expression ${{n}^{2}}+6n+8$ the roots will be
 $\begin{align}
  & \dfrac{-6\pm \sqrt{{{\left( 6 \right)}^{2}}-4\left( 8 \right)}}{2}=\dfrac{-6\pm \sqrt{36-32}}{2} \\
 & \Rightarrow \dfrac{-6\pm \sqrt{4}}{2}=\dfrac{-6\pm 2}{4} \\
 & \Rightarrow -3\pm 1=-4,-2 \\
\end{align}$ .
Therefore factors will be $\left( n+4 \right)\left( n+2 \right)$ . Similarly we can factorize any expression.