Question

How do you simplify $ \dfrac{{{x}^{2}}+5x+6}{{{x}^{2}}-4}? $

Answer 1

Hint: To simplify the given expression we will first factorize both the numerator and denominator of the given fractional expression and then we will cancel the common terms of both the numerator and denominator and at last the term which we obtain after canceling will be the simplified form of the given expression.

Complete step by step answer:
We can see that the given expression $ \dfrac{{{x}^{2}}+5x+6}{{{x}^{2}}-4} $ has quadratic terms in both the numerator and denominator. So, to simplify the given expression we will first factorize both numerator and denominator and then we will cancel the common term of both the numerator and denominator to simplify it.
So, from numerator of $ \dfrac{{{x}^{2}}+5x+6}{{{x}^{2}}-4} $ , we have:
 $ \Rightarrow {{x}^{2}}+5x+6 $
Now, we will factorize it using middle term split:
From the method of middle term split, we know that the middle terms (i.e. term that has degree 1) can be divided into two-term such that the sum of their coefficient is equal to the coefficient of x term, and the product of them is equal to constant in the term in the given polynomial.
So, we can split middle terms 5x as (3x, 2x) as sum of (3x + 2x) = 5x and product of coefficient of 3x and 2x i.e. 6 is equal to constant term in the polynomial $ {{x}^{2}}+5x+6 $ .
 $ \begin{align}
  & \Rightarrow {{x}^{2}}+5x+6 \\
 & \Rightarrow {{x}^{2}}+3x+2x+6 \\
\end{align} $
Now, we will take ‘x’ as common from the first two term and ‘2’ from the last two term, then we will get:
 $ \begin{align}
  & \Rightarrow {{x}^{2}}+3x+2x+6 \\
 & \Rightarrow x\left( x+3 \right)+2\left( x+3 \right) \\
\end{align} $
Now, since we can see that $ \left( x+3 \right) $ is common in both the term so we will take it as common:
 $ \begin{align}
  & \Rightarrow x\left( x+3 \right)+2\left( x+3 \right) \\
 & \Rightarrow \left( x+3 \right)\left( x+2 \right) \\
\end{align} $
Now, we cannot further factorize it, so factor of $ {{x}^{2}}+5x+6 $ is $ \left( x+3 \right)\left( x+2 \right) $ .
Now, from denominator of $ \dfrac{{{x}^{2}}+5x+6}{{{x}^{2}}-4} $ , we have:
 $ \Rightarrow {{x}^{2}}-4 $
Now, we will factorize $ {{x}^{2}}-4 $ and find all its factors.
We can write $ {{x}^{2}}-4 $ as:
 $ \Rightarrow {{x}^{2}}-{{2}^{2}} $
Since, we know that $ \left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right) $
So, we can write $ {{x}^{2}}-{{2}^{2}} $ as:
 $ \Rightarrow {{x}^{2}}-{{2}^{2}}=\left( x-2 \right)\left( x+2 \right) $
So, the factor of $ {{x}^{2}}-4 $ is $ \left( x-2 \right)\left( x+2 \right) $ .
Thus, we can write $ \dfrac{{{x}^{2}}+5x+6}{{{x}^{2}}-4} $ as $ \dfrac{\left( x+3 \right)\left( x+2 \right)}{\left( x-2 \right)\left( x+2 \right)} $ .
We can see that $ \left( x+2 \right) $ is common to both numerator and denominator, so we will cancel it, and then we will get:
 $ \Rightarrow \dfrac{{{x}^{2}}+5x+6}{{{x}^{2}}-4}=\dfrac{\left( x+3 \right)\left( x+2 \right)}{\left( x-2 \right)\left( x+2 \right)} $
 $ \Rightarrow \dfrac{\left( x+3 \right)}{\left( x-2 \right)} $
Hence, simplified form of $ \dfrac{{{x}^{2}}+5x+6}{{{x}^{2}}-4} $ is $ \left( \dfrac{x+3}{x-2} \right) $ .This is our required solution.

Note:
Students are required to note that we can also factorize a quadratic polynomial by first finding the roots of the given quadratic polynomial by equating it to zero and we will use the formula $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ to find the roots and let us suppose that roots of quadratic equation are $ \alpha ,\beta $ , then factor of the quadratic polynomial is $ \left( x-\alpha \right)\left( x-\beta \right) $ .