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# How do you simplify $\dfrac{{{x}^{2}}+5x+4}{{{x}^{2}}-16}$?

Last updated date: 03rd Mar 2024
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Hint: In this problem we need to simplify the given fraction. We can observe that the given fraction consists of two quadratic equations in both numerator and denominator. So, we will consider both numerator and denominator individually. First, we will consider the numerator and factorise the quadratic equation. While factoring a quadratic equation $a{{x}^{2}}+bx+c$ we will split the middle term $bx$ as $px+qx$ where $pq=ac$. So, we will compare the quadratic equation with the standard form of the quadratic equation $a{{x}^{2}}+bx+c$ and follow the above procedure and split the middle term. Now we will take appropriate terms as common and factorise it. Now we will consider the denominator and use the algebraic formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . Now we will substitute both the values in the given fraction and simplify it to get the required result.

Complete step by step solution:
Given that, $\dfrac{{{x}^{2}}+5x+4}{{{x}^{2}}-16}$.
Considering the numerator which is ${{x}^{2}}+5x+4$.
Considering the above equation with $a{{x}^{2}}+bx+c$, then we will get
$a=1$, $b=5$, $c=4$.
Now the value of $ac$ is $ac=1\times 4=4$.
Factors of $4$ are $1$, $2$, $4$. From these factors we can write
\begin{align} & 1\times 4=4 \\ & 1+4=5 \\ \end{align}
Now splitting the middle term in the given quadratic equation, then we will get
$\Rightarrow {{x}^{2}}+5x+4={{x}^{2}}+x+4x+4$
Taking $x$ as common from the first two terms and $4$ as common from the last two terms, then we will get
$\Rightarrow {{x}^{2}}+5x+4=x\left( x+1 \right)+4\left( x+1 \right)$
Now taking $x+1$ as common from the above equation, then we will get
$\Rightarrow {{x}^{2}}+5x+4=\left( x+1 \right)\left( x+4 \right).....\left( \text{i} \right)$
Now considering the denominator which is ${{x}^{2}}-16$.
We can write $16={{4}^{2}}$ in the above equation, then we will get
$\Rightarrow {{x}^{2}}-16={{x}^{2}}-{{4}^{2}}$
Applying the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the above equation, then we will get
$\Rightarrow {{x}^{2}}-16=\left( x+4 \right)\left( x-4 \right)...\left( \text{ii} \right)$
From equations $\left( \text{i} \right)$ and $\left( \text{ii} \right)$ we can write the given fraction as
$\Rightarrow \dfrac{{{x}^{2}}+5x+4}{{{x}^{2}}-16}=\dfrac{\left( x+1 \right)\left( x+4 \right)}{\left( x-4 \right)\left( x+4 \right)}$
Cancelling the term $x+4$ which is in both numerator and denominator, then we will get
$\Rightarrow \dfrac{{{x}^{2}}+5x+4}{{{x}^{2}}-16}=\dfrac{x+1}{x-4}$

Hence the simplified form of the given fraction $\dfrac{{{x}^{2}}+5x+4}{{{x}^{2}}-16}$ is $\dfrac{x+1}{x-4}$.

Note: In the above fraction we have the denominator in the form of ${{a}^{2}}-{{b}^{2}}$ so we have used the algebraic formula and simplified the fraction. In some cases, in the denominator, we may have a quadratic equation. Then we need to factorise the denominator as we did for the numerator and simplify the fraction.