Question

How do you simplify $\dfrac{4{{x}^{2}}-9}{2{{x}^{2}}+11x+12}$ ?

Answer 1

Hint: In this problem, we factorise both the numerator and the denominator. The numerator gets factorised to $\left( 2x+3 \right)\left( 2x-3 \right)$ and the denominator gets factorized to $\left( 2x+3 \right)\left( x+4 \right)$ . Finally, cancelling $\left( 2x+3 \right)$ from the numerator and the denominator, we get our answer.

Complete step by step solution:
The given expression that we have at our disposal is,
$\dfrac{4{{x}^{2}}-9}{2{{x}^{2}}+11x+12}$
If we observe carefully, we can see that the term $4{{x}^{2}}$ is a perfect square and is the square of $2x$ . $9$ is also a perfect square and is the square of $3$ . The expression $4{{x}^{2}}-9$ is thus the subtraction of two squares. We know that the expression of subtraction of squares of the form ${{a}^{2}}-{{b}^{2}}$ can be written as $\left( a+b \right)\left( a-b \right)$ . Here, $a=2x,b=3$ . Thus, rewriting the above expression, we get,
$\Rightarrow \dfrac{\left( 2x+3 \right)\left( 2x-3 \right)}{2{{x}^{2}}+11x+12}....\left( 1 \right)$
Now, we focus on the denominator. It is a quadratic expression. To simplify this quadratic expression, we apply the middle terms method. In this method, in a quadratic expression of the form $a{{x}^{2}}+bx+c$ , we express $b$ as a summation of two numbers $d,e$ such that $de=ac$ . Here, $a=2,b=11,c=12$ . Now, $ac=12\times 2=24$ . $24$ can also be written as $8\times 3$ . Also, $8+3=11$ . Thus,
$\Rightarrow 2{{x}^{2}}+11x+12=2{{x}^{2}}+8x+3x+12$
Taking $2x$ common from the first two terms and $3$ from the last two terms, we get,
$\Rightarrow 2{{x}^{2}}+11x+12=2x\left( x+4 \right)+3\left( x+4 \right)$
This can be rewritten as,
$\Rightarrow 2{{x}^{2}}+11x+12=\left( 2x+3 \right)\left( x+4 \right)$
The expression $\left( 1 \right)$ can thus be modified to,
$\Rightarrow \dfrac{\left( 2x+3 \right)\left( 2x-3 \right)}{\left( 2x+3 \right)\left( x+4 \right)}$
In both the numerator and denominator, $\left( 2x+3 \right)$ is present. So, we cancel them off.
$\Rightarrow \dfrac{\left( 2x-3 \right)}{\left( x+4 \right)}$
Therefore, the given expression can be simplified to $\dfrac{\left( 2x-3 \right)}{\left( x+4 \right)}$ .

Note:
The given expression is a huge expression, so we should be careful not to miss any term. Also, this method requires multiple concepts, we should apply them correctly, especially the middle term method. Finally, we should remember to cancel the $\left( 2x+3 \right)$ from the numerator and the denominator.