Question

How do you simplify \[\dfrac{{{x}^{2}}-2x-8}{{{x}^{2}}+3x-4}\]?

Answer 1

Hint: The given two quadratic equations need to be simplified. The equations can be factored into the multiplication form of two linear equations. This will give us factors of the two equations and we will be able to cancel out common factor if any to find the simplified form of the given expression \[\dfrac{{{x}^{2}}-2x-8}{{{x}^{2}}+3x-4}\].

Complete step by step solution:
We have been given the division of two quadratic equations. We need to find the factor of the equations.
The two quadratic equations are \[{{x}^{2}}-2x-8\] and \[{{x}^{2}}+3x-4\].
 We use vanishing method to solve them.
We find the value of $x=a$ for which the function $f\left( x \right)={{x}^{2}}-2x-8=0$.
We take $x=a=4$. We can see $f\left( 4 \right)={{4}^{2}}-2\times 4-8=16-8-8=0$.
So, the root of the $f\left( x \right)={{x}^{2}}-2x-8$ will be the function $\left( x-4 \right)$. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$.
Therefore, the term $\left( x-4 \right)$ is a factor of the polynomial \[{{x}^{2}}-2x-8\].
So, \[{{x}^{2}}-2x-8=\left( x-4 \right)\left( x+2 \right)\].
For $g\left( x \right)={{x}^{2}}+3x-4$, we take $x=a=1$. We can see $f\left( 1 \right)={{1}^{2}}+3\times 1-4=1+3-4=0$.
So, the root of the $g\left( x \right)={{x}^{2}}+3x-4$ will be the function $\left( x-1 \right)$.
Therefore, \[{{x}^{2}}+3x-4=\left( x-1 \right)\left( x+4 \right)\].
Now we put the factored values for the two equations.
We get \[\dfrac{{{x}^{2}}-2x-8}{{{x}^{2}}+3x-4}=\dfrac{\left( x-4 \right)\left( x+2 \right)}{\left( x-1 \right)\left( x+4 \right)}\].
Now we can see there is no common factor in the denominator and the numerator.
We can’t eliminate any factor from both denominator and the numerator.
Therefore, \[\dfrac{{{x}^{2}}-2x-8}{{{x}^{2}}+3x-4}\] can’t be simplified any further.

Note: We can verify the factors for the functions.
We can validate the other solution of $x=-2$ for $f\left( x \right)={{x}^{2}}-2x-8$. We put the value of $x=-2$ in the equation.
$f\left( -2 \right)={{\left( -2 \right)}^{2}}-2\left( -2 \right)-8=4+4-8=0$.
Similarly, the root value $x=-4$ satisfies $g\left( x \right)={{x}^{2}}+3x-4$.