How do you simplify \[\dfrac{{{x}^{2}}-3x+2}{{{x}^{2}}+5x-6}\]?

Hint: We have two quadratic equations in the denominator and the numerator. We factor them using different processes. We eliminate the common factor from them. We find the simplified form the expression.

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 \[{{x}^{2}}-3x+2\] and \[{{x}^{2}}+5x-6\].
 We use vanishing methods to solve them.
We find the value of $x=a$ for which the function $f\left( x \right)={{x}^{2}}-3x+2=0$.
We take $x=a=1$. We can see $f\left( 1 \right)={{1}^{2}}-3\times 1+2=1-3+2=0$.
So, the root of the $f\left( x \right)={{x}^{2}}-3x+2$ will be the function $\left( x-1 \right)$. 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 $\left( x-1 \right)$ is a factor of the polynomial \[{{x}^{2}}-3x+2\].
So, \[{{x}^{2}}-3x+2=\left( x-1 \right)\left( x-2 \right)\].
For $g\left( x \right)={{x}^{2}}+5x-6$, we take $x=a=1$. We can see $f\left( 1 \right)={{1}^{2}}+5\times 1-6=1+5-6=0$.
So, the root of the $g\left( x \right)={{x}^{2}}+5x-6$ will be the function $\left( x-1 \right)$.
Therefore, \[{{x}^{2}}+5x-6=\left( x-1 \right)\left( x+6 \right)\].
Now we put the factored values for the two equations.
We get \[\dfrac{{{x}^{2}}-3x+2}{{{x}^{2}}+5x-6}=\dfrac{\left( x-1 \right)\left( x-2 \right)}{\left( x-1 \right)\left( x+6 \right)}\].
Now we can see in the denominator and the numerator the common factor is $\left( x-1 \right)$.
We can eliminate the factor from both denominator and the numerator.
So, \[\dfrac{{{x}^{2}}-3x+2}{{{x}^{2}}+5x-6}=\dfrac{\left( x-1 \right)\left( x-2 \right)}{\left( x-1 \right)\left( x+6 \right)}=\dfrac{\left( x-2 \right)}{\left( x+6 \right)}\]. The only condition being $x\ne 1$.

Therefore, the simplified form of \[\dfrac{{{x}^{2}}-3x+2}{{{x}^{2}}+5x-6}\] is \[\dfrac{\left( x-2 \right)}{\left( x+6 \right)}\].

Note: We can also factorise the quadratic equations using grouping method.
In the case of \[{{x}^{2}}-3x+2\], we break the middle term $-3x$ into two parts of $-2x$ and $-x$.
So, ${{x}^{2}}-3x+2={{x}^{2}}-2x-x+2$.
Factorising we get
$\begin{align}
  & {{x}^{2}}-3x+2 \\
 & ={{x}^{2}}-2x-x+2 \\
 & =x\left( x-2 \right)-\left( x-2 \right) \\
 & =\left( x-2 \right)\left( x-1 \right) \\
\end{align}$
Similarly, \[{{x}^{2}}+5x-6={{x}^{2}}+6x-x-6=\left( x+6 \right)\left( x-1 \right)\].