Question 1161407
A quadratic equation that can be factored will end up factored into the from {{{K(x-a)(x-b)=0}}} , where {{{a}}} and {{{b}}} are the solutions, and K is the coefficient of the term with {{{x^2}}} , with an integer nonzero value for {{{K}}}
In practice, factoring works best when {{{K=1}}} while {{{a}}} and {{{b}}} are integers.
However, you can factor quadratic equations that have rational solutions that are not integers.
It would be easy to make one.
Let us make {{{a=1/2}}} and {{{b=2/3}}} .
{{{(x-a)(x-b)=(x-1/2)(x-1/3)=x^2-(5/6)x+1/6}}}
{{{x^2-5/6+1/6=0}}} has non-integer solutions,
but students may not find it too easy to factor.
We like integer coefficients, anyway,
so we multiply both sides of the equal sign times {{{6}}} to get
{{{6x^2-5x+1=0}}} .
You can take any two rational numbers and do the same trick.
That is probably what is expected for the quadratic equation "creation."
You could also start from any product of binomials with integer coefficients where the coefficients of x are not 1.
{{{(2x+7)(3x-1)=0}}} is equivalent to {{{6x^2+19x-7=0}}} and would work.
You can see that the solutions are {{{x=-7/2}}} and {{{x=1/3}}} .
  
There is also the case that if the two solutions are opposites, {{{a}}} and {{{-a}}}, the equation is {{{(x+a)(x-a)=0}}} or {{{x^2 -a^2=0}}} .
Fr {{{a=1/3}}} we would get {{{x^2-1/9=0}}} ,
and present it as the equivalent form
{{{9x^2-1=0}}} , which can be factored as can be factored as 
 
To be a smartalec, I could argue that I can factor {{{x^2 -pi=0}}} into
{{{(x-sqrt(pi))(x+sqrt(pi))=0}}} , to find the non-integer solutions
{{{sqrt(pi)}}} and {{{-sqrt(pi)}}} , but only if the teacher had a sense of humor.