Question 1090260
<p>Every quadratic equation has two solutions; every quadratic expression can be factored.  But nearly always when we are asked to factor a quadratic, we want to factor it "over the integers".  And while every quadratic expression can be factored, very few can be factored over the integers.
<br>
So I think the answer to your question is that your quadratic expression can't be factored.
<br>
I assume you are familiar with the quadratic formula, and with the discriminant, b^2-4ac.  A quadratic expression can be factored over the integers if and only if the discriminant is a perfect square (so that the square root of the discriminant is an integer).
<br>
In your quadratic, b^2-4ac is -15.  So not only can the expression not be factored over the integers; but also the negative value of the discriminant means the zeros of the expression (the roots of the equation) are complex.
<br>
The quadratic formula gives the two roots of your example as
{{{x=(3+-i*sqrt(15))/4}}}
or
{{{x=3/4+-i*sqrt(15)/4}}}
<br>
Then, if you wanted a factored form of your quadratic equation, even though the roots are complex, it would be
<br>
{{{2(x-(3+i*sqrt(15))/4)(x-(3-i*sqrt(15))/4)=0}}}