Question 231639
<font face="Garamond" size="+2">


I'm going to go way out on a limb and presume that you mean a <i>quadratic</i> equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 1\ =\ 0\ \Rightarrow\ (x+1)(x-1)=0]


In fact, there is no such thing as a quadratic equation that cannot be solved by factoring.  Even if you have irrational or complex roots, you still have roots.  And if you have roots, you have factors.  Now, in many cases it may be <i>very</i> difficult to find the factors and you may have to resort to other means to solve the equation.


For example, consider:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 5x\ -\ 2\ =\ 0]


which does not factor over the integers (doesn't factor over the rationals, either).  The quadratic formula tells us that the roots are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{5\ \pm\ \sqrt{33}}{2}]


Meaning that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ \frac{5\ +\ \sqrt{33}}{2}\right)\left(x\ -\ \frac{5\ -\ \sqrt{33}}{2}\right)=x^2\ -\ 5x\ -\ 2]


Deucedly difficult to solve by factoring, but not entirely impossible.


On the other hand, if you did not ask the ENTIRE question, that is, if the question was "...cannot solve by factoring <i>over the real numbers (or rationals, or integers)</i>" then you could give:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 1\ =\ 0\ \Rightarrow\]: No real number factors


as an example for the second half of your question.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>