How to know if equation can be solved by factorising before trying? So, I have core 1 test tomorrow and there is a lot of solving of quadratic equations without calculator and my weakest point is the time I waste in trying to factorise and equation but then it ends up it doesn't factorise or the opposite way in which I use equation while I could've done factorising. Is there any way you can know if equation factorise or not?
 A: Check the discriminant. If it is a perfect square, the radical expression in the quadratic formula becomes an integer (I'm assuming you've cleared fractions before beginning the problem, and all coefficients are integers after that).
Recall that the discriminant for $ax^2 + bx + c = 0$ is $\Delta = b^2-4ac$, and that the solution(s) are $$x = \frac{-b\pm\sqrt{\Delta}}{2a}$$
I'm also assuming you're only interested in factoring if you can do it without using radicals. For example, you could factor $x^2-2$ as $(x+\sqrt{2})(x-\sqrt{2})$, but I'm guessing that doesn't interest you.
A: For any quadratic equation $ax^2+bx+c=0$, $a,b,c\in \mathbb{R}$, there exist solution(s) (i.e., you can factorise) if and only if the discriminant 
$$D=b^2-4ac$$
is higher than or equal to zero.
A: Two ways.  The first is particularly useful if the leading coefficient $a$ (as in $ax^2$) is $1$:  If two factors $(x-m)(x-n)$ exist with $m$ and $n$ integers (or even rational) then $m$ and $n$ wil always be factors the constanc coefficient $c$.
This means you need to try very few combinations, since depending on the sign of $c$ 
of the factors will will always be dictated by $b$.
For example, 
$$ x^2 +6x +10$$
could only have $m,n$ as $(1,10)$ or $(2,5)$; since neither of those add up to $6$ you can't factor this with easy factors.
The other way is to find $b^2-4ac$.  If that is a perfect square, then the equation can be factored nicely.  If not, then at least you are halfway toward finding the roots using the quadratic formula.
A: You can only factorise easily (without involving surds) if the discriminant is a perfect square. Mind you ,if you've already worked the discriminant out, you may as well go on and solve using the formula.
A: You can check if their are easy factors by applying the rational root theorem, which in the special case of 
$$x^2+bx+c$$
says that the only rational roots can be factors of $c$, and in the more general case of
$$ax^2+bx+c$$
says the only rational roots have to be of a form $\dfrac{\text{factor of }c}{\text{factor of }a}$.
Still, it seems that applying the quadratic formula isn't so hard to be fast at in simple cases.
A: For example,
x2+6x+10
x2+6x+10
could only have m,nm,n as (1,10)(1,10) or (2,5)(2,5); since neither of those add up to 66 you can't factor this with easy factors.
The other way is to find b2−4acb2−4ac. If that is a perfect square, then the equation can be factored nicely. If not, then at least you are halfway toward finding the roots using the quadratic formula.
