How to factorize quadratic equations quickly? It takes me more than a minute to quickly factorise this kind of quadratic expression.
$$3n^2 -53n + 232$$
I need to solve them in less than $10$-$15$ seconds. Please tell me a way I can solve them.
 A: $$3n^2 -53n + 232$$
I use a process called "A-C Method", "Grouping", or "Split the Middle Term".
$$3*232=696$$
Now, what two numbers add to -53 but multiply to 696?
$-23 * -30 $? That's $690$. Close.
Now, keep in mind, (a+b)(a-b) is a maximum when b=0. Therefore, since we are at $690$, and wish a higher result, the numbers need to get closer, so we just move by 1, since $690$ was so close to $696$.
$-24 * -29 $? That's it! Second guess.
$$3n^2-24n-29n+232$$  (we split the middle term so the 24 is a nice multiple of 3)
$$(3n^2-24n)-(29n-232)$$ (Now, we group, and be mindful of that minus)
$$3n(n-8)-29(n-8)$$
$$(3n-29)(n-8)$$
Will update/edit if OP needs clarification.
A: $$ax^2+bx+c=a(x-x_1)(x-x_2)$$
where$$x_1=\frac{-b-\sqrt{b^2-4ac}}{2a};\;x_2=\frac{-b+\sqrt{b^2-4ac}}{2a}$$
A: As Raffaele says, it's tricky when the leading coefficient isn't $1$, or a perfect square. Especially if you don't know if the expression is reducible or not. You can tell that by looking at the discriminant, $b^2-4ac$, but if you've gone that far, you might as well eliminate all the guesswork and just use the quadratic formula. ;)
But anyway, assuming it is factorisable, because $3$ is prime, $-53<0$ and $232>0$, we know the factors must be of the form
$$(3n-u)(n-v)$$
with $3v+u=53$ and $uv=232$ for $u,v>0$.
Now $232=8\cdot29$ and any multiple of $29>53$, which eliminates all possibilities except $3v+29=53$. Thus $u=29$ and $v=8$, so the desired factorisation is
$$(3n-29)(n-8)$$

Another approach is to complete the square, but it is a bit tedious, and the numbers may get too large for rapid mental calculation.
We need the leading coeffient to be a square, and the coefficient of the $n$ term to be even. So we have to multiply this expression by $12$. Thus
$$36n^2 - 12\cdot53n + 12\cdot8\cdot29$$
$$(6n)^2 - 2\cdot6\cdot53n + 53^2 - 53^2 + 12\cdot8\cdot29$$
$$(6n-53)^2 - (53^2 - 12\cdot8\cdot29)$$
That constant term looks pretty bad, until we notice that $53=29+24$ and $4\cdot24=12\cdot8$
$$(6n-53)^2 - ((29+24)^2 - 4\cdot24\cdot29)$$
$$(6n-53)^2 - (29-24)^2$$
$$(6n-53)^2 - 5^2$$
$$(6n-53+5)(6n-53-5)$$
$$(6n-48)(6n-58)$$
$$12(n-8)(3n-29)$$
And now we can drop that multiplier of $12$
$$(n-8)(3n-29)$$
As I said, it's a bit tedious, but we got there eventually. ;)
A: I am not sure if this helps you.
If $$ax^2+bx+c=0$$ has a rational solution $\frac p q$ then $p\mid c$ and $q\mid a$.
So if
$$ax^2+bx+c=0$$ then the only integer solutions are divisors of $232$ because $3$ is prime.
We have $232=2^3\cdot 29$. So the divisors are $\{\pm 1, \pm2, \pm4, \pm 8\, \pm 29, \pm2 \cdot 29, \pm4 \cdot 29, \pm 8 \cdot 29\}$
If we have an integer solution $x_1$, the numerator $p$ of the other solution $x_2=\frac p q$ is a divisor of $232$,  too, and it satisfies $$53-3\cdot x_1=p\in D$$. From this we see that 53-3*8=29$ asns so $8$ is the solution.
But all in all I think using the formula
$$x_1=\frac{-b-\sqrt{b^2-4ac}}{2};\;x_2=\frac{-b+\sqrt{b^2-4ac}}{2}$$
is the most efficient way. Calculating $b^2-4ac$ shouldn't be a problem, and calculating the square root of a 4 digit number is also rather simple. It is a two digit number. The left digit can be estimated by comparing the size of the numbr to $10^2,20^2,...$ and the right digit can be estimated by  comparing the squares modulo $10$.
