I have a quick question here.

For an exercise, I was asked to factor:

$$11x^2 + 14x - 2685 = 0$$

How do I figure this out quickly without staring at it forever? Is there a quicker mathematical way than guessing number combinations, or do I have to guess until I find the right combination of numbers?

The answer is:

$$(11x + 179)(x - 15) = 0 $$

You want to know the divisibility rules for small numbers. We know $2685$ is divisible by $5$ because of the last digit and by $3$ because of the sum of the digits. Once you find those factors, divide them out, getting $179$. The rules show it is not divisible by $2,3,5,11$ (or $7$ if you know that one, but it is less common. I like the double the ones digit and subtract, which just gives a yes/no answer). Since you only need to check primes up to the square root of the number, you would just have to trial divide by $13$ and maybe $7$ to find that $179$ is prime. In exercises there will always be small factors. In RSA encryption, not so much.

Here's one suggestion: Look at $11$ and $-2685$. Factorize $2685$ into primes to get $3.5.179$. Then check the following.

$11.(-3)+(895), 11.(-15)+179, 11.(3)+(-895), 11.(15)+(-179)$ and check which yields $14$. The one that yields $14$ is $11.(-15)+179$. So you get $(11x+179)(x-15)$.

I don't know if this method is standard but it seems to work for me.

As mentioned in a comment, you can take all the fun out of this by just using the quadratic equation. But, OK, I'll bite!!!

So we assume that we don't know that formula and that the polynomial factors over the rational numbers (or alternatively, we can't miss a chance of employing twisted logic and factoring integers).

Recall the rational root theorem. Before tackling this also review the divisibility rules mentioned by Ross Millikan, and for good measure open up a list of the first 1000 prime numbers.

When we look at $11x^2+14x−2685=0$ the fact that $2,685$ has at least three factors all distinct from $11$ puts us in a foul mood (it would be nice if $11$ was factor, since the rational root theorem would be easier to apply). So we come up with a 'lazy idea'.

We can substitute $x = z + 1$ into $11x^2+14x−2685=0$ and note after regrouping back into polynomial form, the constant term will decrease and the leading coefficient will remain the same. We can keep doing this and see if life gets any easier.

In general, substituting $x = z + 1$ into $ax^2+bx+c=0$ gives $az^2 +(2a+b)z + (a+b+c)$.

For convenience (and a logical abuse), we will keep using the same variable $z$ as we substitute (it won't matter).

Substitution 1: The equation $11x^2+14x−2685=0$ becomes

$tag 1 11z^2 + 36z -2660$

We see $2$, $5$, and $7$ as factors - reject.

Substitution 2: The equation $11z^2 + 36z -2660$ becomes

$tag 2 11z^2 + 58z -2613$

We see $3$ as a divisor, and when we divide it out we get $871$. The 'easy pickings' divisibility rules are no help, so we check the prime number listing. We see that $871$ is a composite that doesn't include $11$ as a factor - reject.

Substitution 3: The equation $11z^2 + 58z -2613$ becomes

$tag 3 11z^2 + 80z -2544$

Just too many factors - reject.

Substitution 4: The equation $11z^2 + 80z -2544$ becomes

$tag 4 11z^2 + 102z -2453$

The number composite number $2,453$ (see prime list) is not divisible by $2$, $5$ or $3$. With a little amount of work you find that $2,453 = 11 times 223$. THIS IS IT!

Setting up for the rational roots, we are looking at

$quad pm frac {1,11,223,2453}{1,11}$

The number $1$ doesn't work, so we check the next easiest number $pm 11$ and find that $-11$ is a root of equation $text{(4)}$.

Now since the substitution was so simple, we can go back in one step, $-1 -1 -1 -1 = -4$, so that $-15$ is a root of the original equation. In a number of ways you can now get the final answer,

$quad (11x + 179)(x - 15) = 0$