# What is the fastest way to find factors of a number that add up to a sum?

So I am studying for the SAT and am re-learning the grouping method for factoring. The current problem I'm on is $$7m^2 +59m+24$$

With grouping, the traditional way is to find two numbers that are factors of a*c, so in this case 7*24, that add up to 59 (factors of 168 that add up to 59). I can't do this very quickly. Given enough time, I can factor tree it and figure it out, and that what I recall doing when I learned this last year. However, given that I will be under a time constraint during the SAT, I'm wondering if there's a faster way..

I found this link: Here which I found interesting, but it seems to only work in situations where you can factor a number and its square (like 5 and 25). I also found this link: Here from Khan academy, but it only tells me the traditional 'find factors of ac that add up to b'

Anyways, any help is greatly appreciated!

Something that might help would be to use the fact that the product, $168$, is already mostly factored: it's $7 \cdot 24$. In fact, we don't really care that the product is $168$; we can just start working with $7$ and $24$ right away. You can play with these by taking out a factor from one and multiplying it by the other: for example, $\left(7\cdot2, \frac{24}2\right)$.
Now, observe that $59$ is much larger than $7+24$, so we can try borrowing a reasonably large factor from $24$ to multiply by $7$. Of all the possibilities (namely $1,2,3,4,6,8,12,24$), $6$ and $8$ are the likely suspects; the others are clearly too small or too big. Then, since $59 \approx 7\cdot8$, we can try $8$ first, and indeed it works.
In a way, this was a lucky case: $7$ is a prime number, so we didn't have to worry about factoring it (apart from $(1,7\cdot24)$, which clearly isn't the solution). But hopefully this gives you another perspective on the factoring approach.
You could try this $$m=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Given $7m^2+59m+24$ where $a=7,b=59,c=24$