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

Question

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!

Answer 1

You could try this $$m=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Given $7m^2+59m+24$ where $a=7,b=59,c=24$

Answer 2

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.