How do I factorise this difficult quadratic without a calculator? On one of the UKMT maths challenge past papers(Team challenge) it asks you this question:
Factorise $120x^2 + 97x - 84$
That is the whole question. 
I used a calculator and found that you factorise it into $(40x-21)(3x+4)$
Bearing in mind that a calculator is not allowed in the team challenge, how can this be done? Is there a simple trick to this special case?

In this context, "factorise" obviously means that the two bracketed linear terms will have integer coefficients.
Start by listing the possible factor pairs for the first and last terms of the quadratic:
$$\left(\begin{matrix}120\\1\end{matrix}\right),\left(\begin{matrix}60\\2\end{matrix}\right),\left(\begin{matrix}40\\3\end{matrix}\right),\left(\begin{matrix}30\\4\end{matrix}\right),\left(\begin{matrix}20\\6\end{matrix}\right),\left(\begin{matrix}15\\8\end{matrix}\right),\left(\begin{matrix}12\\10\end{matrix}\right)$$
And
$$\left(\begin{matrix}84\\1\end{matrix}\right),\left(\begin{matrix}42\\2\end{matrix}\right),\left(\begin{matrix}28\\3\end{matrix}\right),\left(\begin{matrix}21\\4\end{matrix}\right),\left(\begin{matrix}14\\6\end{matrix}\right),\left(\begin{matrix}12\\7\end{matrix}\right)$$
We have to pick one pair from the first set and one pair from the second set to make up the required linear coefficients.
However, in this case, we require the difference in the products to be $97$ which is an odd number, so we can immediately eliminate any pairs containing only even numbers.
So we are left with, for the first pair, 
$$\left(\begin{matrix}120\\1\end{matrix}\right),\left(\begin{matrix}40\\3\end{matrix}\right),\left(\begin{matrix}15\\8\end{matrix}\right)$$
And for the second pair, 
$$\left(\begin{matrix}84\\1\end{matrix}\right),\left(\begin{matrix}28\\3\end{matrix}\right),\left(\begin{matrix}21\\4\end{matrix}\right),\left(\begin{matrix}12\\7\end{matrix}\right)$$
Now by quick inspection we can further eliminate some pairs, such as $$\left(\begin{matrix}120\\1\end{matrix}\right)$$
since this will clearly result in a product pair which is too large. It is then a matter of checking the remaining pairs, which shouldn't take too long.
