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?
On
The rational root theorem gives you a number of choices. You can cut them down quite a bit by noting that the value of your quadratic is negative at $0$ and positive at $1$, so one of the roots must be in $[0,1]$ That gets rid of half the possibilities because of the sign and many where you take a small factor of $120$.
Added: Following up, maybe you should do a binary search. You can find that $p(\frac 12)=30+\frac {97}2-84\lt 0$, so the root is just a bit larger than $\frac 12$. A little estimation will show $p(\frac 34) \gt 0$. Now if you list the factors of $84$ as $1,2,3,4,6,7,12,14,21,28,42,84$ and $120$ as $1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120$ and look for a pair greater than $\frac 12$ and less than $\frac 34$ you don't have many choices.
Alternately you can just use the quadratic formula. Computing $97^2+4 \cdot 84 \cdot 120=49729$ is not easy, but can be done. Doing $97^2=(100-3)^2=10000-600+9$ will speed things. Then taking the square root can be done by estimation. You have $220^2=48400, 230^2=52900$ and the last digit being $9$ says you want $223^2$
Both approaches are some work, but it is supposed to be a challenge.
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.