You do all your work on lined paper - I don't think there is anything stopping you asking for plain paper (or even graph paper) but your sketches will probably have quite of bit of writing attached/nearby and this is easiest on lined paper - it also means that a graph on it's own won't become detached from the rest.
Graph paper is to be discouraged as STEP wants sketches, not plots.

It certainly isn't obvious why we can eliminate $-4$ straight away. If anyone has a neater argument please let me know!

SPOILERS>>>>

Doing the more obvious eliminations using the hints document leaves four possible values for $k_1$ i.e. $2, 6, 8$ and $-4$.

We have three equations:
$k_1 \times k_2 \times k_3 \times k_4=576= 1 \times 2^6 \times 3^2$ (eqn 1)
$(k_1-1) \times (k_2-1) \times (k_3-1) \times (k_4-1)=175 = 1 \times 5^2 \times 7$ (eqn 2)
$(k_1+1) \times (k_2+1) \times (k_3+1) \times (k_4+1)=1323 = 1 \times 3^3 \times 7^2$ (eqn 3)

If $k_1=-4$ then either one other is negative (so $k_2=-4$ say) or all four are negative. If
all four are negative then they all must equal $-4$, but this contradicts eqn 1.

If two are negative then (using eqn 2) we have (WLOG) $k_3-1=1$ and $k_4-1=7$, so $k_3=2$
and $k_4=8$. This does not satisfy eqn 1 as there are no factors of 3.

So the only possible values of $k_i$ are $2, 6$ and $8$. By equation 1 we must have two $6$'s (so that we have a factor of $3^2$ and the other two must have a factor of $2^4$ between them so one is $2$ and one is $8$.

If I was doing this question I would then want to expand $(x+2)(x+6)(x+6)(x+8)$ to make sure I was correct!