For the first problem, imagine it in this way

So you can begin by writing

and your table should look like

You are searching for the pair from the table which adds together to make $-2k.$

The other two can be treated similarly.

1. $$(x-k{)}^2$$

2. Our table looks like

   $$\begin{array}{c}-2k^2\\ \begin{array}{cc}\pm 2& \mp k^2\\ \pm 2k& \mp k\end{array}\end{array}$$

   We need the pair which adds to make $-k$ and so we get

   $$(x-2k)(x+k)$$

3. We can consider this as a quadratic in terms of $p$, so we get

   $$\begin{array}{c}-6q^2\\ \begin{array}{cc}\pm 6& \mp q^2\\ \pm 3& \mp 2q^2\\ \pm 2& \mp 3q^2\\ \pm 6q& \mp q\\ \pm 3q& \mp 2q\\ \pm 2q& \mp 3q\end{array}\end{array}$$

   We need the pair which sums to $-5q$, so we get

   $$\frac{(3p-6q)}{3}\frac{(3p+q)}{1}=(p-2q)(3p+q)$$