Thank you to everyone who submitted solutions to this problem.
Raphael, from Tangarora College in New Zealand, got us started with this solution to the first part of question 1:
1. a. If the quadratic $x^2+bx+10$ factorises as $(x+r)(x+s)$, then $r$ and $s$ must multiply to give $10$. As these must be integers, they must be factors of $10$.
The pairs of factors that multiply to give 10 are: $(1,10)$; $(-1,-10)$; $(2,5)$ and $(-2, -5)$.
The value of $b$ for each of these pairs is: $11$, $-11$, $7$, $-7$. Therefore these are the values of $b$ that can produce integer factorisations.
Here are Alice from Island school's solutions to questions 1 to 3:
1.a) $x^2+bx+10$
$10$ has 4 factor pairs: $(1,10),(-1,-10),(2,5),(-2,-5)$ giving $b=11,-11,7,-7$ respectively. Therefore there are 4 quadratics of this form that have integer factorisations.
1.b) $x^2+bx+30$
$30$ has 8 factor pairs: $(1,30),(-1,-30),(2,15),(-2,-15),(3,10),(-3,-10),(5,6),(-5,-6)$ giving $b=31,-31,17,-17,13,-13,11,-11$ respectively, so there are 8 such quadratics.
1.c) $x^2+bx-10$
$-8$ has 4 factor pairs: $(1,-8),(-1,8),(2,-4),(-2,4)$ giving $b=-7,7,-2,2$ respectively, so there are 4 such quadratics.
1.d) $x^2+bx-16$
$-16$ has 5 factor pairs: $(1,-16),(-1,16),(2,-8),(-2,8),(4,-4)$ giving $b=-15,15,-6,6,0$ respectively, so there are 5 such quadratics.
Aayush from Island School summarised the general result for any fixed integer $c$:
When $c$ is a fixed integer, the number of quadratics of the form $x^2+bx+c$ with integer coefficients is the same as the number of factor pairs of $c$.
Parts e) and f) are slightly different as the the coefficient of $x^2$ is no longer 1, so we have to consider the more general factorisation $(px+q)(rx+s)$ where $p,q,r,s$ are integers. Expanding this out to $prx^2+(ps+qr)x+qs$ we see that this factorisation is possible when $b$ is the sum of two integers whose product is the product of $a$ and $c$, so now we need to find
the factor pairs of $a c$ not just $c$.
1.e) $2x^2+bx+6$
$12$ has $6$ factor pairs: $(1,12),(-1,-12),(2,6),(-2,-6),(3,4),(-3,-4)$ giving $b=13,-13,8,-8,7,-7$ respectively, so there are 6 such quadratics.
1.f) $6x^2+bx-20$
$-120$ has $16$ factor pairs: $(-1,120),(1,-120),(-2,60),(2,-60),(-3,40),(3,-40),(-4,30),(4,-30),(-5,24),(5,-24),(-6,20),(6,-20),(-15,8),(15,-8),(-10,12),(10,-12)$ giving $b=119,-119,58,-58,37,-37,26,-26,19,-19,14,-14,7,-7,2,-2$ respectively, so there are 16 such quadratics.
Aayush summarised the general result for any fixed integers $a$ and $c$.
When $a$ and $c$ are fixed integers, the number of quadratics of the form $ax^2+bx+c$ with integer coefficients is the same as the number of factor pairs of $a c$. Alice noted that this is not the same as the number of factor pairs of $a$ multiplied by the number of factor pairs of $c$ as that would count repeated quadratics.
2. This time it is the constant that's allowed to vary
2.a) $x^2+6x+c$
In order to get the value of $6x$, when it is factorized in the form of $(x+r)(x+s)$, $r+s$ must equal to $6$.
For $c$ to be positive, both of the addition pairs have to be either positive or negative.
The addition pairs for $6$ are: $(1,5),(2,4),(3,3)$ which give $c=5,8,9$ respectively and so there are 3 such quadratics.
2.b) $x^2-10x+c$
The addition pairs of $-10$ (where both values must be negative) are $(-1,-9),(-8,-2),(-7,-3),(-6,-4),(-5,-5)$ giving $c=9,16,21,24,25$ so there are 5 such quadratics.
2.c) $3x^2+5x+c$
The addition pairs of $5$ are $(1,4),(2,3)$. However to factorize in the form of $(3x+y)(x+z)$, one of the pair must have a factor of 3. In this case only $(2,3)$ has a factor of $3$ so there is 1 such quadratic (and this quadratic has $c=2$).
2.d) $10x^2-6x+c$
The addition pairs for $-6$ are $(-1,-5), (-2,-4),(-3,-3)$. None of these addition pairs are multiples of 10 (when the two terms in the pair are multiplied together) and so there are no quadratics of the form $10x^2-6x+c$ (with $c$ positive) that factorise into integer factors.
3. What about if $c$ is allowed to be a negative integer?
In this case, there would be infinitely many answers to question 2.
For example in 2 c) $3x^2+5x+c$:
Possible addition pairs for $5$ where $c$ is negative are $(6,-1),(9,-4),(12,-7),(15,-10)...$.