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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 $(6,-1),(9,-4),(12,-7),(15,-10)...$.
Factorisable Quadratics
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 $(6,-1),(9,-4),(12,-7),(15,-10)...$.
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