Factorisable quadratics
This will encourage you to think about whether all quadratics can be factorised and to develop a better understanding of the effect that changing the coefficients has on the factorised form.
Problem
Factorisable quadratics
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The quadratic $x^2+4x+3$ factorises as $(x+1)(x+3)$. In both the original quadratic and the factorised form, all of the coefficients are integers.
The quadratic $x^2-4x+3=(x-1)(x-3)$ similarly factorises with all of the coefficients being integers.
How many quadratics of the following forms factorise with integer coefficients? Here, $b$ is allowed to be any integer (positive, negative or zero). For example, in part a, $b$ could be $-7$, since $(x-2)(x-5)=x^2-7x+10$.
a. $x^2+bx+10$b. $x^2+bx+30$c. $x^2+bx-8$d. $x^2+bx-16$e. $2x^2+bx+6$f. $6x^2+bx-20$ -
This time, it is the constant which is allowed to vary.
How many quadratics of the following forms factorise with integer coefficients? Here, $c$ is allowed to be any positive integer.
a. $x^2+6x+c$b. $x^2-10x+c$c. $3x^2+5x+c$d. $10x^2-6x+c$ - What are the answers to question 2 if $c$ is only allowed to be a negative integer?
Generalising
Can you generalise your answers to the above questions?
a. Generalising question 1, if $c$ is a fixed integer, how many quadratics of the form $x^2+bx+c$ factorise with integer coefficients? Here, $b$ is allowed to be any integer.
b. Further generalising question 1, if $a$ and $c$ are fixed integers, with $a$ positive, how many quadratics of the form $ax^2+bx+c$ factorise with integer coefficients? Again, $b$ is allowed to be any integer.
c. How can we generalise question 2 or 3?
This is an Underground Mathematics resource.
Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Getting Started
How many quadratics of the following forms factorise with integer
coefficients? Here, $b$ is allowed to be any integer (positive,
negative or zero).
$$x^2+bx+10$$
If we knew the value of $b$, how would we go about attempting to
factorise the quadratic?
How might this help us to work out which values of $b$ allow us to
factorise the quadratic?
coefficients? Here, $b$ is allowed to be any integer (positive,
negative or zero).
$$x^2+bx+10$$
If we knew the value of $b$, how would we go about attempting to
factorise the quadratic?
How might this help us to work out which values of $b$ allow us to
factorise the quadratic?
This is an Underground Mathematics resource.
Student Solutions
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 $(6,-1),(9,-4),(12,-7),(15,-10)...$.