Multivariate Factorisation using the Factor Theorem

What you will get from this section. You will learn how to factorise expressions involving two or more variables, such as $y^3+xy^2-2x^3$, by treating one variable as the unknown and the others as constants. We can then apply the familiar factor theorem to identify factors that might otherwise be difficult to spot.

TMUA Relevance Score: 6/10

The factor theorem

You already know the factor theorem for a polynomial in a single variable. If $f(x)$ is a polynomial, then

$f(a)=0$ if and only if $(x-a)$ is a factor of $f(x)$.

It turns out that exactly the same idea works when the “constants” are themselves expressions involving another variable — provided we treat that variable as fixed while considering the polynomial in $x$. And really, why wouldn't it?

A motivating example

Consider $x^3+x^2-2$. Notice that $x=1$ is a root, so $(x-1)$ is a factor.

Wait, I meant consider $x^3+kx^2-2k^3$. Notice that $x=k$ is a root, so $(x-k)$ is a factor.

Oh, actually, I mean $x^3+yx^2-2y^3$. Notice that $x=y$ makes the expression zero, so $(x-y)$ is a factor.

Therefore, we may divide by $(x-y)$ to obtain $x^3+yx^2-2y^3=(x-y)(x^2+2xy+2y^2)$.

Multivariate factor theorem

Joking aside, I hope this example demonstrates the key idea. I will refer to this natural extension as the multivariate factor theorem. We can write it more formally as:

Given a polynomial $f(x,y)$ and a polynomial $h(y)$,

$f(h(y),y)=0$ for all $y$

if and only if

$(x-h(y))$ is a factor of $f(x,y)$.

If you are slightly lost in the rigorous notation, it is basically saying this: if you have an expression involving $x$ and $y$, and replacing every $x$ with some expression in $y$, say $h(y)=2y$, makes the whole expression equal to $0$ for all $y$, then $x-h(y)=x-2y$ is a factor of the original expression.

There is, of course, nothing special about treating $x$ as the variable. We could just as well consider our example $x^3+yx^2-2y^3$ as a polynomial in $y$. Substituting $y=x$ makes the expression zero, so $(y-x)$ is a factor.

Why is it useful and how do you apply it?

In some TMUA questions, there are critical steps where you are expected to recognise such a factorisation by inspection in order to continue making progress. A good example is 2017 Paper 1 Question 20, where you eventually need to factorise $a^2-2ad-3d^2$ as $(a+d)(a-3d)$, which is not immediately obvious to many students.

The way to apply the theorem is simple. Treat $a$ as the variable and ask: can you spot a simple expression in $d$ that makes the whole expression zero? If so, you can immediately deduce a factor. In this example, setting $a=-d$ makes the expression zero, so $(a+d)$ must be a factor.

The biggest takeaway I hope you get from this topic is that whenever you see a polynomial involving two variables:

Look for a value of one variable, expressed in terms of the other, that makes the whole expression zero. That gives you a factor.

Type 1 examples: Two-variable expressions where every term has the same total degree

There is a particularly useful trick when every term has the same total degree. For example, in $x^3+yx^2-2y^3$, every term has total degree $3$. Such expressions are sometimes called homogeneous polynomials.

The standard approach to multivariate factorisation is to treat one of $x$ or $y$ as the variable, try to spot a root in terms of the other variable, and hence obtain a factor. This is still my recommended approach.

However, when every term has the same total degree, there is a variation of this method which is a little more mechanical, but makes finding possible roots much more tractable. In our example, we can use the substitution $x=ty$. This turns the two-variable expression into an ordinary polynomial in $t$, multiplied by a power of $y$. We can then use the usual factor theorem to find roots $t=c$, which correspond to factors of the form $(x-cy)$.

Example 1

Consider again $x^3+yx^2-2y^3$. Put $x=ty$. We obtain $(ty)^3+y(ty)^2-2y^3=y^3(t^3+t^2-2)$.

Now we only need to consider $t^3+t^2-2$. Since $t=1$ makes this zero, this corresponds to $x=y$, so $(x-y)$ is a factor. Dividing gives $x^3+yx^2-2y^3=(x-y)(x^2+2xy+2y^2)$.

Example 2

Factorise $a^3+8b^3$.

Every term has total degree $3$, so put $a=tb$. Then $a^3+8b^3=b^3(t^3+8)$. Since $t=-2$ makes $t^3+8=0$, this corresponds to $a=-2b$, so $(a+2b)$ is a factor. Dividing gives $a^3+8b^3=(a+2b)(a^2-2ab+4b^2)$.

When every term has the same total degree, try a substitution such as $x=ty$. It often turns the problem into an ordinary polynomial in $t$, making the possible factors much easier to spot.

Type 2 examples: Two-variable expressions where the terms do not all have the same total degree

When the terms have different total degrees, we go back to the main idea: treat one variable as the variable and look for an expression in the other that makes the whole polynomial zero.

When looking for a root, there is often a particular part of the expression that gives you a clue about what to try.

Example 3

Factorise $y^2-x^2+3y+3x$.

Treat this as a polynomial in $y$. The particular giveaway here is $3y+3x$, which becomes zero when $y=-x$, so this is a natural value to try. Substituting $y=-x$ gives $x^2-x^2-3x+3x=0$, so $(y+x)$ is a factor. Dividing gives $y^2-x^2+3y+3x=(y+x)(y-x+3)$.

Example 4

Factorise $a^2+ab-2b^2+a-b$.

Treat $a$ as the variable. The particular giveaway here is $a-b$, which becomes zero when $a=b$, so this is a natural value to try. Substituting $a=b$ gives $b^2+b^2-2b^2+b-b=0$, so $(a-b)$ is a factor. Dividing gives $a^2+ab-2b^2+a-b=(a-b)(a+2b+1)$.

Example 5

Factorise $x^2+xy-4x-2y+4$.

This time, notice that the root does not necessarily have to involve the other variable. Treating $x$ as the variable, try the simple numerical value $x=2$. Substituting gives $4+2y-8-2y+4=0$ for all $y$, so $(x-2)$ is a factor. Dividing gives $x^2+xy-4x-2y+4=(x-2)(x+y-2)$.

This is still exactly the same theorem: the constant $2$ is simply a constant polynomial in $y$. The root does not have to involve the other variable. An ordinary numerical root is just a special case.

Type 3 examples: Three-variable expressions

Nothing particularly dramatic happens when a third variable appears. We simply treat two variables as fixed and regard the expression as a polynomial in the remaining variable.

The idea is exactly the same as before: choose one variable and look for a simple expression involving the other two that makes the whole polynomial zero. If you find one, it immediately gives you a factor.

Example 6

Factorise $x^2+xy+xz-2y^2-yz$.

Treat $x$ as the variable. Trying $x=y$ gives $y^2+y^2+yz-2y^2-yz=0$, so $(x-y)$ is a factor. Dividing gives $x^2+xy+xz-2y^2-yz=(x-y)(x+2y+z)$.

Admittedly, there is not much here to tell you what to try. However, if a factorisation of this kind appears in a TMUA question, the intended root is almost certainly going to be something reasonably simple to spot, such as $x=y$, $x=-z$ — rather than some horrible expression you would never realistically guess.

Why insist that $h(y)$ is a polynomial?

I restricted $h(y)$ to be a polynomial partly because we are interested in polynomial factorisation, and partly in case some smart student writes to me with a comical example claiming the maths is wrong!

For instance, if we allow $h(y)=|y|$, then substituting $x=|y|$ into $x^2-y^2$ gives $|y|^2-y^2=0$ for every real $y$. So we could write $x^2-y^2=(x-|y|)(x+|y|)$, which is perfectly true — just a rather silly way to factorise a polynomial! Since $|y|$ is not a polynomial, neither factor is a polynomial either.

So requiring $h(y)$ to be a polynomial keeps us firmly in the world of polynomial factorisation — and hopefully saves me from a few emails beginning, “Hi Mr Joe, but what if...”

Summary

The multivariate factor theorem is simply the familiar factor theorem applied while treating the other variables as constants.

Look for a value of one variable, expressed in terms of the others, that makes the whole expression zero. That gives you a factor.

For homogeneous polynomials, where every term has the same total degree, a substitution such as $x=ty$ can make possible factors easier to spot. Otherwise, direct inspection is usually the best approach.

Remember that the root may involve another variable, several other variables, or simply be an ordinary numerical value.

Bonus: Where the factor theorem actually comes from

I could quite happily finish the article here. However, there is one small piece of mathematical housekeeping that I have been mildly obsessed with since the 2017 A-level syllabus came out. 😂 The factor theorem is currently stated as a result to be used, but mathematically it is really just a particularly important special case of the remainder theorem, which used to sit alongside it in the A-level syllabus.

Now, I fully accept that this probably bothers me far more than it reasonably should! But I have an unfortunate tendency to want mathematical results to come with their family tree attached, so please indulge me for one bonus section. The remainder theorem is simple, its proof is short and beautiful, and once we have it, the factor theorem follows as a mere consequence.

The remainder theorem

Let $f(x)$ be a polynomial. The remainder theorem says:

The remainder when $f(x)$ is divided by $(x-a)$ is $f(a)$.

Proof

Let $q(x)$ and $r$ be the quotient and remainder when $f(x)$ is divided by $(x-a)$. Since $(x-a)$ has degree $1$, the remainder $r$ must be a constant. Therefore, $f(x)=(x-a)q(x)+r$, which is true for all real values of $x$.

Now let $x=a$. We immediately obtain $f(a)=r$, and the proof is complete.

The factor theorem

So according to the remainder theorem, for example:

• $f(b)=1$ means the remainder when $f(x)$ is divided by $(x-b)$ is $1$.
• $f(c)=2$ means the remainder when $f(x)$ is divided by $(x-c)$ is $2$.

Similarly, $f(a)=0$ means the remainder when $f(x)$ is divided by $(x-a)$ is $0$. But this is equivalent to saying there is no remainder, which means $(x-a)$ is a factor of $f(x)$.

This is the factor theorem: simply the special case of the remainder theorem where the remainder is $0$!

And our multivariate version follows in exactly the same way

There is also a rather satisfying connection back to the main topic of this article.

Treat $f(x,y)$ as a polynomial in $x$, and suppose we divide it by $(x-h(y))$. We may write $f(x,y)=(x-h(y))q(x,y)+r(y)$, where the remainder $r(y)$ does not involve $x$.

Now substitute $x=h(y)$. The first term disappears, leaving $f(h(y),y)=r(y)$.

So if $f(h(y),y)=0$ for all $y$, then the remainder is identically zero, and therefore $(x-h(y))$ is a factor of $f(x,y)$. Conversely, if $(x-h(y))$ is a factor, then substituting $x=h(y)$ clearly makes the expression zero.

So even the multivariate factor theorem we have been using throughout this article is really just the same remainder theorem wearing slightly fancier clothes.

I feel better now. We may finish. 😄

PDF version — Questions · Solutions

PDF version — Questions · Solutions