If it can be taught by equating coefficients, then it should be

This post was prompted by finding the following part of a question in one of the specimen papers for the new A-level maths specifications

Write $1.8+0.4d-0.002d^2$ in the form

$$A-B(d-C)^2$$

where $A$, $B$ and $C$ are constants to be found.

Heaven help any student who's learnt completing the square as something about halving the $x$ coefficient.

Completing the square

Write $3x^2-6x+11$ in the form $m(x+p)^2+q$

The most commonly taught routine for completing the square is

If the coefficient of $x^2$ isn't $1$ (or worse 'if there's a number on the front') then take it outside a bracket not including the $11$ because reasons (unless there's something on the other side in which case divide everything by $3$?) $$3(x^2-2x)+11$$ Ignore the $3$ for a moment and do completing the square on what's in the bracket, which means you half the $x$ coefficient (or worse 'the number in the middle') and put it in a bracket to be squared. Multiply that out and adjust $$\begin{aligned}(x-1)^2&=x^2-2x+1\\x^2-2x&=(x-1)^2-1\end{aligned}$$ Then deal with the $3$ $$\begin{aligned}3(x^2-2x)&=3((x-1)^2-1)\\ &=3(x-1)^2-3\end{aligned}$$ But there was that $11$ there all along so $$3(x-1)^2-3+11=3(x-1)^2+8$$

I am not exaggerating. And students' rehearsal of this script is usually even more fractured. Their teacher may feel a twinge of guilt and give some small nod to algebraic continuity, so the process might look like this

$$\begin{aligned} 3x^2-6x+11&=3(x^2-2x)+11\\ &=3((x-1)^2-1)+11\\ &=3(x-1)^2-3+11\\ &=3(x-1)^2+8 \end{aligned}$$

but even then, what have students learnt? A series of steps, one of which seems so arbitrary as to be close to magical, for solving exactly one kind of problem. And how much have they understood?

By contrast, the idea of equating coefficients is just an extension of collecting like terms, and depends only on multiplying out and substitution

$$\begin{align*} &&3x^2-6x+11&=m(x+p)^2+q\\ &&&=mx^2+2mpx+mp^2+q\\[1em] &x^2:&3&=m\\[1em] &x:&-6&=2mp\\ &&&=6p\\ &&-1&=p\\[1em] &1:&11&=mp^2+q\\ &&&=3(-1)^2+q\\ &&&=3+q\\ &&8&=q\\[1em] &&3x^2-6x+11&=3(x-1)^2+8 \end{align*}$$

The only disadvantage I can see is that this looks a bit longer. But what students gain is a powerful but transparent technique that can be applied in plenty of other contexts, like...

Polynomial division

Divide $2x^3-3x^2+5x-7$ by $x+3$

Search 'polynomial division' and I guarantee you won't get a result in the first half a dozen pages that doesn't describe the algorithm that looks like long division of numbers. And just as the long division routine has nothing to do with understanding numbers, students can rehearse polynomial long division all day long without understanding anything about polynomials. Or division. In fact, at least the long division algorithm for numbers is accessible to understanding given a thorough grounding in place value and a bit of time.

It is a rare teacher, let alone student, who could fluently explain why the polynomial long division algorithm works. Try it. Find a colleague and ask them to explain why you only divide by the leading term and then multiply the whole thing. And then you write that down and subtract it from what's above it. No, not that. You bring that down afterwards.

I'm not saying it cannot be explained. Of course it can. It's reliable and can be executed fairly quickly with practice. But, just as with completing the square, it brings nothing with it in terms of understanding algebra, or wider applicability.

Given some appreciation of how the properties of quotient and remainder in natural numbers translate to polynomials, which in turn is a matter of thinking about and experimenting with indices, this division can be carried out just as easily by writing it as a multiplication and equating coefficients.

Cubic divided by linear will give quadratic with at most a constant remainder: $$\begin{align*} &&2x^3-3x^2+5x-7&=(x+3)(ax^2+bx+c)+d\\ &&&=ax^3+bx^2+cx+d\\ &&&\qquad +3ax^2+3bx+c\\ &x^3:& 2&=a\\ &x^2:& -3&=b+3a=b+6\implies b=-9\\ &x:& 5&=c+3b=c-27\implies c=32\\ &1:& -7&=d+c=d+32\implies d=-39\\ &&2x^3-3x^2+5x-7&=(x+3)(2x^2-9x+32)-39 \end{align*}$$

Partial fractions

What a waste of opportunities it's been, then, if students reach the second year of their course, and partial fraction decomposition, before they finally get to learn this flexible and powerful technique.

Express $\dfrac{3(x+1)}{(x+2)(x-1)}$ in partial fractions

$$\begin{aligned} \dfrac{3(x+1)}{(x+2)(x-1)}&=\dfrac{A}{x+2}+\dfrac{B}{x-1}\\ 3x+3&=A(x-1)+B(x+2)\\ &=Ax-A+Bx+2B\\ x:3&=A+B\\ 1:3&=-A+2B\\ 6&=3B\implies B=2\text{ and }A=1\\ \dfrac{3(x+1)}{(x+2)(x-1)}&=\dfrac{1}{x+2}+\dfrac{2}{x-1} \end{aligned}$$

In anticipation of claims that it's easier to find $A$ and $B$ via well-chosen substitutions: how do you explain to your more astute students why it's okay to substitute in $x$ values that are asymptotes for this function, and likely prohibited by the question text?