Quotients and remainders

A few months ago, I wrote a post about replacing long division with a coefficient-matching process. That's brilliant for C2, but what happens if you're looking at a C4 question that wants a quotient and a remainder?

Well, it gets a bit more complicated, that's what happens. But it's not that much harder, and certainly no harder that algebraic division.

The way to approach it is to think about what happens when you divide numbers - for example, $18 \div 7$. You get a quotient of 2 and a remainder of 4, because $18 = 2\times 7 + 4$. You could also write $\frac{18}{7} = 2\frac{4}{7}$, which tells you something interesting: when you divide $a$ by $b$, you get the quotient, plus the remainder divided by $b$. In general: $\frac{a}{b} = Q + \frac{R}{b}$, and when you're doing this with algebra, $Q$ and $R$ are polynomials.

A typical question asks you to find the quotient and remainder when you divide $x^4 + 3x^3 + 7x^2 - 10x + 7$ by $x^2 + 3x + 5$.

We can draw some immediate conclusions about the quotient and the remainder: the quotient has to be a quadratic expression, because $x^4 \div x^2 = x^2$. The remainder has to be a linear expression, because if it was a quadratic (or more complicated), we could divide it by the quadratic on the bottom. Them's the rules:

  • The quotient is as complicated as the lead term of the top divided by the lead term of the bottom1
  • The remainder is one order less complicated that what you're dividing by

That means we can say:

$\frac{x^4 + 3x^3 + 7x^2 - 10x + 7}{x^2 + 3x + 5} \equiv (ax^2 + bx + c) + \frac{dx + e}{x^2 + 3x + 5}$.

Let's multiply up the bottom:

${x^4 + 3x^3 + 7x^2 - 10x + 7} \equiv (ax^2 + bx + c)({x^2 + 3x + 5}) + {dx + e}$.

Multiplying out:

x^4 & = & ax^4 \\
3x^3 & = & 3ax^3 & + bx^3 \\
7x^2 & = & 5ax^2 & + 3bx^2 & + cx^2\\
-10x & = & ~ & 5bx & + 3cx & + dx \\
7 & = & ~ & ~ & + 5c & + e \\ \end{array}$

Yikes. Luckily, we can rattle through that quickly:

$a = 1$ from the first row, which means $b=0$ from the second. From the third, $c=2$, and the fourth gives $d = -16$. Lastly, $e = -3$. Boom!

That means the quotient, $ax^2 + bx + c$, is $x^2 + 2$ (the $x$ coefficient, $b$, is zero) and the remainder is $-16x - 3$.

I'm not saying "thou shalt do it this way" - after all, there's more than one way to do it. However, if you're making mistakes with long division, perhaps you'd like to give it a try. Let me know how you get on!


Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. Or, more technically, it's the difference between the orders of the polynomials. []


3 comments on “Quotients and remainders

  • Cav

    Great article thanks. Interestingly, I’ve always preferred long division, but I can see the appeal of equating coefficients, so normally teach both. This year’s cohort of year 12s hated equating coefficients, and all opted for long division as method of choice! It’s bizarre, usually at least one prefers each method!

    • Colin

      It’s almost as if different people have different tastes in maths! Some people, bizarrely, prefer $y=mx+c$… ;o)

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