Polynomial division and Ruffini's rule in Year 3

At a liceo linguistico (the Italian high school focused on modern languages), Year 3 opens with polynomial division, Ruffini’s rule and factorisation. The first test arrives in five to eight weeks, so between the end of October and the middle of November. With 2 hours of maths a week, and eleven hours of languages taking up the rest of the timetable, this block goes by quickly and does not come back: if you let it slide now, in November you will meet it again in a test and nowhere else.

What you can already do, and what is missing

If you reached Year 3 having already seen common factors, the difference of squares and the special trinomial, I will not repeat that part here: you will find it already explained, with exercises, in factorisation and algebraic fractions in Year 2. This post starts where that one stops: what to do when none of the classic techniques works, that is, when the polynomial has no convenient common factor, is not a difference of squares, and is not a trinomial whose two terms fall out at once. In that case the tool is polynomial division, and in particular its fast version: Ruffini’s rule.

Polynomial long division

Dividing one polynomial by another works like dividing numbers: you look for a quotient $Q(x)$ and a remainder $R(x)$ such that

with the degree of $R(x)$ lower than the degree of $B(x)$ (or a zero remainder). The procedure in columns:

1. Order both polynomials by decreasing degree.
2. Put a zero in place of every missing term. If $A(x)=x^3-5x+2$ is missing the second-degree term, you write it as $x^3+0x^2-5x+2$. This is the most frequent mistake in this chapter — skipping a degree shifts all the later calculations and the final remainder is wrong without any single step looking wrong.
3. Divide the first term of $A(x)$ by the first term of $B(x)$: you get the first term of the quotient.
4. Multiply that term by the whole of $B(x)$, change the signs and add to the dividend, as in the division of numbers.
5. Repeat with the partial remainder until its degree is lower than the degree of $B(x)$: what is left is $R(x)$.
6. Always check: recalculate $B(x)\cdot Q(x)+R(x)$ and make sure you get $A(x)$ back.

The remainder theorem

Before dividing in full, there is a shortcut worth knowing: the remainder of the division of $P(x)$ by $(x-a)$ is simply $P(a)$.

This means that to know whether a division by $(x-a)$ will give a zero remainder you do not need to divide: you just substitute $x=a$ in the polynomial and calculate. If the result is zero, the division is exact; if it is not, that number is the remainder, with no further calculation.

Ruffini’s theorem and Ruffini’s rule

Ruffini’s theorem (what English-language textbooks call the factor theorem) says that a polynomial $P(x)$ is divisible by $(x-a)$ if and only if $P(a)=0$, that is, if and only if $a$ is a root of the polynomial. It is the natural consequence of the remainder theorem: zero remainder means exact division.

Ruffini’s rule is the practical scheme for carrying out that division, but only when the divisor is of the first degree, in the form $x-a$ (mind the sign: to divide by $x+3$ you use $a=-3$). You build a grid with the coefficients of $P(x)$ in decreasing order (zeros included for the missing terms, exactly as in long division) and proceed like this:

1. Write the coefficients of $P(x)$ in a row.
2. Write $a$ on the left.
3. The first coefficient comes down unchanged.
4. Each following coefficient is obtained by multiplying the last number found by $a$ and adding it to the coefficient above it.
5. The last number in the bottom row is the remainder: if it is zero, the division is exact and the numbers before it are the coefficients of the quotient, whose degree is one lower than that of $P(x)$.

How to look for roots

This is the step that textbooks explain badly and that makes Ruffini usable instead of a blind guess: the integer roots of a polynomial with integer coefficients have to be looked for among the divisors of the constant term, taken with both signs.

If $P(x)=x^3-2x^2-5x+6$, the constant term is $6$: its divisors (with sign) are $\pm 1,\pm 2,\pm 3,\pm 6$. You do not need to try them all at random: you try each one with the remainder theorem, that is, you substitute it into the polynomial, until one makes $P(x)$ zero. The first one that works is the root to use with Ruffini.

Factorising a cubic polynomial with Ruffini

Once you have found a root $a$, Ruffini divides $P(x)$ by $(x-a)$ and returns a quotient of the second degree. At that point you factorise the quotient with the special trinomial (or with the quadratic formula), and the starting polynomial is the product of $(x-a)$ and that factorisation. You can see it in the second exercise below.

What it is for right away

It is not a chapter on its own: in Year 3 at a languages high school it is needed immediately for the algebraic fractions and rational equations of November-December, where the first step is always to factorise numerator and denominator. If a polynomial does not factorise with the techniques seen in Year 2, the only way forward is the one in this post.

Exercise 1 — long division with a non-zero remainder

Divide $A(x)=2x^3-3x^2+x-4$ by $B(x)=x-2$.

$A(x)$ is already ordered and complete, with no missing terms.

First step. $2x^3:x=2x^2$. I multiply $2x^2\cdot (x-2)=2x^3-4x^2$, change the signs and add:

Partial remainder: $x^2+x-4$ (I bring down the following terms).

Second step. $x^2:x=x$. I multiply $x\cdot (x-2)=x^2-2x$, change the signs and add:

Partial remainder: $3x-4$.

Third step. $3x:x=3$. I multiply $3\cdot (x-2)=3x-6$, change the signs and add:

The degree of $2$ (zero) is lower than the degree of $B(x)$ (one): I stop. Quotient $Q(x)=2x^2+x+3$, remainder $R=2$.

Check:

I expand $(x-2)(2x^2+x+3)=2x^3+x^2+3x-4x^2-2x-6=2x^3-3x^2+x-6$, and adding the remainder $2$ I get exactly $2x^3-3x^2+x-4=A(x)$. Checked.

If you want to test for yourself whether you have understood the method, the practice test on factorisation has more exercises of this kind: the correct answer is free, the guided solution unlocks with the first lesson. It is filed under Year 1 because at the science high school this topic is seen at the end of the first year, but the content is the same: at a languages high school it comes in Year 3, you are not getting the wrong quiz.

Exercise 2 — factorising a cubic polynomial with Ruffini

Factorise $P(x)=x^3+2x^2-5x-6$.

Step 1 — I look for the root among the divisors of the constant term. The constant term is $-6$: the divisors with sign are $\pm 1,\pm 2,\pm 3,\pm 6$.

I try $x=1$: $P(1)=1+2-5-6=-8$. It is not zero.

I try $x=-1$: $P(-1)=-1+2+5-6=0$. It is a root.

Step 2 — I divide with Ruffini’s rule by $(x-(-1))$, so $a=-1$.

Coefficients of $P(x)$ in order: $1,2,-5,-6$.

Going through the calculations: the first coefficient comes down unchanged, $1$. Then $1\cdot (-1)=-1$, added to $2$ gives $1$. Then $1\cdot (-1)=-1$, added to $-5$ gives $-6$. Then $-6\cdot (-1)=6$, added to $-6$ gives $0$: the remainder is zero, confirming that $x=-1$ is a root.

The quotient has coefficients $1,1,-6$, that is, $Q(x)=x^2+x-6$.

Step 3 — I factorise the second-degree quotient with the special trinomial. I look for two numbers $p,q$ with $p+q=1$ and $p\cdot q=-6$: they are $3$ and $-2$. So $x^2+x-6=(x+3)(x-2)$.

Step 4 — final factorisation:

Check by multiplying back. $(x+1)(x+3)=x^2+4x+3$. Then $(x^2+4x+3)(x-2)=x^3+4x^2+3x-2x^2-8x-6=x^3+2x^2-5x-6$: you get exactly the starting polynomial.

If you want to practise on other polynomials of this kind, with automatic marking, there is the practice test on factorisation — same label as before, same content in Year 3 at a languages high school.

For the full picture of what you will face in maths up to the end of a languages high school, functions and limits in Year 5 tells you where this same syllabus gets to two years later. If instead you want to revise what else comes up in the rest of Year 3, there is the Year 3 formula sheet. If before dividing you need to factorise with special products, formulas and exercises are in special products: formulas and worked examples.

For how we work on maths at a languages high school, the reference page is maths.

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