Factorising: Year 2 human sciences

End of September 2026: at the human-sciences high school (liceo delle scienze umane) the review of equations and radicals is wrapping up, and between the end of September and mid-October the new chapter starts, factorising and algebraic fractions. The test comes six to nine weeks from now, so between early November and early December. It is not yet time for revision: it is the time when it is worth arriving with a method, before the teacher speeds up.

At the human-sciences high school maths has 3 hours a week in Year 2, dropping to 2 in the last three years: it is the subject with the most structural failing grades on this track, and not through the students’ fault. With so little time in class, guided practice is scarce, and anyone who doesn’t get home with a clear method builds up a delay that is hard to recover by November.

Why we factorise (and why it is not an exercise for its own sake)

Factorising is not an isolated chapter: it is the tool you need for everything else this year. Two examples are enough to see why.

Simplifying an algebraic fraction. $\frac{x ^{2} - 9}{x ^{2} - 3 x}$ cannot be simplified as long as numerator and denominator remain sums. Once factorised, numerator and denominator become products, and equal factors cancel — we will see shortly how, and this is where most mistakes happen.

Solving an equation with the zero-product property. $x^{2} - 5 x + 6 = 0$ is not solved “by eye”: you factorise the left-hand side into $(x - 2) (x - 3) = 0$ , and from there the equation splits into two first-degree equations, $x - 2 = 0$ or $x - 3 = 0$ . Without factorising, this method does not exist.

If you don’t factorise well, you can’t simplify algebraic fractions or solve fractional equations: they are the rest of this year’s syllabus.

The decision tree: what to look at, in order

The reason factorising seems hard is that the book presents five or six techniques as if they were equivalent alternatives to be recognised by intuition. They are not: there is a fixed order to follow every time, and following it removes almost all the uncertainty.

Taking out the total common factor. Always the first step, on every polynomial, no exceptions. If there is a factor common to all the terms, you take it out before looking at any other technique.
Count the terms that remain.
Two terms → difference of squares $a^{2} - b^{2} = (a - b) (a + b)$ , or sum/difference of cubes $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$ and $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$ .
Three terms → square of a binomial $a^{2} + 2 ab + b^{2} = (a + b)^{2}$ if the pattern fits, otherwise the special trinomial (sum and product).
Four terms → factorising by grouping, in pairs.
If none of the above works → try Ruffini.

This is the only order that makes sense: following it saves you from trying techniques at random, and above all from the most common mistake of all, stopping at the first common factor when in fact the factor you put in brackets can still be factorised.

The special trinomial: where mistakes really happen

With three terms of the form $x^{2} + b x + c$ , the special trinomial looks for two numbers $p$ and $q$ such that:

p + q = b, p \cdot q = c

Then $x^{2} + b x + c = (x + p) (x + q)$ . The problem is not the rule, it is finding the two numbers when the signs differ. Take $x^{2} - 2 x - 15$ : here $c = - 15$ is negative, so $p$ and $q$ must have opposite signs. You need two numbers whose product is $- 15$ (so one positive and one negative) and whose sum is $- 2$ . The pairs with product $- 15$ are $(1, - 15)$ , $(3, - 5)$ , $(5, - 3)$ , $(15, - 1)$ : the one that adds up to $- 2$ is $(3, - 5)$ . So $x^{2} - 2 x - 15 = (x + 3) (x - 5)$ .

The practical rule for not getting the signs wrong: if $c$ is positive, $p$ and $q$ have the same sign as $b$ ; if $c$ is negative, they have opposite signs, and the sign of the number with the larger absolute value follows $b$ .

Factorising by grouping: the warning about the sign

With four terms you group in pairs and take out a factor from each, hoping the same factor is left in brackets:

a x + a y - b x - b y = a (x + y) - b (x + y) = (x + y) (a - b)

The trap is when the factor to take out of the second pair is negative. If you take out a negative number, you change all the signs inside the bracket:

a x - a y - b x + b y = a (x - y) - b (x - y) = (x - y) (a - b)

Here $- b x + b y$ becomes $- b (x - y)$ , not $- b (x + y)$ : this is the point where half the class loses the sign and the final common factor doesn’t work out.

Ruffini, the essential version

If after trying everything else a polynomial can’t be factorised, what is left is division with Ruffini’s rule. In short: you look among the divisors of the constant term for a number $x_{0}$ that makes the polynomial vanish (that is, $P (x_{0}) = 0$ ), then divide the polynomial by $(x - x_{0})$ with the Ruffini scheme, obtaining a polynomial of lower degree to factorise further. It is a solid method, but with 3 hours a week it is not the priority: consolidate the first five steps of the tree first, which cover the great majority of the exercises in the test; leave Ruffini for last in your study.

Domain of existence: before simplifying, not after

An algebraic fraction is defined only where the denominator is different from zero. The domain of existence (C.E.) is found by factorising the denominator and requiring each factor to be different from zero — and it must be done before simplifying, not after. It is mistake number one of this chapter: if you simplify first, you lose track of the value that made the original denominator zero, and that value remains excluded from the domain all the same.

Simplifying an algebraic fraction: you cancel factors, never terms

Let’s go back to $\frac{x ^{2} - 9}{x ^{2} - 3 x}$ . First the C.E.: the denominator factorises as $x (x - 3)$ , so we need $x \neq = 0$ and $x \neq = 3$ .

Then we factorise the numerator too, a difference of squares:

\frac{x ^{2} - 9}{x ^{2} - 3 x} = \frac{( x - 3 ) ( x + 3 )}{x ( x - 3 )} = \frac{x + 3}{x} (x \neq = 0, x \neq = 3)

The factor $(x - 3)$ appears both above and below as a factor of the whole numerator and the whole denominator, so it cancels. What you can never do is cancel a term. A numerical counterexample settles the argument: $\frac{2 + 4}{2} = 3$ . If someone “simplified” by cancelling the 2 inside the sum with the 2 in the denominator they would get $1 + 4 = 5$ , which is wrong. A number inside a sum is not a factor of the whole sum, and the same holds with letters: you simplify only when numerator and denominator are written as products.

Adding algebraic fractions

As with numerical fractions, to add two algebraic fractions you need the lowest common denominator, found by factorising each denominator and taking each factor with the highest exponent with which it appears:

\frac{1}{x} + \frac{1}{x - 3} = \frac{( x - 3 ) + x}{x ( x - 3 )} = \frac{2 x - 3}{x ( x - 3 )} (x \neq = 0, x \neq = 3)

Here too the C.E. is written straight away, looking at the two original denominators, not at the final result.

Three worked exercises

1. Multi-step factorising: common factor and difference of squares

2 x^{3} - 8 x

First step of the tree: taking out the total common factor. The common factor between $2 x^{3}$ and $8 x$ is $2 x$ :

2 x^{3} - 8 x = 2 x (x^{2} - 4)

Inside the bracket a binomial remains: it is a difference of squares, $x^{2} - 2^{2}$ , so it factorises further:

2 x (x^{2} - 4) = 2 x (x - 2) (x + 2)

Stopping at the first step, $2 x (x^{2} - 4)$ , is the most common mistake in this type of exercise: after every common factor you must always recheck whether what is left in brackets can be factorised further.

If you want to practise more exercises of this type with automatic marking, here is the practice test on factorising. It is catalogued under Year 1 because at the science high school factorising starts at the end of the first year, but the exercises are the same: at the human-sciences high school the topic arrives in Year 2, the content doesn’t change.

2. Special trinomial with opposite signs

x^{2} - 2 x - 15

Here $b = - 2$ and $c = - 15$ . Since $c$ is negative, $p$ and $q$ must have opposite signs. I look for two numbers with product $- 15$ and sum $- 2$ : trying the pairs, $3 \cdot (- 5) = - 15$ and $3 + (- 5) = - 2$ both work. So:

x^{2} - 2 x - 15 = (x + 3) (x - 5)

Quick check: redoing the product, $(x + 3) (x - 5) = x^{2} - 5 x + 3 x - 15 = x^{2} - 2 x - 15$ . It works.

3. Simplification with the domain stated

\frac{x ^{2} - x - 6}{x ^{2} - 4}

First the denominator, for the C.E.: $x^{2} - 4 = (x - 2) (x + 2)$ , so $x \neq = 2$ and $x \neq = - 2$ .

Then the numerator, a special trinomial with $b = - 1$ , $c = - 6$ : I look for two numbers with product $- 6$ and sum $- 1$ , they are $2$ and $- 3$ :

x^{2} - x - 6 = (x + 2) (x - 3)

I put it together and cancel the factor $(x + 2)$ , common to numerator and denominator:

\frac{( x + 2 ) ( x - 3 )}{( x - 2 ) ( x + 2 )} = \frac{x - 3}{x - 2} (x \neq = 2, x \neq = - 2)

The C.E. remains $x \neq = 2$ and $x \neq = - 2$ even after the simplification: the value $x = - 2$ has disappeared from the reduced fraction, but it was still a prohibition on the original denominator, so it must be written all the same.

Typical mistakes and how to notice them yourself

Typical mistake	How you notice it
Cancelling terms instead of factors	Try with numbers: $\frac{2 + 4}{2} = 3$ , not $1 + 4 = 5$ . If the “cut” changes the numerical result, it wasn’t a factor
Forgetting the domain of existence	Always ask: for which values does the denominator, once factorised, vanish? It must be written before simplifying
Getting signs wrong in the special trinomial	If $c < 0$ the two numbers have opposite signs; if $c > 0$ they have the same sign as $b$
Stopping at the first common factor	After every step, recount the terms left in brackets and recheck whether they factorise further
Factorising $a^{2} + b^{2}$ like a difference of squares	The sum of two squares cannot be factorised over the real numbers: only the difference $a^{2} - b^{2}$ has a formula

How we work on this in lessons

With factorising I always start from the decision tree written on a sheet of paper, not from the book’s list of formulas: first you count how many terms there are, then you choose the technique, not the other way round. It is a method that holds even on exercises you have never seen, while learning to recognise a pattern “by eye” only works on those identical to the ones done in class.

If the previous chapter, radicals and rationalisation, is not yet fully solid, it is worth closing that one before sinking into factorising: they are different chapters but the precision in calculation they need is the same. The next topic of the year, linear systems, will use the substitution between algebraic fractions that you are learning now, so arriving there with this chapter in order saves you time later.

If you want a hand setting up the method before the November test, you will find the full path on the maths tutoring page, or write to me through the contact page: the first introductory call is free and lasts half an hour.

Factorising and algebraic fractions in Year 2