Factorising: method and worked exercises

In Year 1 of high school, in every kind of liceo, the sequence is almost always this: monomials and polynomials, then special products in November–December, then factorising between December and January, with algebraic fractions after that. The test on factorising falls between January and February, that is thirteen to sixteen weeks from today, 5 October. I am writing this post now, and not in January, for two reasons: anyone who reaches special products in November should already have the map of where it leads, and factorising is where the first-term failing grade is born. In my lessons it is the number one cause of failing grades on the school report at the end of the term, and almost always the problem is not factorising itself, but special products that have not become automatic.

For this reason the order of study from October to January is part of the guide (you will find it at the end). The formulas for special products are explained in special products: formulas and exercises and I won’t repeat them. Here is the method: in what order to try the criteria, when to stop and how to check.

What factorising means

Factorising a polynomial means writing it as a product of polynomials of lower degree, the factors. It is the inverse operation of special products and of multiplying polynomials. The result is correct if, multiplying the factors, you get back to the starting polynomial: this is the final check and you can always do it, in two minutes, even in a test.

The rule to keep in mind is one: factorise as far as you can. A polynomial is fully factorised when no factor can be factorised any further. Stopping halfway costs marks even if the steps you did are right.

The decision tree

Faced with a polynomial, don’t guess: try the criteria always in the same order.

Taking out a common factor. Is there a factor common to all the terms (a number, a letter, or both)? Take it out. It always goes first, because it simplifies everything else.
Difference of squares (or of cubes). Two terms with a minus sign between them, both squares? Then $A^{2} - B^{2} = (A + B) (A - B)$ . For cubes: $A^{3} - B^{3} = (A - B) (A^{2} + A B + B^{2})$ and $A^{3} + B^{3} = (A + B) (A^{2} - A B + B^{2})$ .
Square of a binomial. Three terms, two squares and the double product: $A^{2} + 2 A B + B^{2} = (A + B)^{2}$ or $A^{2} - 2 A B + B^{2} = (A - B)^{2}$ .
Quadratic trinomial $x^{2} + s x + p$ . Look for two numbers that add up to $s$ and multiply to $p$ : $x^{2} + s x + p = (x + m) (x + n)$ with $m + n = s$ and $m \cdot n = p$ .
Factorising by grouping. Four or more terms: group them in pairs, take out a factor from each group and look for a factor common to everything.
Ruffini. For polynomials of degree three or higher when none of the criteria above works. For now I only mention it: I talk about it in polynomial division and Ruffini’s rule, because in Year 1 you usually don’t get that far.

What I see, what I try

What I see	What I try
A factor present in every term	Taking out a common factor
Two terms with a minus between them, two squares	Difference of squares
Two terms with a plus between them	Sum of squares: it cannot be factorised. Sum of cubes: it can
Three terms, two squares and a double product	Square of a binomial
Three terms of the form $x^{2} + s x + p$	Sum and product
Four terms	Factorising by grouping
Degree three or more with no pattern	Ruffini

Taking out a common factor: watch the number

When taking out a common factor you take the greatest common divisor of the coefficients and the common letters with the lowest exponent. Example: in $6 x^{3} y - 12 x^{2} y^{2}$ the common number is $6$ (GCD of $6$ and $12$ ), the letter $x$ appears with minimum exponent $2$ and the $y$ with minimum exponent $1$ . You take out $6 x^{2} y$ :

6 x^{3} y - 12 x^{2} y^{2} = 6 x^{2} y (x - 2 y)

For calculating the GCD with prime factors you can review GCD and LCM with prime factorisation. The check: divide each original term by the factor taken out and verify that you get the terms in brackets.

Factorising by grouping: where signs get lost

Factorising by grouping works when, after grouping, the brackets that remain are equal. If they are equal except for the sign, you have made a sign mistake or you need to take out a minus. Example:

x^{3} - x^{2} - 4 x + 4

I group the first two and the last two terms. From the first two I take out $x^{2}$ : $x^{2} (x - 1)$ . From the last two I take out $- 4$ : $- 4 x + 4 = - 4 (x - 1)$ . The minus sign taken out changes the signs inside the bracket, and this is where many people go wrong, writing $- 4 (x + 1)$ . With the bracket $(x - 1)$ common:

x^{2} (x - 1) - 4 (x - 1) = (x - 1) (x^{2} - 4) = (x - 1) (x - 2) (x + 2)

The last step is a difference of squares: remember to stop only after checking every factor. Quick check: $(x - 1) (x^{2} - 4) = x^{3} - 4 x - x^{2} + 4$ , which rearranged is $x^{3} - x^{2} - 4 x + 4$ .

Typical mistakes

Forgetting the numerical common factor. In $2 x^{3} - 8 x$ the student takes out $x$ and writes $x (2 x^{2} - 8)$ , forgetting that $2$ is common to both terms. The polynomial is not factorised fully.
Getting the signs wrong when grouping. When you take out a negative term, all the signs inside the bracket change.
Factorising a sum of squares. $x^{2} + 9$ cannot be factorised over the real numbers: the formula $A^{2} + B^{2}$ has no factorisation. The same rule holds for $x^{2} + 4$ inside $x^{4} - 16$ , which stays as it is.
Stopping halfway. After the first step a factor may itself be a difference of squares or a trinomial.
Confusing the square of a binomial with the trinomial $x^{2} + s x + p$ . In the first, the two numbers are equal. If $s^{2} = 4 p$ you are looking at a square: $x^{2} + 6 x + 9 = (x + 3)^{2}$ , not $(x + 3) (x + 3)$ written at random, even though the value is the same.
Not doing the final check. Multiplying the factors takes two minutes and catches most sign errors.

Exercise 1: two criteria in sequence

Factorise $2 x^{3} - 8 x$ .

First criterion, taking out a common factor. The common number is $2$ (GCD of $2$ and $8$ ) and the common letter is $x$ . I take out $2 x$ :

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

Inside the bracket there is a difference of squares, because $x^{2} - 4 = x^{2} - 2^{2}$ :

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

The factors $2 x$ , $x - 2$ , $x + 2$ cannot be factorised further. Check: $(x - 2) (x + 2) = x^{2} - 4$ , and $2 x (x^{2} - 4) = 2 x^{3} - 8 x$ . Correct.

Exercise 2: the trinomial with sum and product

a) Factorise $x^{2} - 5 x + 6$ .

I look for two numbers with sum $- 5$ and product $+ 6$ . The product is positive, so the numbers have the same sign, and since the sum is negative they are both negative. The pairs of divisors of $6$ are $(1, 6)$ and $(2, 3)$ : with a minus sign, the sum $- 5$ is given by $- 2$ and $- 3$ . So

x^{2} - 5 x + 6 = (x - 2) (x - 3)

Check: $(x - 2) (x - 3) = x^{2} - 3 x - 2 x + 6 = x^{2} - 5 x + 6$ .

b) Factorise $x^{2} + 2 x - 15$ .

I look for two numbers with sum $+ 2$ and product $- 15$ . The product is negative, so the numbers have opposite signs. The pairs of divisors of $15$ are $(1, 15)$ and $(3, 5)$ . With opposite signs and sum $+ 2$ , $5$ and $- 3$ work ( $5 - 3 = 2$ , $5 \cdot (- 3) = - 15$ ). So

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

Check: $(x + 5) (x - 3) = x^{2} - 3 x + 5 x - 15 = x^{2} + 2 x - 15$ . Correct.

Practical rule: if the product is negative, the number with the larger absolute value takes the sign of the sum.

To get comfortable with these patterns on more cases you can use the Year 1 maths test on factorising and algebraic fractions: the correct answer is free, the guided solution is unlocked with your first lesson.

Exercise 3: factorising by grouping with letters

Factorise $a x + a y + b x + b y$ .

Four terms: factorising by grouping. I group the first two and the last two:

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

Now the two brackets are equal, $(x + y)$ is a factor common to everything:

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

Check: $(x + y) (a + b) = a x + b x + a y + b y$ , which matches the starting polynomial apart from the order of the terms.

Exercise 4: a case with several steps

Factorise $3 x^{3} - 12 x^{2} + 12 x$ .

Step one, taking out a common factor: the common number is $3$ and the common letter is $x$ , so I take out $3 x$ :

3 x^{3} - 12 x^{2} + 12 x = 3 x (x^{2} - 4 x + 4)

Step two, I look at the trinomial in brackets: $x^{2}$ and $4 = 2^{2}$ are squares and the middle term is $- 2 \cdot x \cdot 2 = - 4 x$ , so it is a square of a binomial:

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

Final result:

3 x^{3} - 12 x^{2} + 12 x = 3 x (x - 2)^{2}

Check: $(x - 2)^{2} = x^{2} - 4 x + 4$ , and multiplying by $3 x$ I get $3 x^{3} - 12 x^{2} + 12 x$ .

A second example with two differences of squares in a row: $x^{4} - 16$ . First $x^{4} = (x^{2})^{2}$ and $16 = 4^{2}$ , so $(x^{2} - 4) (x^{2} + 4)$ . Then $x^{2} - 4 = (x - 2) (x + 2)$ , while $x^{2} + 4$ is a sum of squares and stops here:

x^{4} - 16 = (x - 2) (x + 2) (x^{2} + 4)

How to use the months from October to January

October, with monomials. Train yourself to recognise squares and double products: if you see $9 a^{2}$ you immediately think $(3 a)^{2}$ . It seems a detail, and it is the part that saves the most time in January.

November–December, with special products. Make the formulas for the square of a binomial, difference of squares and cube of a binomial automatic, in both directions: knowing how to expand $(x + 3)^{2}$ and recognising $x^{2} + 6 x + 9$ are the same skill read from two sides. Redo the exercises on special products until the signs come out without thinking.

December–January, with factorising. Write the tree at the top of the page and apply it in order, even when you already “see” the solution: it is the safest way not to skip a common factor. Then do five mixed-type factorisations every day, with a final check.

To find out what awaits you in class in the following months, in the first maths test you will find how the first tests are set up, while the overview of the year is in the Year 1 syllabus and formula sheet. Factorising comes back in Year 2 with algebraic fractions: if you want to see where it leads, there is factorising and algebraic fractions for the human-sciences high school. The whole subject is on the maths page.

Before January

The practice test on factorising is available in season, from October to May: Year 1 maths test on factorising and algebraic fractions. My advice is to try it once when you reach special products, to see what you are missing, and again two weeks before the test.

If special products are not yet automatic and you don’t want to reach January with the failing grade already written, write to me through the contact page: the free 30-minute introductory call is where I tell you which type of factorising to start from.

Factorising polynomials: method and worked exercises