“Mathematics is the cheapest science. Unlike physics or chemistry, it does not require any expensive equipment. All one needs for athematics is a pencil and paper.” - George Polya
By their GCSE or National 5 exams, children in the UK will need to have an understanding of factorising in maths and as maths is a compulsory subject, everyone will have to do it.
For those who dislike or even hate maths, factorising can be a pain, but in this article, we’ll look at some simple ways to do it.
What Is Factorisation?
Factorisation is the process of transforming two or more factors, which can make expressions easier to work out. If doing this with mental arithmetic, you’ll want to group the common factors together and then add up the others.
Let’s look at 200 x 25 + 425 x 25.
We can simplify this as follows:
200 x 25 + 425 x 25
= 25 x (200+425)
= 25 x 625
In this case, 25 is the highest common factor. You just have to add up the remaining terms and multiply them by 25 to solve.
Let’s imagine we’d like to calculate the difference between the areas of two circles. This means you’ll need to calculate the area of both circles and subtract one from the other. A = (π x R²) - (π x R²) = (π x 25²) - (π x 15²).
(π x 25²) - (π x 15²) results in you having to calculate two squares and then multiply them both by pi (roughly 3.14) before you can subtract one area from the other.
This means: A = π x (25² - 15²)
Fortunately, we know that the difference of two squares is:
a² - b² = (a - b) (a + b)
A = π (25 - 15) x (25 + 15),
= π x 10x40
= π x 400 or 400π.
If we take π as 3.14, the difference between these two circles is 1,256cm².
Factorising algebraic expressions into the products of factors is very useful in these kinds of formulae.
Factorising Literal Equations
When it comes to solving equations, you need to know which algebraic rules to apply.
To factorise an expression into various products, you need to find a common factor so we’re going to look for a term that will allow us to multiply the first by the second. 4x+20 is equal to 2(2x+10).
Factorising allows you to break down the equation.
There are two ways to factorise:
- Special products
If you want to know the result of f(x) = 0, we know that the product must be zero if one of the factors is 0.
You can rewrite f(x) = 0 as y(x) x g(x) = 0 and then you can find a solution for y(x) = 0 or g(x) = 0.
Let’s look at another example.
Let’s imagine in your end of year test you have to solve the following equation: 4x² = 64. This isn’t the easiest sum to do in your head, but you could replace x by 1, 2, 3, etc. until you find the answer for 4x² = 64.
If you use a common product, it’ll be much easier to solve.
4x²=64 is the same as 4x² - 64 = 0.
Thus, f(x) is 4x² - 64. If f(x) is equal to 0, then 4x² - 64 is also equal to zero.
We can use the difference of two squares.
You can factorise this to (2x-8) (2x+8) = 0.
This then becomes 2x-8 = 0 or 2x + 8 = 0.
Now, 2x-8 = 0 becomes x = 8/2 and 2x+8 = becomes x = -8/4. The equation has two solutions: -4 and 4.
Remember that Descartes’ rule of signs when solving quadratic equations and linear equations can help. After simplification, the number of sign changes can tell you how many positive, negative, and imaginary zeroes you’ll have.
You also need to pay attention to the brackets when writing out polynomials and then there’s the order of operations that you have to be careful with.
Always check your results at the end as it can be annoying to drop marks in an exam for silly mistakes or using the wrong sign.
Factorising with Several Common Factors
When there’s only one common factor, the solution is fairly simple.
But what happens if the common factor includes two terms?
Here’s a quick exercise:
Find the common factor and factorise the following expression: (2x - 1) (x + 3) - (4x - 5) (x + 3).
The expression takes the form (ax + ...) + (ax + ...).
Here, the common factor is (x+3) with two terms. To factorise, you need to do much the same as with a single term, but you’ll want to use square brackets to isolate the terms.
Here are the results:
The common factor is (x+3)
Use the principle of distributivity and pay attention to the rule of signs.
- A = (2x - 1) (x + 3) - (4x - 5) (x + 3)
- A = (x + 3) [(2x – 1) – (4x – 5)]
- A = (x + 3) (2x – 1 – 4x + 5)
- A = (x + 3) (– 2x + 4)
For A to be equal to 0, (x+3) = 0 or (-2x + 4) = 0.
The two solutions are therefore x = -3 or x =2.
Special products are a particular type of equation.
We can use them to factorise.
There are three ways to do this so you should learn them off by heart:
- (a+b)² = a² + 2ab + b²
- (a-b)² = a² - 2ab + b²
- (a+b) (a-b) = a² - b²
These are particularly useful with quadratic equations, which you’ll start seeing a lot of as you study maths during school.
The sum of two squares is equal to the square of the first plus double the first term multiplied by the second plus the square of the second.
The second states that the difference between the two terms is equal to the difference of the first term squared and double the product of the first and second terms plus the square of the second term. And, finally, the last one states that the product of the sum of the two terms minus their difference is equal to the first term squared minus the second term squared.
How can you factorise a² + 6a + 9?
Answer: a² + 6a + 9 = a² + 2a x 3 + 3², d'où a² + 6a + 9 = (a+3)²
How can you factorise x² - 81?
Let’s look for a value for x where the square is equal to 81; x = 9.
By using the third method above, we see that x² - 81 = (x + 9) (x - 9).
Thanks to factorisation, you can solve this equation with whole numbers, fractions, and square roots.
Now let’s see how things let a little trickier when it comes to factorising a quadratic equation.
How to Factorise a Quadratic Equation
Students tend to learn how to factorise quadratic equations later in their schooling.
This involves rewriting the quadratic equation in the standard form, which means we’ll have a, b, and c, where a ≠ 0 and Δ represents the discriminant of the polynomial equation ax2 + bx + c.
If x1 and x2 are the roots of a quadratic polynomial ax2 + bx + c, they can be factorised as z (x-x1) (x-x2).
If x0 is the only root of the quadratic polynomial ax² + bx + c, then it can be factorised as (x-x0)².
Since the product of a (x-x0)² = a(x-x0)(x-x0), we know that x0 is a double root.
We can deduce the following:
- If ∆ = 0, ax² + bx + c has a double real root x(0) = - (b/2a) and ax² + bx + c = a (x - x0)² and for all real x, ax2 + bx + c = a (x - x0)².
- If ∆ < 0, the polynomial ax² + bx + c cannot be factorised in ℝ.
- If ∆ > 0, ax² + bx + c has two real distinct roots, (x1) = (-b - √∆)/2a and (x2) = (-b + √∆)/2a and for all real x, ax² + bx + c = a (x-x1) (x-x2).
- If c = 0, the factorised form of the expression ax² + bx + c becomes x (ax + b).
If you’re struggling with this, it’d be a good idea to work on factorising outside of class with each method.
To study maths even more, you might also want to learn how to factorise using your calculator.
Factorising can allow you to find solutions more quickly as long as you don’t trip up over all the brackets!
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