“The difference between the poet and the mathematician is that the poet tries to get his head into the heavens while the mathematician tries to get the heavens into his head.” - G.K. Chesterton
In the UK, we rank 11th in terms of mathematics performance on the PISA test amongst OCDE subjects as maths is a subject that every student has to take as it’s obligatory throughout their schooling. From reception to year 12, you have to learn about maths and that includes expanding polynomials.
When it comes to maths, especially at GCSE, these types of questions appear regularly and whenever you see a bracket, it's a good idea to know how to multiply out the expressions both inside and outside of it.
In this article, we’ll be looking at distributivity, double distribution, and special products.
Maths is one of the most important classes in school. However, a lot of students don’t like it and algebra, after all, is hardly self-explanatory. However, it’s something you’ll have to study quite a bit of, especially in secondary school.
The distributive property allows us to expand brackets and make things simpler. While a bracket can be useful for showing that an operation needs to be completed before everything outside of it, with multiple brackets, it can be easier to expand the brackets by multiplying out the terms (known as expanding).
You can use this to expand the multiplication of several like terms into a sum such as x(y + z) = xy + xz.
There are two ways to do this:
- Distributive property of multiplication
- Double distribution
Distribution gives us an answer as a sum (or subtraction) of terms. There are many different ways to write out a maths equation and during your time with algebra, you'll see this many times.
Still not clear?
Here's an example of a number being multiplied and how many different ways you can get to the same answer.
- 10 x 25 = 10 x (20 + 5) = 10 x 20 + 10 x 5 = 200 + 50 = 250.
- So 10 x 25 = 10 x (20 + 5): 10 x 25 = 10 x (20 x 5) is effectively the same as 10 x (35 -10)
- 10 x 25 = 10 x (35 - 10) = 10 x 35 - 10 x 10 = 350 - 100 = 250
- 10 x 25 = 10 x (27 - 5 + 3) = 10 x 27 - 10 x 5 + 10 x 3 = 270 - 50 + 30 = 250
The result can allow us to make equations simpler and often easier to do in our heads. It also allows us to get rid of brackets, which is want you need to do if you want to simplify a maths equation.
An equation with a factor multiplied by a group in the form of the sum is the same as the sum as the factor individually multiplied by the members of the group.
This is often shown as a(b + c) = ab + ac.
The same is true of subtraction: a(b - c) = ab - ac.
This technique can be used to help us solve polynomials.
Remember, there's very rarely a single way to write algebraic expressions and when it comes to finding a solution, especially when you see a bracket, the longer or expanded version may be key to answering the question.
In some cases, a mathematical expression is more complicated and includes second-degree polynomials or quadratic polynomials.
So how can you resolve a quadratic polynomial equation with several coefficients?
It’s easier than you might think.
For the function f(x) = (x - 1)(2x + 3), you need to group the terms and transform the products into the sum of two terms.
It works like this: (a + b) x (c + d) = ac + ad + bc + bd.
- (a + b) (c - d) = ac - ad + bc - bd
- (a - b) (c + d) = ac + ad - bc - bd
- (a - b) (c - d) = ac - ad - bc + bd
Don’t forget any of the terms and make sure you remember the rule of signs.
- f(x) = (x -1)(2x + 3)
- = 2x² + 3x - 2x + (-1 x 3)
- = 2x² + 3x - 2x - 3
- = 2x² + 3x - 2x - 3
This gives us f(x) = 2x² + 3x - 2x - 3
Can you do solve the following:
- f(x1) = (x + 3)(2x + 1)
- f(x2) = (5 + x)(3x − 2)
- f(x3) = (6 − 5x)(7 − 4x)
Here are the answers (don’t peek!):
- f(x1) = 2x2 + x + 6x + 3 = 2x² + 7x + 3
- f(x2) = 15x – 10 + 3x2 − 2x = 3x² + 13x - 10
- f(x3) = 20x² + 42 − 24x − 35x = 20x² - 59x + 42.
Distributing these types of equations are good mental gymnastics.
How to Simplify Algebraic Expressions
With the types of equations we saw earlier, we can end up with a lot of addition and subtraction with a common variable (often x).
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This means that you can simplify them. With a common factor, simplifying is quite easy. You just add or subtract the terms with the same factor. You can then write them in increasing or decreasing powers. This will allow you to carry out the operations more simply. Let’s group all the x, x2, and unknowns.
- 2x + 12 + 4x² - 6x + 4x² + 2
- = 8x² - 4x + 14
Better yet, let’s look at the following expression (3x + 1) (2x + 4). With these brackets, we can simplify the whole thing by expanding them.
- = (3x + 1) (2x + 4)
- = 6x² + 12x + 2x + 4,
- = 6x² + 14x + 4.
This is called expansion as you take what’s in the brackets and express them individually, which can make them easier to work with. Rewriting the expression into a simpler form is known as reduction. After you’ve done this a few times, it’ll become second nature to you.
Essentially, you're multiplying out the terms in the brackets and grouping them so that you're left with expressions in the form of addition or subtraction and the common terms grouped. You’ll be able to quickly put equations into forms you can work with.
Algebra will have never seemed so simple.
You’ll want to learn more about special products for some algebraic expressions. This is a particular case when you can expand something with a common factor to get the expressions outside of the brackets by multiplying.
You can expand these cases with polynomial variables.
It can also be used for solving a quadratic equation. Remember that quadratic equations, however, have two answers and, in some cases, have both a positive and negative number as the solution.
In this case, the variables a and b can be whole numbers, real numbers, or complex numbers.
The special products are as follows:
- (a + b)² = a² + 2ab + b²
- (a - b)² = a² - 2ab + b²
- (a + b) (a - b) = a² - b²
The second is a particular version of the first case.
The square of a binomial and the product of the sum and difference of two binomials are special products.
With the equals sign, both sides of the equation are the same. They can be transformed as follows:
For example: f(x) = (2x - 3)² + (x + 5) (3 - x).
We have to start by expanding the special product (2x - 3)².
- (2x - 3)² = (2x)² - 12x + 3²,
- = 4x² - 12x + 9.
- f(x) = 4x² - 12x + 9 + (x + 5) (3 - x)
We can then expand the second term using the distributivity.
- (x + 5) (3 - x) = x (3 - x) + 5 (3 - x) = 3x - x² + 15 - 5x = -x² - 2x + 15
- f(x) = 4x² - 14x + 24 - x² - 2x + 15 = 3x² - 14x + 24
Pay particular attention to the brackets, especially during the first step with the special products and make sure you’re careful when following the rules. Put away your calculator and practise expanding polynomials yourself. You’ll soon get the hang of everything.
If you need more help with expanding brackets or simplifying maths equations, consider getting in touch with a private tutor. No matter what level you are or what you're stuck on, you should be able to find a suitable tutor on Superprof.
Remember that a lot of the tutors offer the lesson for free so try a few out before deciding which will be right for you.
It's more than just about finding the answer to maths questions, you need to understand how you get to the solution, especially if you're studying for your GCSE or A Level maths exam.
Fortunately, a private tutor can help you with revision, provide you with examples to work on, and show you different ways to understand the questions and problems you'll see. After all, when it comes to learning maths, practice makes perfect!
If you can't find any suitable tutors in your local area, you can always learn with an online tutor. While learning hands-on subjects is often better with a face-to-face tutor, online tutors are great for maths and other academic subjects.
It's always a good idea to outline your requirements before you start looking for tutors. On the Superprof website, you can see what experience they have, what their other students have to say about them, and how much they charge each hour. Before you start getting in touch with tutors and arranging free lessons, we recommend that you narrow down your search to tutors that meet your requirements.