Topic 3 Algebraic expressions

An algebraic expression is an expression which is made up of variables and constants, along with algebraic operations (addition, subtraction, etc.).

3.1 Collecting terms

One of the things you might want to do when given an algebraic expression is to collect like terms.

Like terms are terms which have the same set of letters. E.g.

• $a$ and $2a$ are like terms as both are just $a$ with a number in front of it ($a$ implictly means $1\times a$, and $2a$ means $2\times a$).
• $5a$ and $7a$ are like terms for the same reason.
• $5a$ and $7a^2$ are not like terms, as $a^2$ is $a $a$
• $5a$ and $5ab$ are similarly not like terms.
• $ab$ and $ba$ are like terms, as the order of multiplication of numbers doesn’t matter ($2\times 3$ is the same as $3\times 2$) and both $ab$ and $ba$ mean $a\times b$

When given an expression such as $5a+4+3a+6$ the first step of collecting the terms is to put like terms together; in this example that means we reorganise it to $5a+3a+4+6$.

The second step is to do the additions $5a+3a$ becomes $8a$, $4+6$ becomes $10$. Hence the whole expression becomes $8a+10$.

You do need to be careful with minus signs — they need to be kept with their term. E.g. $5a+4-3a+6$ becomes $5a-3a+4+6$ and so ends up as $2a+10$.

Don’t forget that if something is raised to a power, it is a different term, so $5a+4a^2-3a+6a^2$ becomes $5a-3a+4b+6b$ and so ends up as $2a+10a^2$.

3.2 Expanding brackets

An algebraic expression will often contain a term looking like $5(a+b)$; if $a$ and $b$ were numbers, we’d add them together and then multiply by $5$ — remember that in the order of operations brackets come first. As we don’t know what $a$ and $b$ are, this isn’t an option, but if we have other terms in the expression which also involve $a$ or $b$ we might want to expand the bracket.

Expanding the bracket means that we take each number in the bracket and multiply by the number outside. E.g.

• $5(a+b)$ becomes $5a+5b$
• $5(a-b)$ becomes $5a-5b$
• $-5(a+b)$ becomes $-5a-5b$
• $-5(a-b)$ becomes $-5a+5b$

The terms inside the brackets can be more complicated. E.g.

• $5(a+3b)$ is $5a+5\times 3b$ which becomes $5a+15b$
• $5(2a+3b)$ becomes $10a+15b$

Or even just numbers. E.g.

• $5(a+3)$ becomes $5a+15$

3.3 Factorising

Factorising is the reverse of expanding brackets, and is often used when you have several terms involving the same factor and you want to write them all as one term.

E.g. If I have $3a+ba$ I could write this as $(3+b)a$.

You will have probably noticed that this is very similar to collecting like terms, which we studied above — collecting like terms is really just a special case of this, so the steps are similar.

1. Look at all the terms and decide which terms have a common factor
2. Choose a common factor to “factorise out”
3. Write this factor outside a pair of brackets (the side doesn’t matter)
4. Divide each term involving that factor by the factor, and put it inside the brackets.

E.g. Let’s take a look at $3a+ba$

1. There is a common factor of $a$ in both terms
2. We only have one common factor, so choose to factorise the $a$ out.
3. We now start with $(...)a$
4. If we divide $3a$ by $a$ we get $3$ and if we divide $+ba$ by $a$ we get $+b$ so we put $3+b$ inside the brackets to get $(3+b)a$.

You can always double check by multiplying the brackets out again.

More examples:

$3a+3b$

1. We have one common factor, $3$
2. Choose to factorise out $3$
3. $3(...)$
4. $3a$ divided by $3$ is $a$, $+3b$ divided by $3$ is $+b$ so we get $3(a+b)$

$3ab+3ac$

1. We have one common factor, $3a$
2. Choose to factorise out $3a$
3. $3a(...)$
4. $3ab$ divided by $3a$ is $b$, $+3ac$ divided by $3a$ is $+c$ so we get $3a(b+c)$

$a+a^2$

1. We have one common factor, $a$
2. Choose to factorise out $a$
3. $a(...)$
4. $a$ divided by $a$ is $1$, $+a^2$ divided by $a$ is $+c$ so we get $a(1+a)$

$3ab+3ac+3bc$

1. We have choices of common factors:
   • $3$ is common to all of them
   • $3a$ is common to the first two
   • $3b$ is common to the first and third term
   • $3c$ is common to the second and third term
2. Choose to factorise out $3$
3. $3(...)$
4. $3ab$ divided by $3$ is $ab$, $+3ac$ divided by $3$ is $+ac$, $+3bc$ divided by $3$ is $+bc$ so we get $3(ab+ac+bc)$

We could no look at the terms inside the bracket and do some more factorising. E.g. we could get $3\left(a\right(b+c)+bc)$.

3.4 Further Resources

• Khan academy
  • Combining like terms ⧉
• Corbett Maths
  • Collecting like terms - video ⧉
  • Collecting like terms - questions ⧉
  • Expanding brackets - vodeo ⧉
  • Expanding brackets - questions ⧉
  • Factorisation - video ⧉
  • Factorisation - questions ⧉