Simplify expressions by collecting like terms, expand single and double brackets, and factorise using a common factor

What this dot point is asking

SEAB wants you to simplify algebraic expressions by collecting like terms, expand brackets (single and double), and factorise by taking out a common factor. Algebra is the language of the rest of the course - equations, graphs and formulae all rely on these manipulations - so accuracy here pays off everywhere.

The answer

Like terms

Like terms have exactly the same letters raised to the same powers, for example $3x$ and $5x$, or $2x^2$ and $-7x^2$. You can only add or subtract like terms:

Terms such as $3x$ and $3x^2$ are not like terms and cannot be combined.

Expanding single brackets

Multiply the term outside the bracket by each term inside:

Take care with signs: a negative outside the bracket flips the sign of every term inside.

Expanding double brackets

Each term in the first bracket multiplies each term in the second. A reliable order is First, Outer, Inner, Last (FOIL):

Factorising with a common factor

Factorising is the reverse of expanding. Look for the highest common factor of all the terms and write it outside a bracket:

Always take out the highest common factor so the expression is factorised completely. A quick way to find it is to take the largest number that divides every coefficient, and the lowest power of each letter that appears in every term. In $5x^2 - 10x$, the number factor is $5$ and the letter factor is $x$, giving $5x$.

Substituting values into an expression

Algebra becomes a number once you replace each letter with a value. To evaluate $2a + 3b$ when $a = 4$ and $b = 5$, substitute and work out: $2 \times 4 + 3 \times 5 = 8 + 15 = 23$. Use brackets when you substitute a negative number, for example $a^2$ with $a = -3$ becomes $(-3)^2 = 9$. Simplifying an expression before substituting usually makes the arithmetic shorter and safer.

Examples in context

Example 1. Perimeter of a rectangle. A rectangle has length $x + 5$ and width $x$. Its perimeter is $2(x + 5) + 2x = 2x + 10 + 2x = 4x + 10$. Expanding and collecting like terms turns a worded measurement into a tidy formula you can evaluate for any $x$.

Example 2. Simplifying before substituting. To evaluate $3(2x - 1) + 4(x + 2)$ when $x = 5$, simplify first to $10x + 5$, then substitute: $10 \times 5 + 5 = 55$. Simplifying before substituting is faster and less error-prone than substituting into the long form.

Try this

• Cue. Simplify $7a + 3b - 2a + b$. Collect like terms: $5a + 4b$.
• Cue. Expand $(x + 5)(x - 3)$. FOIL gives $x^2 - 3x + 5x - 15 = x^2 + 2x - 15$.
• Cue. Factorise $12y^2 - 8y$. The highest common factor is $4y$, so $4y(3y - 2)$.