Terms with exactly the same variable part, such as three x and minus five x, or x squared and seven x squared. You add their coefficients to collect them.

Why is x plus 3 all squared not x squared plus 9?

Because x plus 3 all squared equals x plus 3 times x plus 3, which is x squared plus three x plus three x plus 9, giving x squared plus six x plus 9. The two cross terms give the extra six x.

How do I expand a plus b times a minus b?

It is the difference of squares: a plus b times a minus b equals a squared minus b squared. The middle terms plus a b and minus a b cancel.

Terms with exactly the same variable part, such as three x and minus five x, or x squared and seven x squared. You add their coefficients to collect them.

Why is x plus 3 all squared not x squared plus 9?

Because x plus 3 all squared equals x plus 3 times x plus 3, which is x squared plus three x plus three x plus 9, giving x squared plus six x plus 9. The two cross terms give the extra six x.

How do I expand a plus b times a minus b?

It is the difference of squares: a plus b times a minus b equals a squared minus b squared. The middle terms plus a b and minus a b cancel.

Year 11 Methods (Unit 1 & 2) Quadratics

Expanding And Collecting Like Terms

Q: In what order should I write the final answer?

In descending powers of x: the x squared term first, then the x term, then the constant.

Q: Can I collect x squared and x together?

No. They have different powers of x, so they are unlike terms and stay separate in the answer.

20 practice questions 1 video lesson Theory + worked examples

Master expanding and collecting like terms for Queensland Year 11 Mathematical Methods (QCAA). Expanding removes brackets using the distributive law, while collecting like terms adds together the terms that share the same variable part.

You will learn to expand a single bracket, multiply two binomials, handle perfect squares and the difference of two squares, then simplify — essential groundwork for factorising and quadratics.

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Expanding And Collecting Like Terms - Video - Expanding (multiplying out) double brackets Watch

Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), expanding means removing brackets with the distributive law, and collecting like terms means adding the terms that share the same variable part. This page shows how to expand a single bracket, a binomial product \((ax+b)(cx+d)\), a perfect square and a difference of squares, then simplify by collecting like terms.

To expand is to multiply out brackets. The distributive law \(a(b+c)=ab+ac\) multiplies every term inside the bracket by the term outside. For two brackets, every term in the first is multiplied by every term in the second.

Like terms have exactly the same variable part (for example \(3x\) and \(-5x\), or \(x^2\) and \(7x^2\)). To collect like terms you add or subtract their coefficients; unlike terms such as \(x^2\) and \(x\) cannot be combined.

Two products come up so often they are worth memorising: the perfect square \((a+b)^2=a^2+2ab+b^2\) and the difference of squares \((a+b)(a-b)=a^2-b^2\).

Multiply everything by everything, then tidy. Expand each bracket fully first, and only then collect the like terms into a single simplified expression.

Expanding \((x+1)(x+3)\) gives \(x^2+4x+3\) — the same parabola, cutting the axis at \(x=-3\) and \(x=-1\).

Area model: a rectangle \((x+3)\) by \((x+2)\) splits into four areas that add to \(x^2+5x+6\).

The distributive law and the two special products:

\[a(b+c)=ab+ac\]

a (b + c) = a b + a c

\[(a+b)^2=a^2+2ab+b^2\]

{(a + b)}^{2} = a^{2} + 2 a b + b^{2}

\[(a+b)(a-b)=a^2-b^2\]

(a + b) (a - b) = a^{2} - b^{2}

Watch the middle term. \((a+b)^2\ne a^2+b^2\); the expansion has the extra \(2ab\). The difference of squares is the one product with no middle term.

How to expand and collect like terms

Multiply out each bracket with the distributive law, so every term inside is multiplied by the term(s) outside.
Group the like terms — those with the same variable part (\(x^2\) with \(x^2\), \(x\) with \(x\), numbers with numbers).
Combine each group by adding or subtracting the coefficients, then write the answer in descending powers of \(x\).

Example 1 — Single brackets and collecting

Expand and simplify \(3(2x+5)-2(x-4)\).

Solution

Expand — distribute across each bracket:

\(3(2x+5)-2(x-4)\)

\(=\)

\(6x+15-2x+8\)

Collect like terms — the \(x\) terms, then the numbers:

	\(=\)	\((6x-2x)+(15+8)\)
	\(=\)	\(4x+23\)

\(3(2x+5)-2(x-4)=4x+23\).

4 x + 23

Example 2 — A binomial product

Expand \((x+5)(x-2)\).

Solution

Multiply every term in the first bracket by every term in the second:

\((x+5)(x-2)\)	\(=\)	\(x\cdot x + x\cdot(-2) + 5\cdot x + 5\cdot(-2)\)
	\(=\)	\(x^2-2x+5x-10\)

Collect the two \(x\) terms:

\(=\)

\(x^2+3x-10\)

\((x+5)(x-2)=x^2+3x-10\).

x^{2} + 3 x - 10

Example 3 — Product plus a perfect square

Expand and simplify \((2x+1)(x-4)+(x+3)^2\).

Solution

Expand the product \((2x+1)(x-4)\):

\((2x+1)(x-4)\)	\(=\)	\(2x^2-8x+x-4\)
	\(=\)	\(2x^2-7x-4\)

Expand the perfect square \((x+3)^2=x^2+2(3)x+3^2\):

\((x+3)^2\)

\(=\)

\(x^2+6x+9\)

Add the two results and collect like terms:

	\(=\)	\((2x^2-7x-4)+(x^2+6x+9)\)
	\(=\)	\(3x^2-x+5\)

\((2x+1)(x-4)+(x+3)^2=3x^2-x+5\).

3 x^{2} - x + 5

Example 4 — Difference of squares in a mix

Expand and simplify \(3(x-2)(x+2)-(2x-1)^2\).

Solution

Use the difference of squares on \((x-2)(x+2)=x^2-2^2\), then multiply by \(3\):

\(3(x-2)(x+2)\)	\(=\)	\(3(x^2-4)\)
	\(=\)	\(3x^2-12\)

Expand \((2x-1)^2=(2x)^2-2(2x)(1)+1^2\):

\((2x-1)^2\)

\(=\)

\(4x^2-4x+1\)

Subtract — distribute the minus sign carefully, then collect:

	\(=\)	\((3x^2-12)-(4x^2-4x+1)\)
	\(=\)	\(3x^2-12-4x^2+4x-1\)
	\(=\)	\(-x^2+4x-13\)

\(3(x-2)(x+2)-(2x-1)^2=-x^2+4x-13\).

- x^{2} + 4 x - 13

Common pitfalls

Forgetting the middle term of a square. \((x+3)^2\) is \(x^2+6x+9\), not \(x^2+9\). Squaring a bracket always gives three terms.

Sign slips when subtracting a bracket. A minus in front of a bracket flips every sign inside: \(-(4x^2-4x+1)=-4x^2+4x-1\).

Combining unlike terms. \(x^2\) and \(x\) are not like terms, so \(x^2+3x\) cannot be simplified to \(4x^2\) or \(4x\).

Frequently asked questions

What does it mean to expand an expression?

To expand is to multiply out the brackets using the distributive law, so \(a(b+c)=ab+ac\), turning a product into a sum of terms.

What are like terms?

Terms with exactly the same variable part, such as \(3x\) and \(-5x\), or \(x^2\) and \(7x^2\). You add their coefficients to collect them.

Why is \((x+3)^2\) not \(x^2+9\)?

Because \((x+3)^2=(x+3)(x+3)=x^2+3x+3x+9=x^2+6x+9\). The two cross terms give the extra \(6x\).

How do I expand \((a+b)(a-b)\)?

It is the difference of squares: \((a+b)(a-b)=a^2-b^2\). The middle terms \(+ab\) and \(-ab\) cancel.

In what order should I write the final answer?

In descending powers of \(x\): the \(x^2\) term first, then the \(x\) term, then the constant.

Can I collect \(x^2\) and \(x\) together?

No. They have different powers of \(x\), so they are unlike terms and stay separate in the answer.

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Expanding And Collecting Like Terms

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Theory