Expanding Binomial Products
A binomial product such as \((5x + 3)(2x - 1)\) is expanded by multiplying each term in the first bracket by each term in the second, then collecting like terms. This NSW Year 9 Advanced maths subtopic builds fluency with these products (NESA outcome MA5‑ALG‑C‑01).
You will expand pairs with negative terms and coefficients other than \(1\), such as \((5 - 2x)(2 - 3x)\), square a binomial like \((3a - 2b)^2\), and simplify a difference of two products. You will find the area left on a tile with a hole cut in it, match coefficients to find unknowns, and prove that \((n + 1)^2 - (n - 1)^2 = 4n\) to evaluate \(101^2 - 99^2\) mentally. It is part of Year 9 Advanced Algebra: Expanding and Factorising.
Theory
To expand a binomial product such as \((a + b)(c + d)\), multiply each term in the first bracket by each term in the second, then collect like terms. This Year 9 Advanced maths page covers negative terms, squared brackets, expressions with two products, area problems and matching coefficients.
A binomial is an expression with two terms, such as \(2x - 7\). A binomial product is two binomials multiplied: \((2x - 7)(x + 3)\).
Expanding uses the distributive law twice. Every term in the first bracket multiplies every term in the second, so there are \(4\) products. Usually the two middle products are like terms and combine.
Each term keeps its sign. In \((6 - x)(2 - x)\) the terms are \(6\), \(-x\), \(2\) and \(-x\), and \(-x \times (-x) = +x^2\).
A squared bracket means the bracket times itself: \((m + 4)^2 = (m + 4)(m + 4)\) \(= m^2 + 8m + 16\). It is not \(m^2 + 16\).
The general product:
Special products worth recognising:
How to expand a binomial product
- Split the first bracket: each of its terms (with its sign) multiplies the whole second bracket.
- Expand both parts to get four products.
- Collect like terms, usually the two middle ones, and write the highest power first.
- Check by substituting a value such as \(x = 1\) into the question and the answer.
Harder expansions
- Subtracting a product: expand it first, then subtract it inside brackets so every sign changes.
- Area with a hole: area \(=\) outer shape \(-\) hole, with each area written as an expanded product.
- Unknown coefficients: expand, collect and match terms of the same power.
- Prove, then use: \((n + 3)^2 - (n - 3)^2 = 12n\), so \(53^2 - 47^2 = 12 \times 50 = 600\) without a calculator.
Multiply each term of the first bracket by the whole second bracket, keeping each sign with its term.
| \((3y - 4)(5 - 2y)\) | \(\) | \(\) |
| \(\) | \(=\) | \(3y(5 - 2y) - 4(5 - 2y)\) |
| \(\) | \(=\) | \(15y - 6y^2 - 20 + 8y\) |
| \(\) | \(=\) | \(\boldsymbol{-6y^2 + 23y - 20}\) |
\(-4 \times (-2y) = +8y\). Write the answer with the highest power first. Check with \(y = 1\): \((-1)(3) = -3\) and \(-6 + 23 - 20 = -3\).
Squaring means multiplying the bracket by itself, so there are four products, two of them the same.
| \((2p + 5q)^2\) | \(=\) | \((2p + 5q)(2p + 5q)\) |
| \(\) | \(=\) | \(4p^2 + 10pq + 10pq + 25q^2\) |
| \(\) | \(=\) | \(\boldsymbol{4p^2 + 20pq + 25q^2}\) |
Pattern check: first squared \((2p)^2 = 4p^2\), twice the product \(2 \times 2p \times 5q = 20pq\), last squared \((5q)^2 = 25q^2\).
Expand each part separately, then subtract the whole of the second one, inside brackets.
| \((x - 4)(x + 6)\) | \(=\) | \(x^2 + 6x - 4x - 24 = x^2 + 2x - 24\) |
| \((x + 1)^2\) | \(=\) | \(x^2 + 2x + 1\) |
| \(\text{Difference}\) | \(=\) | \(x^2 + 2x - 24 - (x^2 + 2x + 1)\) |
| \(\) | \(=\) | \(x^2 + 2x - 24 - x^2 - 2x - 1\) |
| \(\) | \(=\) | \(\mathbf{-25}\) |
Every \(x\)-term cancels, so the value is \(-25\) whatever \(x\) is. Check with \(x = 0\): \((-4)(6) - 1 = -25\).
Expand the left side and collect, then match the \(x^2\)-terms and the \(x\)-terms.
| \((ax + 5)(x - 2)\) | \(=\) | \(ax^2 - 2ax + 5x - 10\) |
| \(\) | \(=\) | \(ax^2 + (5 - 2a)x - 10\) |
| \(x^2\text{-terms: } a\) | \(=\) | \(\mathbf{4}\) |
| \(x\text{-terms: } b\) | \(=\) | \(5 - 2a = 5 - 8 = \mathbf{-3}\) |
Check: \((4x + 5)(x - 2) = 4x^2 - 8x + 5x - 10 = 4x^2 - 3x - 10\). The constants (\(-10\)) already match.
Common pitfalls
Frequently asked questions
How do you expand two brackets?
Multiply each term in the first bracket by each term in the second bracket, giving four products, then collect like terms. For example \((x + 2)(x + 7) = x^2 + 7x + 2x + 14\) \(= x^2 + 9x + 14\).
What is FOIL in maths?
FOIL stands for First, Outer, Inner, Last: the four products you make when expanding two binomials. It is the same as multiplying each term by each term.
How do you expand a bracket squared?
Write it as the bracket times itself and expand, or use the pattern: first term squared, plus twice the product of the terms, plus last term squared. \((x + 6)^2 = x^2 + 12x + 36\).
Why is (a + b) squared not equal to a squared plus b squared?
Because squaring the bracket makes four products, and two of them are ab. Those two middle terms give 2ab, so (a + b) squared = a squared + 2ab + b squared.
What is the difference of two squares?
It is the result of multiplying a sum by a difference of the same two terms: (a + b)(a - b) = a squared minus b squared. The middle terms cancel.