Expanding, Factorising and Simplifying

To expand $(2x - 3)(x + 5)$, multiply each term in the first bracket by each term in the second: $2x^2 + 10x - 3x - 15$, then collect like terms to get $2x^2 + 7x - 15$.

Three products appear so often that you should recognise them rather than expand them each time: $(a+b)^2 = a^2 + 2ab + b^2$, $(a-b)^2 = a^2 - 2ab + b^2$, and $(a+b)(a-b) = a^2 - b^2$.

The middle term of a perfect square comes from the two identical cross-products. In $(x+3)^2$ you get $3x$ twice, which is $6x$. That is the term people lose.

Expand by multiplying every term in one bracket by every term in the other, then collect like terms. Learn the three special products on sight: $(a\pm b)^2 = a^2 \pm 2ab + b^2$ and $(a+b)(a-b) = a^2-b^2$.

Pause, copy the three special products and one expansion showing where the middle term of $(x+3)^2$ comes from, into your book.

We just saw how to expand brackets and recognise the three special products. That raises a question: given the expanded form, how do you get the brackets back? This card answers it → work through a fixed order, HCF first, then a special form, then a trinomial.

Step 1, HCF. Always look for a common factor first. $3x^2 - 12 = 3(x^2 - 4)$, and only now does the difference of squares appear: $3(x-2)(x+2)$.

Step 2, special form. Two terms with a minus between two squares is a difference of squares. A three-term expression matching $a^2 \pm 2ab + b^2$ is a perfect square.

Step 3, trinomial. For $x^2 + bx + c$, find two numbers that multiply to $c$ and add to $b$. For $x^2 + 7x + 10$ they are 2 and 5, giving $(x+2)(x+5)$.

Factorise in order: HCF first, then check for a difference of squares or perfect square, then factorise the trinomial. When the leading coefficient is not 1, split the middle term using two numbers that multiply to $ac$ and add to $b$.

Pause, copy the three-step factorising order, the trinomial rule (two numbers multiplying to $c$ and adding to $b$), and the split-the-middle-term method for $2x^2 + 7x + 3$, into your book.

We just saw the standard factorising order for two and three term expressions. That raises a question: what do you do with four terms, where none of those forms fits? This card answers it → group the terms in pairs and factorise each pair, which exposes a common bracket.

With four terms, group them in pairs: $x^3 + 2x^2 + 5x + 10 = x^2(x + 2) + 5(x + 2)$. The bracket $(x+2)$ is now a common factor, giving $(x+2)(x^2+5)$.

If the pairing does not produce a matching bracket, try a different pairing before concluding it does not factorise.

Every factorisation can be checked by expanding it. If the expansion does not return the original expression, the factorisation is wrong. This is a complete check, not a partial one.

Four terms with no common factor: group in pairs, factorise each pair, then take out the common bracket. Always check a factorisation by expanding it, which either returns the original expression or proves you wrong.

Pause, copy the grouping method with the worked $x^3 + 2x^2 + 5x + 10 = (x+2)(x^2+5)$, the sign warning when the second pair starts with a minus, and the expand-to-check rule, into your book.

1. Expand $(3x + 2)(x - 4)$.
2. Expand $(2y - 5)^2$.
3. Factorise $5a^2 - 20$.
4. Factorise $x^2 - 2x - 15$.