Expanding and Factorising Quadratics

Step 1: Multiply the First terms.

• $2x \times x = 2x^2$

Step 2: Multiply the Outside terms.

• $2x \times 4 = +8x$

Step 3: Multiply the Inside terms.

• $-3 \times x = -3x$

Step 4: Multiply the Last terms.

• $-3 \times 4 = -12$

Step 5: Write out the full sum and collect like terms.

• $2x^2 + 8x - 3x - 12$

• $2x^2 + 5x - 12$

2. The Grid Method (Area Model) This method uses a 2x2 grid, placing one binomial across the top and the other down the side.

Expand and simplify $(x + 3)(x + 2)$

Step 1: Draw a grid and multiply the rows by the columns.

$\times$	$x$	$+2$
$x$	$x^2$

Step 2: Write out the terms from the grid and collect like terms (often found on the diagonal).

• $x^2 + 2x + 3x + 6$

• $x^2 + 5x + 6$

Expanding Products of Three Binomials (Higher Tier)

You can multiply two expressions easily, but try multiplying three without breaking it down into stages. Expanding three binomials is a Higher Tier requirement that relies on the distributive law, the rule used to multiply a single term or bracket by every term inside another bracket.

The process requires two stages. First, expand two binomials to create a quadratic. Then, systematically distribute this resulting quadratic into the third binomial, which requires 6 separate multiplications before final simplification.

Expand and simplify $(x + 2)(x - 3)(2x + 1)$

Step 1: Stage 1 — Expand and simplify the first two brackets.

• $(x + 2)(x - 3) = x^2 - 3x + 2x - 6$

Step 2: Stage 2 — Multiply the resulting quadratic by the third bracket.

• $(x^2 - x - 6)(2x + 1)$

Step 3: Distribute every term systematically.

• $x^2(2x) = 2x^3$
• $x^2(1) = x^2$

Step 4: Collect all like terms to form your final polynomial (an expression consisting of variables and coefficients with non-negative powers).

• $2x^3 + (x^2 - 2x^2) + (-x - 12x) - 6$

Factorising Unitary Quadratics

How do you reverse the process of expanding brackets to find the original factors? To factorise an expression means to rewrite it as a product of its factors (this process is known as factorisation).

A monic/unitary quadratic is a quadratic where the leading coefficient (the numerical value multiplied by the $x^2$ variable) is exactly 1, written in the form $x^2 + bx + c$. To factorise these, use the Factor Pair Rule: you must find two integers ($p$ and ) that multiply to give the and add to give the coefficient .

Factorise $x^2 - 5x + 6$

Step 1: Identify $b$ and $c$.

• $b = -5$
• $c = 6$

Step 2: Find factors of $c$ that add to $b$. Because $c$ is positive and $b$ is negative, both factors must be negative.

• Factors of $6$: $(-1 \times -6)$ and $(-2 \times -3)$
• Sum check: $-2 + -3 = -5$

Step 3: Write the factors in double brackets.

• $(x - 2)(x - 3)$

Factorising Non-Unitary Quadratics

Every time you encounter a non-unitary quadratic (a quadratic where the leading number isn't 1), you need a highly systematic approach to break it apart. For expressions in the form $ax^2 + bx + c$ where $a \neq 1$, use the .

This method involves splitting the middle term $bx$ into two separate terms, allowing you to use factorising by grouping — a technique where you split a four-term expression into two pairs and extract the highest common factor from each.

Factorise $6x^2 - 7x - 3$

Step 1: Identify $a$, $b$, and $c$, then calculate $ac$.

• $a = 6$, $b = -7$, $c = -3$
• $a \times c = 6 \times -3 = -18$

Step 2: Find two integers that multiply to $-18$ and add to $b$ ($-7$).

• The factors are $-9$ and $2$.

Step 3: Decompose the middle term ($-7x$) into $-9x$ and $+2x$.

• $6x^2 - 9x + 2x - 3$

Step 4: Group the four terms into two pairs and factorise each pair.

• First pair: $3x(2x - 3)$
• Second pair: $+1(2x - 3)$

Step 5: Factor out the common bracket.

• $(3x + 1)(2x - 3)$

Factorising the Difference of Two Squares (DOTS)

The difference of two squares is one of the only algebraic rules that allows you to factorise an expression with just two terms. The Difference of Two Squares (DOTS) applies when a binomial consists of a perfect square (a term that is the square of another term) subtracted from another perfect square.

The rule to memorise is:

$$a^2 - b^2 = (a + b)(a - b)$$

Factorise fully $25y^2 - 9$

Step 1: Verify both terms are perfect squares separated by a subtraction sign.

• $25y^2 = (5y)^2$
• $9 = (3)^2$

Step 2: Apply the DOTS formula substituting $a = 5y$ and $b = 3$.

• $(5y + 3)(5y - 3)$

Factorise fully $2x^2 - 50$

Step 1: Look for a Highest Common Factor (HCF) first.

• Factor out $2$: $2(x^2 - 25)$

Step 2: Apply DOTS to the expression inside the bracket.

• $x^2 = (x)^2$ and $25 = (5)^2$