Solving Quadratic Equations by Factorising Revision Notes | GCSE AQA Higher Maths

1. Why Factorising Solves Quadratics

You have already practised expanding brackets — turning $(x+2)(x+3)$ into $x^2 + 5x + 6$. Factorising is that process in reverse: you start with $x^2 + 5x + 6$ and work back to $(x+2)(x+3)$.

The reason this matters for solving equations is a powerful rule: if two things multiply together to give zero, then at least one of them must be zero. So once a quadratic is written as $(x+2)(x+3) = 0$, you can immediately read off the solutions.

This section covers the full method: how to find the right factors, how to write the factorised form, and how to extract the solutions. Every step is examinable, so each one earns marks.

2. The Factorising Method

To factorise $x^2 + bx + c$, you need two numbers $p$ and $q$ such that $p \times q = c$ (they multiply to the constant term) and $p + q = b$ (they add to the coefficient of $x$). Finding this pair is the core skill.

A reliable approach is to list factor pairs of $c$, then check which pair adds to $b$. With practice this becomes quick, but writing out the pairs systematically avoids errors under exam pressure.

Notice that $p$ and $q$ can be negative. The signs are just as important as the magnitudes — getting a sign wrong changes both the sum and the product, so always check both conditions before writing the factorised form.

Worked example: Solving a Quadratic by Factorising

Solve $x^2 + 5x + 6 = 0$

1. $x^2 + 5x + 6 = 0$ — already in the correct form. — The factorisation method requires zero on the right-hand side. If it is not, rearrange first. Here it is, so proceed. [Setup — no mark yet]
2. Factor pairs of 6: $(1, 6)$, $(2, 3)$, $(-1, -6)$, $(-2, -3)$ Check sums: $1+6=7$, $2+3=5$ ✓ — The pair $(2, 3)$ satisfies both conditions: $2 \times 3 = 6$ and $2 + 3 = 5$. [M1 — correct method for finding factors]
3. $(x + 2)(x + 3) = 0$ — Replace $p$ and $q$ with 2 and 3. You can verify by expanding: $(x+2)(x+3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6$ ✓ [A1 — correct factorisation]
4. $x + 2 = 0 \quad \Rightarrow \quad x = -2$ $x + 3 = 0 \quad \Rightarrow \quad x = -3$ — Each bracket is set equal to zero separately. Both values are valid solutions. [A1 for each correct solution]

Answer: $x = -2$ or $x = -3$

3. Working With Negative Signs

Negative values of $b$ or $c$ are where most errors creep in. The sign rules are straightforward once you see the pattern: if $c$ is positive, both $p$ and $q$ have the same sign (both positive or both negative); if $c$ is negative, $p$ and $q$ have opposite signs.

Worked example: Factorising With a Negative Constant

Solve $x^2 + x - 6 = 0$

1. $b = 1$, $c = -6$ — The constant is negative, so $p$ and $q$ will have opposite signs.
2. $(1, -6)$: sum $= -5$ $(-1, 6)$: sum $= 5$ $(2, -3)$: sum $= -1$ $(-2, 3)$: sum $= 1$ ✓ — The pair $(-2, 3)$ gives product $-6$ and sum $1$. [M1 — correct method]
3. $(x - 2)(x + 3) = 0$ — Verify by expanding: $x^2 + 3x - 2x - 6 = x^2 + x - 6$ ✓ [A1 — correct factorisation]
4. $x - 2 = 0 \Rightarrow x = 2$ $x + 3 = 0 \Rightarrow x = -3$ — [A1 for each correct solution]

Answer: $x = 2$ or $x = -3$

4. Equations That Need Rearranging First

Not every quadratic arrives neatly as $x^2 + bx + c = 0$. Exam questions frequently present quadratics in a different arrangement — terms on both sides, or the equation set equal to a non-zero number. The factorisation method only works when one side is exactly zero, so rearranging is a compulsory first step.

The strategy is always the same: collect all terms onto one side (usually the left), leaving zero on the right, then factorise. Resist the temptation to factorise before rearranging — it leads to incorrect solutions every time.

Worked example: Solving a Quadratic That Requires Rearrangement

Solve $x^2 + 3x = 10$

1. $x^2 + 3x - 10 = 0$ — Subtract 10 from both sides. The equation must equal zero before factorising. [M1 — correct rearrangement]
2. Factor pairs of $-10$: $(1,-10)$, $(-1,10)$, $(2,-5)$, $(-2,5)$ Sums: $-9$, $9$, $-3$, $3$ ✓ — The pair $(-2, 5)$ works: $-2 \times 5 = -10$ and $-2 + 5 = 3$. [M1 — correct factors]
3. $(x - 2)(x + 5) = 0$ — Expand to verify: $x^2 + 5x - 2x - 10 = x^2 + 3x - 10$ ✓ [A1 — correct factorisation]
4. $x - 2 = 0 \Rightarrow x = 2$ $x + 5 = 0 \Rightarrow x = -5$ — [A1 for each correct solution]

Answer: $x = 2$ or $x = -5$

Worked example: Rearranging When Terms Are on Both Sides

Solve $x^2 - 2x = 3x + 18$

1. $x^2 - 2x - 3x - 18 = 0$ $x^2 - 5x - 18 = 0$ — Subtract $3x$ and subtract $18$ from both sides. Collect like terms. [M1 — correct rearrangement]
2. Factor pairs of $-18$: $(1,-18)$, $(-1,18)$, $(2,-9)$, $(-2,9)$, $(3,-6)$, $(-3,6)$ Sums: $-17$, $17$, $-7$, $7$, $-3$, $3$ None sum to $-5$. — None of the integer factor pairs of $-18$ sum to $-5$. This quadratic does not factorise over the integers, so factorising is not the right method here. On an Higher-tier paper you would move to the quadratic formula or completing the square after checking the rearrangement is correct.

Answer: This quadratic does not factorise over the integers, so after checking the rearrangement you would solve it using a different method such as the quadratic formula.

Worked example: Rearranging When Terms Are on Both Sides (Factorisable Version)

Solve $x^2 - 2x = 3x - 4$

1. $x^2 - 2x - 3x + 4 = 0$ $x^2 - 5x + 4 = 0$ — Subtract $3x$ and add $4$ to both sides. Collect like terms carefully, paying attention to signs. [M1 — correct rearrangement]
2. Factor pairs of $4$: $(1,4)$, $(-1,-4)$, $(2,2)$, $(-2,-2)$ Sums: $5$, $-5$ ✓, $4$, $-4$ — The pair $(-1, -4)$ works: $(-1) \times (-4) = 4$ and $(-1) + (-4) = -5$. [M1 — correct factors]
3. $(x - 1)(x - 4) = 0$ — Expand to verify: $x^2 - 4x - x + 4 = x^2 - 5x + 4$ ✓ [A1 — correct factorisation]
4. $x - 1 = 0 \Rightarrow x = 1$ $x - 4 = 0 \Rightarrow x = 4$ — [A1 for each correct solution]

Answer: $x = 1$ or $x = 4$

5. Visual Connection: Roots and the Graph

The solutions you find by factorising are called the roots of the quadratic. Graphically, they are the $x$-coordinates where the parabola crosses the $x$-axis. Seeing this connection helps you sense-check your answers: if you solve $x^2 - 5x + 6 = 0$ and get $x = 2$ and $x = 3$, the graph should cross the $x$-axis at those two points.

A quadratic can have two distinct roots (crosses the $x$-axis twice), one repeated root (touches the $x$-axis once), or no real roots (never crosses). The factorisation method can solve the distinct-root case and the repeated-root case whenever the quadratic factorises, and both types appear in GCSE questions.

6. Deliberate Practice

The method is now in place. These problems cover the full range of question types you will meet: straightforward factorising, negative signs, and equations needing rearrangement. Attempt each one before checking the solution.

7. Exam Technique and Common Mistakes

GCSE Higher exam questions on this topic typically award marks for the factorised form and then for each correct solution. That means even if you make an arithmetic slip finding the factors, you can still earn method marks — but only if your working is visible. Always write out the factorised form as a separate line before solving.

The most common source of lost marks is forgetting to rearrange before factorising. If you see a quadratic with terms on both sides, or set equal to a non-zero number, write the rearrangement step explicitly — examiners award a mark for it.

Solving quadratic equations by factorising — including those that require rearrangement — is a core GCSE Higher algebra skill. The method is always: rearrange to zero, find the factor pair, write the factorised form, apply the zero product property, state both solutions.

GCSE factorising questions usually begin with monic quadratics, including equations that already equal zero, equations that must be rearranged first, and repeated-root cases such as $x^2-6x+9=(x-3)^2$. The essential routine is always the same: write the equation in the form $x^2+bx+c=0$, identify the factor pair with the correct product and sum, write the brackets, then set each factor equal to zero.