Solving Quadratic Equations by Factorising

We can factorise the left-hand side: \( x^2 + 5x + 6 = (x + 2)(x + 3) \). Setting each factor equal to zero gives \( x + 2 = 0 \), so \( x = -2 \), and \( x + 3 = 0 \), so \( x = -3 \). Substituting back: \( (-2)^2 + 5(-2) + 6 = 4 \; – \; 10 + 6 = 0 \) ✓ and \( (-3)^2 + 5(-3) + 6 = 9 \; – \; 15 + 6 = 0 \) ✓. Both solutions work, and no others exist.

The reason there are exactly two solutions comes from the zero product property: if \( AB = 0 \), then \( A = 0 \) or \( B = 0 \). Since the quadratic factorises into two distinct linear factors, each factor generates one solution — no more, no fewer. The “quadratics only have one answer” misconception often arises because students stop after finding the first solution, or because they confuse quadratics with linear equations.

Substituting \( x = -3 \): \( (-3)^2 \; – \; 9 = 9 \; – \; 9 = 0 \) ✓. Many students recognise that \( x = 3 \) is a solution (since \( 3^2 = 9 \)) but forget the negative root. The equation factorises as \( (x \; – \; 3)(x + 3) = 0 \) using the difference of two squares. Setting \( x + 3 = 0 \) gives \( x = -3 \).

A common error is to reason “\( x^2 = 9 \), so \( x = 3 \)” — the “only the positive square root” misconception. Squaring a negative number also gives a positive result, so both \( x = 3 \) and \( x = -3 \) satisfy \( x^2 = 9 \).

We need to factorise a non-monic quadratic (where the \( x^2 \) coefficient is greater than 1). We look for numbers that multiply to \( 2 \times 3 = 6 \) and add to 7. These are 6 and 1. Splitting the middle term: \( 2x^2 + 6x + x + 3 = 0 \). Factorising by grouping: \( 2x(x + 3) + 1(x + 3) = 0 \), which gives \( (2x + 1)(x + 3) = 0 \).

Setting \( x + 3 = 0 \) gives \( x = -3 \). Setting \( 2x + 1 = 0 \) gives \( 2x = -1 \), so \( x = -0.5 \). Many students struggle here because they expect factors to always look like \( (x \pm a) \), forgetting that the coefficient of \( x \) in the bracket can be greater than 1.

Factorising: \( x^2 \; – \; 10x + 25 = (x \; – \; 5)(x \; – \; 5) = (x \; – \; 5)^2 = 0 \). This gives \( x \; – \; 5 = 0 \), so \( x = 5 \) is the only solution. We can verify: \( 5^2 \; – \; 10(5) + 25 = 25 \; – \; 50 + 25 = 0 \) ✓.

Students often assume that every quadratic must produce two different answers. Here, the quadratic is a perfect square trinomial, so both factors are identical, producing a repeated root.

Example: \( x^2 \; – \; 7x + 10 = 0 \)

Another: \( 2x^2 \; – \; 14x + 20 = 0 \)

Creative: \( (x \; – \; 2)(x \; – \; 5) = 0 \) — left in factored form, this is already a valid equation with the required solutions.

Trap: \( x^2 + 7x + 10 = 0 \) — a student might think “2 and 5 add to 7 and multiply to 10, so this works.” But this factorises as \( (x + 2)(x + 5) = 0 \), giving \( x = -2 \) and \( x = -5 \), not \( x = 2 \) and \( x = 5 \). The “sign confusion” misconception strikes: the factors contain +2 and +5, but the solutions are their negatives.

Example: \( x^2 \; – \; 3x = 0 \)

Another: \( x^2 + 8x = 0 \)

Creative: \( 5x^2 \; – \; x = 0 \) — a non-monic example. Factorises as \( x(5x \; – \; 1) = 0 \), giving \( x = 0 \) and \( x = \frac{1}{5} \).

Trap: \( x(x + 4) = 5 \) — a student might see \( x \) as a factor and think “setting \( x = 0 \) gives a solution.” But substituting \( x = 0 \) gives \( 0 \times 4 = 0 \neq 5 \). The equation does not equal zero, so the zero product property cannot be applied. This targets the “factorising when RHS ≠ 0” misconception.

Example: \( x^2 + 7x + 12 = 0 \) — solutions \( x = -3 \) and \( x = -4 \)

Another: \( x^2 + 8x + 15 = 0 \) — solutions \( x = -3 \) and \( x = -5 \)

Creative: \( x^2 + 2x + 1 = 0 \) — this gives \( (x + 1)^2 = 0 \), so \( x = -1 \) is a repeated negative root. Both “solutions” are negative (even though they’re the same).

Trap: \( x^2 \; – \; 5x + 6 = 0 \) — a student might think “the \( -5x \) makes it negative, so the solutions should be negative.” But this factorises as \( (x \; – \; 2)(x \; – \; 3) = 0 \), giving \( x = 2 \) and \( x = 3 \) — both positive. This is the “negative coefficient means negative solutions” misconception: the sign of the \( x \)-coefficient does not directly determine the sign of the solutions.

Example: \( x^2 + 3x + 1 = 0 \) — the only integer pairs with product 1 are \( (1, 1) \) and \( (-1, -1) \), which sum to 2 and −2 respectively. Neither gives 3, so no integer factorisation exists.

Another: \( x^2 \; – \; 5x + 1 = 0 \) — same pairs with product 1, sums of 2 and −2. Neither gives −5.

Creative: \( x^2 + x \; – \; 1 = 0 \) — integer pairs with product −1 are \( (1, -1) \), which sums to 0, not 1. This needs the quadratic formula, giving irrational solutions.

Trap: \( x^2 + 16 = 0 \) — this is a Sum of Two Squares. Unlike the difference of two squares (\( x^2 – 16 \)), this cannot be factorised using real numbers. Students often try to write \( (x+4)(x+4) \) (which gives \( x^2+8x+16 \)) or \( (x+4)(x-4) \) (which gives \( x^2-16 \)). It has no real solutions at all.

This is true for many quadratics — for example, \( x^2 \; – \; x \; – \; 6 = 0 \) factorises as \( (x \; – \; 3)(x + 2) = 0 \), giving two distinct solutions (often called roots) \( x = 3 \) and \( x = -2 \). However, repeated roots provide the counterexample: \( x^2 \; – \; 6x + 9 = 0 \) factorises as \( (x \; – \; 3)^2 = 0 \), but has only one distinct solution, \( x = 3 \).

The misconception is that factorising always produces two different answers. A perfect square trinomial factorises into two identical brackets, yielding a single repeated root.

If \( x^2 \; – \; 5x + 6 = 0 \) has solutions \( x = 2 \) and \( x = 3 \), then multiplying both sides by 2 gives \( 2x^2 \; – \; 10x + 12 = 0 \). Checking: \( 2(2^2) \; – \; 10(2) + 12 = 8 \; – \; 20 + 12 = 0 \) ✓ and \( 2(3^2) \; – \; 10(3) + 12 = 18 \; – \; 30 + 12 = 0 \) ✓. The solutions are unchanged.

Multiplying (or dividing) both sides of an equation by any non-zero constant is a valid operation that preserves all solutions. The “different-looking equation means different solutions” misconception leads students to think the two equations are fundamentally different, but they are equivalent.

Factorising: \( x^2 + bx = x(x + b) = 0 \). By the zero product property, \( x = 0 \) or \( x + b = 0 \) (giving \( x = -b \)). Since \( x \) is always a common factor, \( x = 0 \) is always a solution. For example: \( x^2 + 3x = 0 \to x(x + 3) = 0 \to x = 0 \) or \( x = -3 \); and \( x^2 \; – \; 10x = 0 \to x(x \; – \; 10) = 0 \to x = 0 \) or \( x = 10 \).

The key point is that the absence of a constant term guarantees \( x \) is a factor, which guarantees \( x = 0 \) as a solution (root). Students who divide by \( x \) instead of factorising will miss this solution entirely.

Many quadratic equations factorise neatly — for example, \( x^2 + x \; – \; 12 = 0 \) factorises as \( (x + 4)(x \; – \; 3) = 0 \), giving \( x = -4 \) and \( x = 3 \). But consider \( x^2 + x \; – \; 1 = 0 \): the integer pairs multiplying to −1 are \( (1, -1) \), which sum to 0, not 1. No integer factorisation exists.

This equation requires the quadratic formula or completing the square. The “all quadratics can be factorised” misconception leads students to assume that every quadratic can be cracked by finding a factor pair, but factorising only works when the solutions are rational numbers with a convenient form.

The factorisation \( (x + 4)(x \; – \; 3) = 0 \) is correct. However, the student fell for the “signs in the brackets are the solutions” trap. They just read off the numbers 4 and -3. But we need to find the value of \( x \) that makes the bracket equal to zero.

The correct solutions are \( x = -4 \) and \( x = 3 \). Always perform a sanity check: substituting \( x = 4 \) gives \( 4^2 + 4 \; – \; 12 = 8 \neq 0 \), proving it wrong.

The answer is correct, but the method is dangerous. You cannot divide by \( x \) unless you know \( x \neq 0 \). By dividing by \( x \), you are effectively assuming \( x \) is not zero, which instantly deletes the valid solution \( x = 0 \). You are removing a factor, and therefore removing a solution.

The correct approach is to factorise: \( x^2 \; – \; 4x = x(x \; – \; 4) = 0 \). This systematic method finds both solutions (\( x=0, x=4 \)) without relying on luck to spot the missing one.

The student has applied the “zero product property” to a non-zero product. The logic “if \( AB = 0 \) then \( A=0 \) or \( B=0 \)” ONLY works for zero. If \( AB = 8 \), there are infinite possibilities (e.g., \( 2 \times 4 \), \( 1 \times 8 \), \( 0.5 \times 16 \)).

You must rearrange to equal zero before factorising.
Correct method: \( x^2 + 2x \; – \; 8 = 0 \), then \( (x + 4)(x \; – \; 2) = 0 \), giving \( x = -4 \) or \( x = 2 \).

The student fell for the “coefficient becomes the solution” trap. When they reached \( 2x = 0 \), they just saw the “2” and wrote \( x = 2 \). But if \( 2 \times \text{something} = 0 \), that something must be 0.

The correct solutions are \( x = 0 \) and \( x = 5 \). Sanity Check: Substitute the student’s answer \( x = 2 \) into the original equation: \( 2(2^2) \; – \; 10(2) = 8 \; – \; 20 = -12 \neq 0 \).