Factorising into Double Brackets

If we expand \( (x + 7)(x + 10) \) we get \( x^2 + 10x + 7x + 70 = x^2 + 17x + 70 \), which is not the same as \( x^2 + 7x + 10 \). The error comes from the “direct placement” misconception — thinking the numbers in the brackets are just the coefficient of \( x \) and the constant.

We need two numbers that multiply to give 10 and add to give 7. The correct factorisation is \( (x + 2)(x + 5) \). This can be visualised using an area model:

Expanding \( (x + 3)(x + 4) \) gives \( x^2 + 4x + 3x + 12 = x^2 + 7x + 12 \). This has \( +7x \), not \( -7x \). Expanding \( (x \; – \; 3)(x \; – \; 4) \) gives \( x^2 \; – \; 4x \; – \; 3x + 12 = x^2 \; – \; 7x + 12 \) ✓. The constant term is \( +12 \) in both cases because negative × negative = positive, but the crucial difference is the sign of the middle term.

When the constant is positive and the coefficient of \( x \) is negative, both numbers in the brackets must be negative. This gives a positive product (\( (-3) \times (-4) = +12 \)) and a negative sum (\( (-3) + (-4) = -7 \)).

\( x^2 \; – \; 25 \) can be written as \( x^2 + 0x \; – \; 25 \), so we need two numbers that multiply to give \( -25 \) and add to give 0. The pair \( +5 \) and \( -5 \) works: \( 5 \times (-5) = -25 \) and \( 5 + (-5) = 0 \). So \( x^2 \; – \; 25 = (x + 5)(x \; – \; 5) \). Expanding confirms: \( x^2 \; – \; 5x + 5x \; – \; 25 = x^2 \; – \; 25 \) ✓.

This is a difference of two squares — any expression of the form \( a^2 \; – \; b^2 \) factorises as \( (a + b)(a \; – \; b) \).

Expanding \( (2x + 4)(x + 5) \) gives \( 2x^2 + 10x + 4x + 20 = 2x^2 + 14x + 20 \) — this does match the original expression. However, the factorisation is not complete because the first bracket still contains a common factor: \( 2x + 4 = 2(x + 2) \).

The correct approach is to extract the common factor first: \( 2x^2 + 14x + 20 = 2(x^2 + 7x + 10) = 2(x + 2)(x + 5) \).

Example: \( x^2 + 3x + 2 = (x + 1)(x + 2) \)

Another: \( x^2 + 8x + 15 = (x + 3)(x + 5) \)

Creative: \( x^2 + 100x + 99 = (x + 1)(x + 99) \) — uses an extreme factor pair, but still works.

Trap: \( x^2 + 5x \; – \; 6 \) — a student might offer this because most of the terms are positive, but \( (x + 6)(x \; – \; 1) \) has one positive and one negative number in the brackets. When the constant is negative, the signs in the brackets must be different.

Example: \( x^2 + 2x \; – \; 8 = (x + 4)(x \; – \; 2) \)

Another: \( x^2 \; – \; x \; – \; 12 = (x \; – \; 4)(x + 3) \)

Creative: \( x^2 \; – \; 99x \; – \; 100 = (x \; – \; 100)(x + 1) \)

Trap: \( x^2 \; – \; 5x + 6 \) — a student might think the subtraction sign means the constant must be negative, but \( x^2 \; – \; 5x + 6 = (x \; – \; 2)(x \; – \; 3) \) has a positive constant.

Example: \( x^2 + 7x + 12 = (x + 3)(x + 4) \)

Another: \( x^2 + 13x + 12 = (x + 1)(x + 12) \)

Creative: \( x^2 \; – \; 8x + 12 = (x \; – \; 2)(x \; – \; 6) \) — same constant of 12, but negative middle term means both bracket numbers are negative.

Trap: \( x^2 + 12x + 12 \) — a student might assume that if the constant is 12, you can just put 12 as the coefficient of \( x \) too. But to factorise this, we’d need two integers that multiply to 12 and add to 12. No factor pair sums to 12.

Example: \( 2x^2 + 7x + 3 = (2x + 1)(x + 3) \)

Another: \( 3x^2 + 10x + 8 = (3x + 4)(x + 2) \)

Creative: \( 6x^2 + x \; – \; 1 = (3x \; – \; 1)(2x + 1) \) — the coefficient of \( x \) is just 1.

Trap: \( 2x^2 + 8x + 6 \) — You might think this counts, but all three numbers are divisible by 2. It simplifies to \( 2(x^2 + 4x + 3) \), so it violates the rule “does not have a common factor”.

This is sometimes true. For example, \( x^2 + 5x + 6 = (x + 2)(x + 3) \). But \( x^2 + 3x + 1 \) cannot be factorised into brackets with integer values, because no pair of integers multiplies to 1 and adds to 3.

This is always true. Factorising is the exact reverse of expanding. If expanding \( (x + a)(x + b) \) produces \( x^2 + px + q \), then by definition, \( x^2 + px + q \) factorises back to \( (x + a)(x + b) \).

This is sometimes true. If \( c \) is positive, \( m \) and \( n \) must share the same sign. The sign of the middle term (\( b \)) determines whether they are both positive or both negative.

• If \( b > 0 \), both are positive: \( x^2 + 5x + 6 = (x+2)(x+3) \)
• If \( b < 0 \), both are negative: \( x^2 - 5x + 6 = (x-2)(x-3) \)

Students who think “positive constant means positive brackets” will always fail when the middle term is negative.

The sum of two squares never factorises into double brackets (over the real numbers). Consider \( x^2 + 9 \). We would need numbers that multiply to \( +9 \) and add to 0. The pairs for 9 are (1,9) summing to 10, or (-1,-9) summing to -10, or (3,3) summing to 6, or (-3,-3) summing to -6. None sum to 0.

The student has made the “direct placement” misconception. They simply placed the visible numbers into the brackets. The correct factorisation is \( (x + 2)(x + 7) \), because \( 2 \times 7 = 14 \) and \( 2 + 7 = 9 \).

The student is confusing “no written coefficient” with zero. When an \( x \) term exists but has no number in front of it, the coefficient is 1 (because \( 1 \times x = x \)).

If the coefficient were 0, the \( x \) term would not be there at all (e.g. \( x^2 – 12 \)). We need numbers that multiply to -12 and add to 1. The answer is \( (x+4)(x-3) \).

The student has the correct answer but faulty reasoning (the “ignore the middle term” misconception). It only worked here by luck. If the expression was \( x^2 + 5x + 3 \), their logic would still give \( (x+1)(x+3) \), which expands to \( x^2 + 4x + 3 \), not \( 5x \).

The student correctly identified the numbers 2 and 9 but failed to check the signs.

Because the constant is positive and the middle term is negative, both numbers must be negative. Correct answer: \( (x-2)(x-9) \).

The student skipped the common factor check. While \( (3x + 3)(x + 3) \) does expand to the original expression, the first bracket \( (3x+3) \) can still be simplified. The fully factorised form is \( 3(x+1)(x+3) \).