Expanding Double Brackets

Substitution proves this quickly: let \( x = 1 \), then \( (1+3)(1+5) = 24 \), but \( 1^2 + 15 = 16 \).

Visually (see diagram), this error is like calculating the area of the top-left and bottom-right corners but ignoring the other two. The correct expansion includes every combination: \( x^2 + 5x + 3x + 15 = x^2 + 8x + 15 \).

Try \( x = 2 \): \( (2 + 3)^2 = 25 \), but \( 2^2 + 9 = 13 \). The error comes from “distributing” the square to each term, as though squaring works like doubling. But squaring means multiplying the bracket by itself.

\( (x + 3)^2 = (x + 3)(x + 3) = x^2 + 3x + 3x + 9 = x^2 + 6x + 9 \). You must include the sum of the two cross-multiplications (\( 3x + 3x \)).

The constant term comes from multiplying the two number parts: \( (-2) \times (-7) = +14 \). The key rule is that negative × negative = positive.

The full expansion is \( x^2 \; – \; 9x + 14 \). While the middle term (\( -9x \)) is negative because you are adding two negative amounts (\(-2x\) and \(-7x\)), the constant is positive because you are multiplying them.

Expand both. \( (2x + 1)(x + 3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3 \).
And \( (x + 1)(2x + 3) = 2x^2 + 3x + 2x + 3 = 2x^2 + 5x + 3 \).

They share the same \( x^2 \) and constant terms, but the middle terms differ. By swapping the constants (1 and 3) between the brackets, you change which number gets multiplied by \( 2x \) and which gets multiplied by \( x \).

Example: \( (x + 3)(x \; – \; 3) = x^2 \; – \; 9 \)

Another: \( (x + 7)(x \; – \; 7) = x^2 \; – \; 49 \)

Creative: \( (2x + 1)(2x \; – \; 1) = 4x^2 \; – \; 1 \) — this works for non-monics too.

Trap: \( (x + 3)(x \; – \; 4) \) — this gives \( x^2 \; – \; x \; – \; 12 \), which has a \( -x \) term. Just having one \(+\) and one \(-\) isn’t enough; the numbers must be identical (difference of two squares).

Example: \( (x + 2)(x \; – \; 5) = x^2 \; – \; 3x \; – \; 10 \)

Another: \( (x \; – \; 1)(x + 4) = x^2 + 3x \; – \; 4 \)

Creative: \( (x \; – \; 7)(x + 1) = x^2 \; – \; 6x \; – \; 7 \) — constant is \( -7 \), equal to one of the original numbers.

Trap: \( (x \; – \; 2)(x \; – \; 5) = x^2 \; – \; 7x + 10 \) — constant is \( +10 \). Students assume two minuses make the answer negative, but for multiplication, negative × negative = positive.

Example: \( (x + 3)(x \; – \; 2) = x^2 + x \; – \; 6 \) — since \( 3 + (-2) = 1 \).

Another: \( (x + 5)(x \; – \; 4) = x^2 + x \; – \; 20 \)

Creative: \( (2x + 1)(x + 0) = 2x^2 + x \). This also has a coefficient of 1 for x!

Trap: \( (x + 1)(x + 1) = x^2 + 2x + 1 \) — coefficient of x is 2. Students often confuse the number 1 in the bracket with the resulting coefficient.

Example: \( (2x + 1)(x + 2) = 2x^2 + 5x + 2 \)

Another: \( (3x + 1)(x + 1) = 3x^2 + 4x + 1 \)

Creative: \( (100x + 1)(x + 1) = 100x^2 + 101x + 1 \) — extreme coefficients.

Trap: \( (2x + 3)(x \; – \; 1) \) — having positive x terms isn’t enough; the constant must be positive too, which requires both number parts to have the same sign.

This is always true for this specific form. Expanding gives: \( x^2 + (a+b)x + ab \).

Note for deeper thinking: This rule breaks immediately if the coefficient of \( x \) is not 1. For example, in \( (2x+1)(x+3) \), the middle term is \( 7x \), not \( (1+3)x \). The “sum of constants” rule is a special case for monic quadratics only.

Usually true. \( (x + 1)(x + 2) = x^2 + 3x + 2 \) (3 terms).

FALSE case: When \( a = -b \), the middle terms cancel out. \( (x + 4)(x \; – \; 4) = x^2 \; – \; 16 \) (only 2 terms). This is the difference of two squares.

The \( x^2 \) term is always positive. The constant term is \( a \times b \), and since negative × negative = positive, the constant is also positive. Only the middle term is negative.

Example: \( (x \; – \; 2)(x \; – \; 3) = x^2 \; – \; 5x + 6 \). Two of the three terms are positive!

TRUE case: \( (x + 2)(x + 3) = x^2 + 5x + 6 \) — all present.

FALSE case 1: \( (x + 2)(x \; – \; 2) = x^2 \; – \; 4 \) — no \( x \) term.

FALSE case 2: \( x(x + 5) = x^2 + 5x \) — if we allow \( a=0 \), there is no constant term.

The student only multiplied the “matching” pairs (firsts and lasts). They missed the two cross-multiplications: \( x \times 3 = 3x \) and \( 4 \times x = 4x \). The correct expansion requires every term in the first bracket to multiply every term in the second.

Correct answer: \( x^2 + 7x + 12 \).

This “double the number” shortcut is dangerous. It only works for simple cases like \( (x+a)^2 \). It fails immediately if there is a coefficient in front of x.

For example, if the student tries to “double the 4” for \( (2x + 4)^2 \), they would get \( 4x^2 + 8x + 16 \). But the actual middle term is \( 16x \) (from \( 8x + 8x \)). The middle term comes from summing the cross-products, not just doubling the constant.

The student made a sign error on the constant term. They treated the multiplication as \( 4 \times 3 \), ignoring the negative sign on the 4.

The constant comes from \( (-4) \times (+3) = -12 \).
Correct answer: \( x^2 \; – \; x \; – \; 12 \).

The student thinks \( x \times x = 2x \). This confuses multiplication with addition. \( x \times x = x^2 \).

Expanding double brackets with \( x \) terms in both must produce a quadratic (highest power is \( x^2 \)). If there is no \( x^2 \), the first multiplication step is wrong.