Products of brackets in more than one variable

The method for \((a+b)(c+d)\) works for any terms, including terms in two variables and brackets with three terms: multiply every term in the first bracket by every term in the second bracket.

Why: each term must meet each other term once, so the number of products is the number of terms in the first bracket times the number in the second. A two-term bracket times a three-term bracket gives \(2\times3=6\) products before collecting.

Multiplication frame for \((x+2y)(3x-y+4)\)

Method

1. Count the products you expect, such as \(2\times3=6\).
2. Fill a multiplication frame or draw arrows, keeping each sign with its term.
3. Collect like terms; here only the \(xy\)-terms combine.
4. Write the answer tidily: squared terms first, then products such as \(xy\), then single letters, then numbers.

Example

Expand and simplify \((x+2y)(3x-y+4)\).

\(\begin{aligned} (x+2y)(3x-y+4) &= 3x^2-xy+4x+6xy-2y^2+8y \\ &= 3x^2+5xy-2y^2+4x+8y \end{aligned}\)

Expand and simplify \((3p-2q)(p+4q)\).

\(\begin{aligned} (3p-2q)(p+4q) &= 3p^2+12pq-2pq-8q^2 \\ &= 3p^2+10pq-8q^2 \end{aligned}\)

Check: count the terms before collecting (six in the first example). With \(x=y=1\), \((x+2y)(3x-y+4)=3\times6=18\) and \(3+5-2+4+8=18\). A common error is to forget the products with the constant \(+4\).