Multiplying two binomials and beyond

To multiply $(a + b)(c + d)$, treat the first bracket as a single thing and distribute it across the second:

$(a + b)(c + d) = (a+b) \cdot c + (a+b) \cdot d = ac + bc + ad + bd.$

So we get four terms: every term of the first bracket times every term of the second. The trick is to do this without missing a pair.

Concept

FOIL. A widely used memory device for (F)irst, (O)uter, (I)nner, (L)ast , the four products in $(a+b)(c+d)$:

• First: $a \cdot c$
• Outer: $a \cdot d$
• Inner: $b \cdot c$
• Last: $b \cdot d$

Sum: $ac + ad + bc + bd$. (Note the order in the sum doesn't matter, but you should not skip any of the four.)

Example: $(x + 3)(x + 5)$.

• F: $x \cdot x = x^2$.
• O: $x \cdot 5 = 5x$.
• I: $3 \cdot x = 3x$.
• L: $3 \cdot 5 = 15$.

Sum: $x^2 + 5x + 3x + 15 = x^2 + 8x + 15$.

Beyond two binomials. The same idea works for any polynomial product. Each term in the first bracket multiplies each term in the second bracket. If the brackets have $m$ and $n$ terms, you get $m \cdot n$ products before collecting like terms.

Example: $(x + 2)(x^2 - 3x + 4)$.

• $x \cdot x^2 = x^3$.
• $x \cdot (-3x) = -3x^2$.
• $x \cdot 4 = 4x$.
• $2 \cdot x^2 = 2x^2$.
• $2 \cdot (-3x) = -6x$.
• $2 \cdot 4 = 8$.

Collect: $x^3 + (-3x^2 + 2x^2) + (4x - 6x) + 8 = x^3 - x^2 - 2x + 8$.

Triple brackets. Multiply two at a time. For $(x+1)(x+2)(x+3)$, first do $(x+1)(x+2) = x^2 + 3x + 2$, then multiply by $(x+3)$.

Geometric picture. Just as $a(b+c)$ corresponds to cutting a rectangle once, $(a+b)(c+d)$ corresponds to cutting a rectangle once horizontally and once vertically , four sub-rectangles, areas $ac, ad, bc, bd$. This is the "box method", a great visual aid.

Worked examples

Example 1. Expand $(x + 4)(x - 2)$.

• $x \cdot x = x^2$.
• $x \cdot (-2) = -2x$.
• $4 \cdot x = 4x$.
• $4 \cdot (-2) = -8$.
• Sum: $x^2 + 2x - 8$.

Example 2. Expand $(2a + 3)(3a - 5)$.

• $2a \cdot 3a = 6a^2$.
• $2a \cdot (-5) = -10a$.
• $3 \cdot 3a = 9a$.
• $3 \cdot (-5) = -15$.
• Sum: $6a^2 - a - 15$.

Example 3. Expand $(x - 3)(x^2 + 2x - 4)$.

• $x \cdot x^2 = x^3$, $x \cdot 2x = 2x^2$, $x \cdot (-4) = -4x$.
• $-3 \cdot x^2 = -3x^2$, $-3 \cdot 2x = -6x$, $-3 \cdot (-4) = 12$.
• Sum: $x^3 - x^2 - 10x + 12$.

Example 4. Use the box method to expand $(a + 2b)(3a - b)$.

• Sum: $3a^2 + (-ab + 6ab) - 2b^2 = 3a^2 + 5ab - 2b^2$.

Try it yourself

1. Expand $(x + 6)(x + 7)$.
2. Expand $(2x - 1)(3x + 4)$.
3. Expand $(a - 5)(a + 5)$.
4. Expand $(x^2 + 1)(x + 2)$.
5. Expand $(p + q)(p - q + 1)$.
6. Expand $(x + 1)(x + 2)(x + 3)$.
7. Expand $(2a + 3b)^2$ by FOIL (or your favourite method).
8. The length of a rectangle is $(x + 5)$ and the width is $(x - 3)$. Find its area.

Activity

Box drawings. For each binomial product above, draw the corresponding $2 \times 2$ (or $2 \times 3$) box, label the rows and columns, fill in the cells, then sum. You will rarely "miss a term" again once your eye is trained to fill every cell of a box.