Expanding double brackets — FOIL method — Mathematics

Before this

Expanding brackets · coming soon
Collecting like terms · coming soon
Rationalising the denominator

This leads to

Expanding triple brackets
Factorising quadratics

A shortcut that does not work

Here is a calculation that many students get wrong, and not only students:

(x + 3) (x + 5)

The tempting answer is to multiply the first parts and the last parts: $x \times x = x^{2}$ and $3 \times 5 = 15$ , so $x^{2} + 15$ . It looks neat. It is wrong.

You can prove that it is wrong in ten seconds. Put a number in place of $x$ , say $x = 10$ . Then $(x + 3) (x + 5) = 13 \times 15 = 195$ . But $x^{2} + 15 = 100 + 15 = 115$ . The two are not equal, so they are not the same expression. Something is missing, and this lecture finds it.

Every term meets every term

You already know how to expand a single bracket: $3 (x + 5) = 3 x + 15$ . The 3 multiplies everything inside the bracket. Two brackets work the same way. Treat the whole first bracket as one number that multiplies the second bracket:

(x + 3) (x + 5) = x (x + 5) + 3 (x + 5)

Now each part is a single bracket:

x (x + 5) + 3 (x + 5) = x^{2} + 5 x + 3 x + 15

There are four products, not two. Each term in the first bracket multiplies each term in the second. The two in the middle, $5 x$ and $3 x$ , are like terms, so they collect:

(x + 3) (x + 5) = x^{2} + 8 x + 15

Check with $x = 10$ : $100 + 80 + 15 = 195$ . It matches $13 \times 15$ . The tempting answer missed the middle terms, the 80.

A picture of the four products

Why must there be four? Think of $(x + 3) (x + 5)$ as the area of a rectangle with height $x + 3$ and width $x + 5$ . Split each side into its two parts, and the rectangle splits into four pieces.

The rectangle with sides x + 3 and x + 5 splits into four pieces: x², 5x, 3x and 15. The tempting answer x² + 15 keeps only two corners and forgets the two strips.

The whole area is the sum of the four pieces: $x^{2} + 5 x + 3 x + 15$ . The answer $x^{2} + 15$ keeps the top-left square and the bottom-right corner and throws away the two strips. That is it. That is the whole idea. Expanding two brackets means adding up every piece of the rectangle: each term times each term.

You have done this before without the name. In the lectures on surds, you expanded brackets such as $(3 + 2) (3 - 2)$ in exactly this way: four products, then collect.

FOIL: a checklist, not a reason

Many books use the word FOIL to make sure none of the four products is forgotten:

F — First: the first term of each bracket, $x \times x = x^{2}$ .
O — Outer: the two terms on the outside, $x \times 5 = 5 x$ .
I — Inner: the two terms on the inside, $3 \times x = 3 x$ .
L — Last: the last term of each bracket, $3 \times 5 = 15$ .

FOIL is useful, but it is only a checklist. The reason is the rectangle: every term meets every term. Some teachers draw arcs joining the pairs, and some use a grid. They are all the same idea. Hold on to the reason, because FOIL stops working when a bracket has three terms, and the rectangle never does. The next lecture uses exactly that.

Signs travel with their terms

A minus sign belongs to the term after it. In $(2 x - 1)$ , the second term is $- 1$ , not 1. Every product that uses it carries the minus sign.

Try the grid below. Choose the numbers in $(x + b) (x + d)$ , including negative ones, and watch the four pieces.

Explore — the four products of two brackets

Move b and d. Watch the four pieces of the grid, the expansion, and the number check with x = 10. Look for the rule behind the middle term and the last term.

b: 3

d: 5

Four expansions, each with its grid of products and a check at x = 10:

Brackets	Four products	Expanded
(x + 3)(x + 5)	x², 5x, 3x, 15	x² + 8x + 15
(x − 2)(x + 6)	x², 6x, −2x, −12	x² + 4x − 12
(x − 4)(x − 4)	x², −4x, −4x, 16	x² − 8x + 16
(x + 5)(x − 5)	x², −5x, 5x, −25	x² − 25

In every row, the middle term is (b + d)x and the last term is b × d. Check the first row with x = 10: 13 × 15 = 195 and 100 + 80 + 15 = 195.

Did you find the pattern? For $(x + b) (x + d)$ , the middle term is always $(b + d) x$ , and the last term is always $b \times d$ . That is no accident: the two strips are $b x$ and $d x$ , and the corner is $b d$ . And when $b$ and $d$ are opposites, such as 5 and −5, the middle terms cancel completely. You will use both facts again when you learn to factorise.

Stop and think

Is $(x + 3) (x + 5) = x^{2} + 15$ ever true?

Show the answer

For one value only. The correct expansion is $x^{2} + 8 x + 15$ , which equals $x^{2} + 15$ only when $8 x = 0$ , that is, when $x = 0$ . An expansion must be true for every value of $x$ . That is why one number check, with a value such as $x = 10$ , is enough to catch the mistake: a correct expansion passes every check, and a wrong one fails almost all of them. (Avoid checking with $x = 0$ or $x = 1$ , which can hide mistakes.)

Worked examples

Worked Example 1 — Two positive brackets

Expand $(x + 3) (x + 5)$ .

Four products: $x^{2}$ , $5 x$ , $3 x$ , $15$ . Collect: $x^{2} + 8 x + 15$ .

Check with $x = 10$ : $13 \times 15 = 195$ and $100 + 80 + 15 = 195$ .

Worked Example 2 — A negative term and a coefficient

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

Four products: $2 x \times x = 2 x^{2}$ , $2 x \times 4 = 8 x$ , $(- 1) \times x = - x$ , $(- 1) \times 4 = - 4$ .

So $2 x^{2} + 8 x - x - 4 = 2 x^{2} + 7 x - 4$ . The minus sign stayed with the 1 in both products that used it.

Check with $x = 10$ : $19 \times 14 = 266$ and $200 + 70 - 4 = 266$ .

Worked Example 3 — The middle terms vanish

Expand $(3 x - 2) (3 x + 2)$ .

Four products: $9 x^{2}$ , $6 x$ , $- 6 x$ , $- 4$ . The middle terms cancel: $6 x - 6 x = 0$ .

So $(3 x - 2) (3 x + 2) = 9 x^{2} - 4$ .

Check with $x = 10$ : $28 \times 32 = 896$ and $900 - 4 = 896$ . This pattern, the same two terms with opposite signs between them, has its own lecture later in this topic. The surd conjugates followed the same pattern.

Worked Example 4 — A squared bracket

Expand $(x - 4)^{2}$ .

A square means the bracket times itself: $(x - 4) (x - 4)$ . Four products: $x^{2}$ , $- 4 x$ , $- 4 x$ , $16$ . The last one is $(- 4) \times (- 4) = + 16$ .

So $(x - 4)^{2} = x^{2} - 8 x + 16$ .

Check with $x = 10$ : $6^{2} = 36$ and $100 - 80 + 16 = 36$ .

Worked Example 5 — Brackets with surds

Expand and simplify $(2 + 3) (1 + 3)$ .

Four products: $2$ , $23$ , $3$ , $3 \times 3 = 3$ .

Collect: $2 + 3 = 5$ , and $23 + 3 = 33$ . So the answer is $5 + 33$ .

The method does not care whether the terms are letters, numbers or surds. Every term meets every term.

Now look for a mistake in someone else's work.

Spot the error — a sign that goes missing

A student expands $(2 x - 3) (x + 4)$ . Exactly one line contains a mistake. Tap it.

A student's expansion of (2x − 3)(x + 4). Line 3 is wrong.

Line 1. First: 2x × x = 2x². Correct.
Line 2. Outer: 2x × 4 = 8x. Correct.
Line 3. Inner: 3 × x = 3x. Wrong: the term is −3, so the product is −3x.
Line 4. Last: −3 × 4 = −12. Correct.
Line 5. Collecting: 2x² + 8x + 3x − 12 = 2x² + 11x − 12. The collecting is done correctly, but it carries the mistake from line 3. The right answer is 2x² + 5x − 12.

Three classic mistakes

Trap 1 — First times first, last times last

$(x + 3) (x + 5)$ is not $x^{2} + 15$ . That uses two of the four products and forgets the two strips of the rectangle. Always write all four products before collecting.

Trap 2 — Losing a minus sign

In $(2 x - 3) (x + 4)$ , the second term is $- 3$ , not 3. Every product that uses it carries the minus sign: $- 3 \times x = - 3 x$ and $- 3 \times 4 = - 12$ . If both terms are negative, the product is positive, as in $(- 4) \times (- 4) = 16$ .

Trap 3 — Squaring each term

$(x + 4)^{2}$ is not $x^{2} + 16$ . Write it as $(x + 4) (x + 4)$ and expand: $x^{2} + 8 x + 16$ . A number check shows it: with $x = 10$ , $1 4^{2} = 196$ , but $100 + 16 = 116$ .

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Expand and simplify $(x + 6) (x - 2)$ .

Hint 1
Write the four products. The second term of the second bracket is $- 2$ .

Hint 2
$x^{2}$ , $- 2 x$ , $6 x$ , $- 12$ .

Full answer
$x^{2} - 2 x + 6 x - 12 = x^{2} + 4 x - 12$ . Check with $x = 10$ : $16 \times 8 = 128$ and $100 + 40 - 12 = 128$ .
Expand and simplify $(2 x + 1) (3 x - 5)$ .

Hint 1
The first product is $2 x \times 3 x = 6 x^{2}$ .

Hint 2
The middle products are $- 10 x$ and $3 x$ .

Full answer
$6 x^{2} - 10 x + 3 x - 5 = 6 x^{2} - 7 x - 5$ . Check with $x = 10$ : $21 \times 25 = 525$ and $600 - 70 - 5 = 525$ .
Expand $(x + 5)^{2}$ . Then use your answer to work out $10 5^{2}$ without a calculator.

Hint 1
Write $(x + 5) (x + 5)$ and expand.

Hint 2
$105 = x + 5$ when $x = 100$ .

Full answer
$(x + 5)^{2} = x^{2} + 10 x + 25$ . With $x = 100$ : $10000 + 1000 + 25 = 11025$ . So $10 5^{2} = 11025$ .
A square garden has sides of $x$ metres. It is redesigned: one pair of sides is made 3 m longer, and the other pair 3 m shorter. Does the area get bigger, smaller, or stay the same? By how much? Does the answer depend on $x$ ?

Hint 1
The new area is $(x + 3) (x - 3)$ . Expand it.

Hint 2
Compare your answer with the old area, $x^{2}$ .

Full answer
$(x + 3) (x - 3) = x^{2} - 3 x + 3 x - 9 = x^{2} - 9$ . The new area is always 9 m² smaller, whatever the size of the garden (as long as $x$ is more than 3). Check with $x = 10$ : $13 \times 7 = 91$ , and $100 - 9 = 91$ . In the picture, the new rectangle gains a 3 by $(x - 3)$ strip and loses a 3 by $x$ strip, a difference of a 3 by 3 square.

The bigger idea

Expanding brackets is the distributive law, the same law that makes $3 (x + 5) = 3 x + 15$ and makes long multiplication work. When you work out $13 \times 15$ as $10 \times 10 + 10 \times 5 + 3 \times 10 + 3 \times 5$ , you are expanding $(10 + 3) (10 + 5)$ . Algebra just lets the 10 be any number.

This unit is about reading algebra in both directions. Expanding turns a product of brackets into a sum of terms. The next lectures extend it to three brackets, and then run it backwards: given $x^{2} + 8 x + 15$ , find the two brackets it came from. The pattern you found in the explorer, middle term $b + d$ and last term $b d$ , is the key that makes that possible.

Key concepts in this lesson

To expand two brackets, multiply each term in the first bracket by each term in the second, giving four products, then collect like terms.
The four products are the four pieces of a rectangle whose sides are the two brackets.
FOIL (first, outer, inner, last) is a checklist for the four products, and the reason behind it is that every term must meet every term.
A sign belongs to the term after it, so a negative term makes every product that uses it follow the sign rules.
For (x + b)(x + d), the expansion is x² + (b + d)x + bd, and a number check such as x = 10 quickly confirms an expansion.