Expanding triple brackets step by step — Mathematics

Before this

Expanding double brackets
Collecting like terms · coming soon
Volume of cuboids · coming soon

This leads to

Factorising quadratics
The factor theorem · coming soon
Solving cubic equations · coming soon

Three brackets, and a checklist that runs out

Here is the next step up:

(x + 1) (x + 2) (x + 3)

Three brackets look like a new problem. The tempting move is the same one that failed last time: multiply the first terms and the last terms. That gives $x \times x \times x = x^{3}$ and $1 \times 2 \times 3 = 6$ , so $x^{3} + 6$ .

You already know how to test that. Put $x = 10$ . Then the brackets are $11 \times 12 \times 13 = 1716$ . But $x^{3} + 6 = 1000 + 6 = 1006$ . The two are not equal. So a lot is missing, 710 of it.

The last lecture also gave you a warning. FOIL is a checklist for four products: first, outer, inner, last. It works for two brackets with two terms each. With three brackets there is no "inner" and no "outer" any more. The checklist simply runs out.

The reason behind FOIL does not run out. Every term must meet every term. This lecture shows how to keep that rule when there are three brackets, and why the answer comes out the way it does.

The old skill, used twice

Start with a fact about ordinary multiplication. To work out $4 \times 5 \times 6$ , you do not multiply three numbers at once. You multiply two of them, $4 \times 5 = 20$ . Then you multiply the result by the third: $20 \times 6 = 120$ . It does not matter which pair you choose first. $5 \times 6 = 30$ , and $4 \times 30 = 120$ too.

Brackets are numbers in disguise, so the same is true for them. This means that three brackets can be expanded in two steps:

Step 1. Expand any two of the brackets. You get a quadratic: an expression with $x^{2}$ , $x$ and a number.
Step 2. Multiply that quadratic by the third bracket. Then collect like terms.

Step 1 is exactly the last lecture. For $(x + 1) (x + 2)$ , the four products are $x^{2}$ , $2 x$ , $x$ and $2$ . They collect to $x^{2} + 3 x + 2$ . So now the problem is:

(x^{2} + 3 x + 2) (x + 3)

This is where FOIL fails. The first bracket has three terms. Which ones are "first", "outer" and "inner"? The checklist has no answer.

The rectangle with three rows

The last lecture promised that the rectangle never stops working. Here it is. Draw a grid. Put the three terms of the quadratic down the side, one for each row. Put the two terms of the third bracket across the top, one for each column. Then fill each cell with the product of its row and its column.

Step 2 as a grid. The rows are the three terms of x² + 3x + 2. The columns are the two terms of x + 3. Three rows times two columns make six cells, so there are six products.

Three rows and two columns make six cells. So Step 2 has exactly six products. The grid makes it hard to lose one, because an empty cell is easy to see. Here they are in a line:

x^{3} + 3 x^{2} + 3 x^{2} + 9 x + 2 x + 6

Now collect like terms. There is one $x^{3}$ . The two $x^{2}$ terms make $6 x^{2}$ . The two $x$ terms make $11 x$ . The number is 6.

(x + 1) (x + 2) (x + 3) = x^{3} + 6 x^{2} + 11 x + 6

Check with $x = 10$ : $1000 + 600 + 110 + 6 = 1716$ . It matches $11 \times 12 \times 13$ . Look at that number again: 1716. The tempting answer $x^{3} + 6$ kept the 1000 and the 6. It threw away the 600 and the 110, the two middle terms.

That is it. That is the whole idea. Three brackets are two expansions in a row. In the second one, each term of the quadratic meets each term of the third bracket, and a grid keeps count.

Expanding three brackets

1. Expand two of the brackets and collect, to get a quadratic.

2. Multiply each term of the quadratic by each term of the third bracket (six products), then collect like terms.

Three brackets of the form $(x + a)$ always expand to a cubic: an expression whose highest power is $x^{3}$ .

Why the answer is a cubic: a box of eight blocks

Two brackets made a rectangle, because a rectangle has two sides. Three brackets need a shape with three edges: a box. Think of $(x + 1) (x + 2) (x + 3)$ as the volume of a box. Its width is $x + 3$ , its height is $x + 2$ and its depth is $x + 1$ .

A box with width x + 3, height x + 2 and depth x + 1. Cutting every edge at the x mark splits it into eight blocks. There is one x³ cube and three blocks of x² (x by x by 3, x by x by 2, x by x by 1). There are three blocks of x (x by 2 by 3, x by 1 by 3, x by 1 by 2) and one small block of 1 × 2 × 3 = 6.

Cut each edge into its two parts, as in the picture. The box splits into $2 \times 2 \times 2 = 8$ blocks. Each block takes one part from each edge. So each block is one choice: $x$ or the number, from each of the three brackets.

Choose $x$ three times: one block, $x \times x \times x = x^{3}$ . This is why the answer is a cubic.
Choose $x$ twice and one number: three blocks, $1 x^{2}$ , $2 x^{2}$ and $3 x^{2}$ . Together, $(1 + 2 + 3) x^{2} = 6 x^{2}$ .
Choose $x$ once and two numbers: three blocks, $2 \times 3 \times x$ , $1 \times 3 \times x$ and $1 \times 2 \times x$ . Together, $(6 + 3 + 2) x = 11 x$ .
Choose three numbers: one block, $1 \times 2 \times 3 = 6$ .

These are the same four terms as the two-step method gave: $x^{3} + 6 x^{2} + 11 x + 6$ . The two-step method is just a tidy way to visit all eight blocks. Step 1 joins some of them early, which is why Step 2 has six products and not eight.

Now try it with your own numbers. Change the three brackets and watch the six products in the grid.

Explore — the six products of Step 2

Choose a, b and c in (x + a)(x + b)(x + c). Step 1 expands the first two brackets. The grid shows the six products of Step 2. Look for a rule behind the x² term, the x term and the number.

a: 1

b: 2

c: 3

Four expansions of (x + a)(x + b)(x + c), each checked with x = 10:

Brackets	Step 1	Expanded
(x + 1)(x + 2)(x + 3)	x² + 3x + 2	x³ + 6x² + 11x + 6
(x − 2)(x + 1)(x + 4)	x² − x − 2	x³ + 3x² − 6x − 8
(x + 2)(x + 2)(x + 2)	x² + 4x + 4	x³ + 6x² + 12x + 8
(x + 3)(x − 3)(x − 5)	x² − 9	x³ − 5x² − 9x + 45

In every row, the x² term is a + b + c, the x term is ab + bc + ca, and the number is abc. Check the first row with x = 10: 11 × 12 × 13 = 1716 and 1000 + 600 + 110 + 6 = 1716.

Did you see the pattern? For $(x + a) (x + b) (x + c)$ , the box picture explains every term:

(x + a) (x + b) (x + c) = x^{3} + (a + b + c) x^{2} + (ab + b c + c a) x + ab c

The $x^{2}$ term collects the three blocks that use one number, so its coefficient is the sum of the three numbers. The $x$ term collects the three blocks that use two numbers, so its coefficient is the sum of the products of pairs. The number at the end uses all three numbers, so it is their product.

These facts give you two fast checks on any answer. The number at the end must be the product of the three numbers in the brackets. The $x^{2}$ coefficient must be their sum. In the first example, $1 \times 2 \times 3 = 6$ and $1 + 2 + 3 = 6$ . Both match. The checks do not replace the working. They catch most slips before you hand in an answer.

Stop and think

Does the sum check work for $(2 x + 1) (x + 2) (x + 3)$ ? Is the $x^{2}$ coefficient $1 + 2 + 3 = 6$ ?

Show the answer

No. Expand it: Step 1 gives $(2 x + 1) (x + 2) = 2 x^{2} + 5 x + 2$ . Step 2 gives $2 x^{3} + 6 x^{2} + 5 x^{2} + 15 x + 2 x + 6 = 2 x^{3} + 11 x^{2} + 17 x + 6$ . The $x^{2}$ coefficient is 11, not 6.

The box shows why. The $x^{2}$ blocks now use $2 x$ from the first bracket. The blocks $2 x \times x \times 3$ and $2 x \times x \times 2$ count twice as much, so the coefficient is $2 \times 3 + 2 \times 2 + 1 = 11$ . The sum rule is true only when every bracket starts with a plain $x$ .

The product check still works: $1 \times 2 \times 3 = 6$ . The number at the end only ever uses the numbers in the brackets. Check with $x = 10$ : $21 \times 12 \times 13 = 3276$ and $2000 + 1100 + 170 + 6 = 3276$ .

Before you expand: what is $(x + 2)^{3}$ ?

A cube of a bracket is three brackets in disguise. Make a guess first.

Predict, then reveal — cubing a bracket

What is $(x + 2)^{3}$ when it is expanded? Choose before you work it out.

The expansion of (x + 2)³. Many students write x³ + 8, but that cubes each term and loses six of the eight blocks.

(x + 2)³ = (x + 2)(x + 2)(x + 2). Step 1: (x + 2)(x + 2) = x² + 4x + 4. Step 2: x³ + 2x² + 4x² + 8x + 4x + 8 = x³ + 6x² + 12x + 8.

Check with x = 10: 12³ = 1728, and 1000 + 600 + 120 + 8 = 1728.

The numbers 1, 6, 12, 8 are not random. Divide out the powers of 2 and you get 1, 3, 3, 1: one block with no 2, three blocks with one 2, three with two 2s and one with three. Those counts, 1, 3, 3, 1, appear every time you cube a bracket. You will meet them again later, as a row of Pascal's triangle.

Worked examples

Worked Example 1 — A negative number in a bracket

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

Step 1. $(x - 2) (x + 1) = x^{2} + x - 2 x - 2 = x^{2} - x - 2$ . The minus sign stays with the 2 in every product that uses it.

Step 2. Multiply $x^{2} - x - 2$ by $x + 4$ . The six products are $x^{3}$ , $4 x^{2}$ , $- x^{2}$ , $- 4 x$ , $- 2 x$ and $- 8$ .

Collect. The $x^{2}$ terms are $4 x^{2} - x^{2}$ . Read $- x^{2}$ as $- 1 x^{2}$ : the 1 is there even when it is not written. So $4 x^{2} - 1 x^{2} = 3 x^{2}$ . The $x$ terms are $- 4 x - 2 x = - 6 x$ .

Answer: $x^{3} + 3 x^{2} - 6 x - 8$ .

Checks. Sum: $- 2 + 1 + 4 = 3$ . Product: $- 2 \times 1 \times 4 = - 8$ . With $x = 10$ : $8 \times 11 \times 14 = 1232$ , and $1000 + 300 - 60 - 8 = 1232$ .

Worked Example 2 — Choosing the best pair first

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

You may expand any pair first. Look for a pair that makes Step 1 short. $(x + 3)$ and $(x - 3)$ are the same terms with opposite signs, so their middle terms cancel: $(x + 3) (x - 3) = x^{2} - 9$ .

Step 2. $(x^{2} - 9) (x - 5)$ has only four products: $x^{3}$ , $- 5 x^{2}$ , $- 9 x$ and $+ 45$ . The last is $(- 9) \times (- 5)$ , and two negatives make a positive.

Answer: $x^{3} - 5 x^{2} - 9 x + 45$ .

Checks. Sum: $3 - 5 - 3 = - 5$ . Product: $3 \times (- 5) \times (- 3) = 45$ . With $x = 10$ : $13 \times 5 \times 7 = 455$ , and $1000 - 500 - 90 + 45 = 455$ .

Worked Example 3 — A bracket that starts with 2x

Expand $(2 x - 1) (x + 3) (x - 2)$ .

Step 1. $(2 x - 1) (x + 3) = 2 x^{2} + 6 x - x - 3 = 2 x^{2} + 5 x - 3$ .

Step 2. Multiply by $x - 2$ . The six products are $2 x^{3}$ , $- 4 x^{2}$ , $5 x^{2}$ , $- 10 x$ , $- 3 x$ and $+ 6$ .

Collect. $- 4 x^{2} + 5 x^{2} = x^{2}$ , and $- 10 x - 3 x = - 13 x$ . Answer: $2 x^{3} + x^{2} - 13 x + 6$ .

Checks. The sum check does not apply here, because one bracket starts with $2 x$ . The product check still works: $(- 1) \times 3 \times (- 2) = 6$ . The leading term is $2 x \times x \times x = 2 x^{3}$ . With $x = 10$ : $19 \times 13 \times 8 = 1976$ , and $2000 + 100 - 130 + 6 = 1976$ .

Worked Example 4 — The volume of a box

A storage box is $x$ cm high. It is 2 cm wider than it is high, and 5 cm longer than it is high. Write its volume as an expanded expression. Then find the volume when $x = 10$ .

The volume is height × width × length $= x (x + 2) (x + 5)$ .

Step 1. $(x + 2) (x + 5) = x^{2} + 7 x + 10$ .

Step 2. Multiply by $x$ : every term meets the single term $x$ . So the volume is $x^{3} + 7 x^{2} + 10 x$ cm³.

There is no number at the end. That makes sense: the first bracket is just $x$ , so the product check gives $0 \times 2 \times 5 = 0$ .

When $x = 10$ : $1000 + 700 + 100 = 1800$ cm³. Check directly: $10 \times 12 \times 15 = 1800$ cm³.

Three classic mistakes

Trap 1 — Cubing each term

$(x + 2)^{3}$ is not $x^{3} + 8$ . Write it as three brackets and expand: $x^{3} + 6 x^{2} + 12 x + 8$ . Cubing each term keeps two of the eight blocks of the box. The number check catches it at once: $1 2^{3} = 1728$ , not 1008.

Trap 2 — Losing a product or a sign in Step 2

Step 2 has six products, and it is easy to write five. Draw the grid, or count your products before you collect. Signs travel with their terms here too. In Worked Example 1, the term $- x$ gives $- x \times x = - x^{2}$ and $- x \times 4 = - 4 x$ . When you collect, remember that $- x^{2}$ means $- 1 x^{2}$ .

Trap 3 — Using the middle bracket twice

Some students expand $(x + 1) (x + 2)$ and then $(x + 2) (x + 3)$ , and multiply the two answers. That uses $(x + 2)$ twice. It is four brackets, not three, and the answer contains $x^{4}$ . Each bracket is used exactly once: expand one pair, then multiply by the bracket that is left. A quick warning sign: three brackets of the form $(x + a)$ can never give a power higher than $x^{3}$ .

Try it yourself

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

Expand and simplify $(x + 1) (x + 4) (x - 3)$ .

Hint 1
Step 1: expand $(x + 1) (x + 4)$ first.

Hint 2
Step 1 gives $x^{2} + 5 x + 4$ . Now multiply each of its three terms by $x$ and by $- 3$ .

Full answer
Six products: $x^{3}$ , $- 3 x^{2}$ , $5 x^{2}$ , $- 15 x$ , $4 x$ , $- 12$ . Collect: $x^{3} + 2 x^{2} - 11 x - 12$ . Sum check: $1 + 4 - 3 = 2$ . Product check: $1 \times 4 \times (- 3) = - 12$ . With $x = 10$ : $11 \times 14 \times 7 = 1078$ , and $1000 + 200 - 110 - 12 = 1078$ .
Expand $(x - 1)^{3}$ .

Hint 1
Write it as $(x - 1) (x - 1) (x - 1)$ . Step 1 is $(x - 1)^{2}$ .

Hint 2
$(x - 1)^{2} = x^{2} - 2 x + 1$ . Multiply it by $x - 1$ . Watch the signs: a negative times a negative is positive.

Full answer
Six products: $x^{3}$ , $- x^{2}$ , $- 2 x^{2}$ , $+ 2 x$ , $x$ , $- 1$ . Collect: $x^{3} - 3 x^{2} + 3 x - 1$ . The counts 1, 3, 3, 1 appear again, with alternating signs. Sum check: $- 1 - 1 - 1 = - 3$ . Product check: $(- 1)^{3} = - 1$ . With $x = 10$ : $9^{3} = 729$ , and $1000 - 300 + 30 - 1 = 729$ .
Expand and simplify $(3 x + 1) (x - 2) (x + 2)$ . Which pair should you expand first, and why?

Hint 1
Look for two brackets with the same terms and opposite signs.

Hint 2
$(x - 2) (x + 2) = x^{2} - 4$ . Now multiply $3 x + 1$ by $x^{2} - 4$ .

Full answer
Expand $(x - 2) (x + 2)$ first, because its middle terms cancel: $x^{2} - 4$ . Then $(3 x + 1) (x^{2} - 4)$ has four products: $3 x^{3}$ , $- 12 x$ , $x^{2}$ , $- 4$ . In order: $3 x^{3} + x^{2} - 12 x - 4$ . Product check: $1 \times (- 2) \times 2 = - 4$ . With $x = 10$ : $31 \times 8 \times 12 = 2976$ , and $3000 + 100 - 120 - 4 = 2976$ .
A cube has edges of $x$ cm. Every edge is made 1 cm longer. By how much does the volume grow? Use your answer to find $1 1^{3} - 1 0^{3}$ without working out either cube.

Hint 1
The new volume is $(x + 1)^{3}$ . The increase is $(x + 1)^{3} - x^{3}$ .

Hint 2
Expand $(x + 1)^{3}$ with the box picture: one $x^{3}$ , three blocks of $x^{2}$ , three blocks of $x$ , one block of 1.

Full answer
$(x + 1)^{3} = x^{3} + 3 x^{2} + 3 x + 1$ . So the volume grows by $3 x^{2} + 3 x + 1$ cm³. In the picture, the new cube is the old cube plus three thin slabs ( $x$ by $x$ by 1), three thin rods ( $x$ by 1 by 1) and one small corner cube. With $x = 10$ : $300 + 30 + 1 = 331$ . Check: $1 1^{3} - 1 0^{3} = 1331 - 1000 = 331$ .

The bigger idea

There was never a separate method for three brackets. There is one law: every term meets every term. Two brackets make a rectangle with four pieces. Three brackets make a box with eight. Four brackets would make 16 pieces, in a shape we can no longer draw, and the law would still hold. Mathematics often works like this. A rule that seems to belong to one picture is really a rule about structure, and it keeps working after the picture runs out.

The patterns you found, sum, sum of pairs and product, also work backwards. If you know that a cubic is $x^{3} + 6 x^{2} + 11 x + 6$ , its number 6 tells you which numbers could sit in the brackets. Later, the factor theorem uses exactly this to solve cubic equations. First, the next lecture runs the two-bracket case backwards: given a quadratic, find the two brackets it came from.

Key concepts in this lesson

To expand three brackets, expand any two of them to get a quadratic, then multiply that quadratic by the third bracket and collect like terms.
Multiplying a three-term quadratic by a two-term bracket gives six products, and a grid with three rows and two columns makes sure none is lost.
Three brackets are the three edges of a box, which splits into eight blocks, one for each way of choosing x or the number from each bracket.
For (x + a)(x + b)(x + c), the x² coefficient is a + b + c, the x coefficient is ab + bc + ca and the number is abc.
The sum rule fails when a bracket starts with 2x or 3x, but the number at the end is always the product of the numbers in the brackets.
A cubed bracket such as (x + 2)³ is three brackets, never the cube of each term.