POP

G6.5 Expanding Triple Brackets

Image · 914 × 270

01 — Explanation

How do I expand three linear brackets?

Never try to multiply all three at once. Expand any two of them first (grid method is safest — a row/column for each term, multiply every cell, then collect), getting a trinomial. Keep that result in brackets, then multiply the trinomial by the remaining bracket (a 3×2 grid), and collect like terms one final time.

Expand $(x+2)(x+4)(x+1)$ :

$(x+2)(x+4) = x^2+6x+8$ .
$(x^2+6x+8)(x+1) = x^3+x^2+6x^2+6x+8x+8$ .
Collect: $x^3+7x^2+14x+8$ .

Exam Tip: always re-bracket the result of step 1 before multiplying by the third bracket — treating $x^2+6x+8$ as a single trinomial, not three separate terms to multiply individually and sloppily.

02 — Explanation

How do I expand a squared or cubed bracket?

$(x-1)^2$ means $(x-1)(x-1)$ — write it out in full before expanding. Never jump to $x^2-1$ ; the middle term is real: $(x-1)(x-1)=x^2-2x+1$ . Similarly $(x+2)^3=(x+2)(x+2)(x+2)$ , expanded in the usual two steps to give $x^3+6x^2+12x+8$ .

Common Mistake: $(x-a)^2 \ne x^2-a^2$ — that's a different identity entirely (difference of two squares, which needs two different brackets, not one squared).

A quick check for $(a+b)^3$ : Pascal's row $1,3,3,1$ gives $(a+b)^3 = a^3+3a^2b+3ab^2+b^3$ directly, without expanding by hand.

03 — Explanation

How can I spot a shortcut before expanding mechanically?

Scan for a difference-of-two-squares pair first — it collapses two brackets into one term instantly. For $(x+4)(x+5)(x-4)$ , pair $(x+4)(x-4) = x^2-16$ rather than expanding left to right; then $(x^2-16)(x+5) = x^3+5x^2-16x-80$ — fewer terms to track than going via the trinomial $x^2+9x+20$ .

04 — Explanation

How do I use this to find the volume of a cuboid?

Volume $=$ length $\times$ width $\times$ height, so three algebraic side lengths multiply into exactly this kind of expansion.

Cuboid with sides $(x+2)$ , $(x+5)$ , $(2x-1)$ cm — show the volume is $2x^3+13x^2+13x-10$ :

$(x+2)(x+5) = x^2+7x+10$ .
$(x^2+7x+10)(2x-1) = 2x^3+14x^2+20x-x^2-7x-10$ .
Collect: $2x^3+13x^2+13x-10$ . ✓.

Exam Tip: for "show that" questions the answer is already given — every mark is for visible method (expand pair → re-bracket → expand with third bracket → collect), not the final line alone.

05 — Explanation

How do I find an unknown coefficient in an identity?

An identity ( $\equiv$ ) holds for every value of $x$ , which opens a shortcut: the constant term of the fully expanded cubic is just the product of the three brackets' constant terms.

Given $(x+3)(x+a)(x+7) \equiv x^3+15x^2+71x+105$ , find $a$ : constants multiply to the constant term: $3 \times a \times 7 = 105 \Rightarrow 21a=105 \Rightarrow a=5$ .

(Double-check by equating the $x^2$ coefficient instead: $(x+3)(x+7)=x^2+10x+21$ , so the $x^2$ term overall is $(10+a)x^2$ , and $10+a=15 \Rightarrow a=5$ — same answer, more general method if the constant trick doesn't apply.)

Exam Tip: the constant-term shortcut only finds one unknown directly from the given constant — if two unknowns are needed, you'll have to equate at least one more coefficient (usually $x^2$ or $x$ ) to get a second equation.

06 — Worked example

Worked exam-style question

Given $(x-2)(x+b)(2x+3) \equiv 2x^3-3x^2-17x+30$ , find $b$ .

Use the constant-term shortcut: $(-2)(b)(3) = 30$ .
$-6b = 30 \Rightarrow b = -5$ .

Whenever an identity question only asks for one unknown and gives the constant term, check the constant-term shortcut first — it's almost always faster than a full expansion.