2.6 - Expanding and simplifying brackets

Expand longer products without losing a term or a sign. You will multiply a quadratic by a linear expression, handle three brackets and combine expansions, choosing a useful order and collecting only like terms.

Multiplying a quadratic by a linear expression

Why does $(x + 2)(x + 3)(x - 5)$ have no $x^2$ term when expanded? There are plenty of squared terms along the way, so their disappearance needs explaining. We will return to this after building a reliable way to expand longer products.

You already know

The distributive law multiplies every term in a bracket. Collect terms only when their letters and powers match. In a product, $x \times x^2 = x^3$ ; in a sum, $x + x^2$ stays as two unlike terms.

A quadratic has highest power $2$ ; a linear expression has highest power $1$ . Multiplying a three-term quadratic by a two-term linear expression produces six products before you collect anything. It is the same distributive law as before, with a little more bookkeeping.

A three-term expression times a two-term expression

For real numbers or algebraic expressions $A$ , $B$ , $C$ , $D$ and $E$ ,

(A + B + C)(D + E) = AD + BD + CD + AE + BE + CE

The letters may represent signed terms; every term in one bracket multiplies every term in the other.
Not on the formulae sheet: learn it.

You can organise the products in a grid or in rows. The grid below shows the first example's six products. Each entry includes its sign.

Multiply by	$x^2$	$2x$	$3$
$2x$	$2x^3$	$4x^2$	$6x$
$1$	$x^2$	$2x$	$3$

Worked example 1: six products

Question. Expand and simplify $(x^2 + 2x + 3)(2x + 1)$ .

Plan. Multiply the quadratic by $2x$ , then by $1$ , and collect like terms.

\begin{aligned} (x^2 + 2x + 3)(2x + 1) &= 2x(x^2 + 2x + 3) + (x^2 + 2x + 3) && \text{split the linear factor} \\ &= 2x^3 + 4x^2 + 6x + x^2 + 2x + 3 && \text{six products} \\ &= 2x^3 + 5x^2 + 8x + 3 && \text{collect equal powers} \end{aligned}

Why this step? Keeping the two rows separate until all six products are written makes an omitted term much easier to spot.

Answer. $2x^3 + 5x^2 + 8x + 3$ .

Check. At $x = 1$ , the original gives $6 \times 3 = 18$ ; the expansion gives $2 + 5 + 8 + 3 = 18$ . Its leading term is also $x^2 \times 2x = 2x^3$ .

Change only the constant in the linear factor from $+1$ to $-1$ . Every product in the second row now changes sign.

Worked example 2: a negative row

Question. Expand and simplify $(x^2 + 2x + 3)(2x - 1)$ .

Plan. Multiply by $2x$ and by $-1$ before collecting terms.

\begin{aligned} (x^2 + 2x + 3)(2x - 1) &= 2x(x^2 + 2x + 3) - (x^2 + 2x + 3) && \text{distribute} \\ &= 2x^3 + 4x^2 + 6x - x^2 - 2x - 3 && \text{negate all three terms} \\ &= 2x^3 + 3x^2 + 4x - 3 && \text{collect equal powers} \end{aligned}

Why this step? Subtracting the quadratic subtracts its whole value, so all three terms receive a minus sign.

Answer. $2x^3 + 3x^2 + 4x - 3$ .

Check. At $x = 1$ , both forms give $6$ : $6(2 - 1) = 2 + 3 + 4 - 3$ . At $x = 0$ , both give $-3$ .

A substitution is a useful error detector, although agreement at one value alone does not prove two expressions are identical. The distributive working establishes the result for every value.

Watch out

Wrong: $-(x^2 + 2x + 3) = -x^2 + 2x + 3$ . The minus sign is multiplication by $-1$ , giving $-x^2 - 2x - 3$ .

Now change the coefficients. In this check, one group of like terms cancels completely.

Expanding three linear factors

With three factors, multiply any two first, simplify that result and then multiply by the remaining factor.

The associative law allows the choice of pair. Usually you choose the pair with the easiest coefficients or a useful cancellation. Do not try to multiply all three factors in one leap: it is too easy to miss mixed products.

Two familiar products are especially helpful. Multiplying $(a + b)(a - b)$ gives $a^2 - ab + ab - b^2$ , so the middle terms cancel. Squaring $(a + b)$ , however, means multiplying $(a + b)(a + b)$ , where the middle terms add.

Useful two-bracket identities

For any real numbers $a$ and $b$ ,

\begin{aligned} (a + b)(a - b) &= a^2 - b^2 \\ (a + b)^2 &= a^2 + 2ab + b^2 \\ (a - b)^2 &= a^2 - 2ab + b^2 \end{aligned}

An identity is an equality true for every allowed value of its variables; the symbol $\equiv$ may be used to emphasise this.
Not on the formulae sheet: learn it.

These are shortcuts justified by distribution, not extra rules for changing powers. In particular, a square of a sum includes the middle term.

Worked example 3: multiply two, then the third

Question. Expand and simplify $(x + 2)(x + 3)(x + 5)$ .

Plan. Multiply the first two factors, then multiply their quadratic by $x + 5$ .

\begin{aligned} (x + 2)(x + 3) &= x^2 + 3x + 2x + 6 && \text{four products} \\ &= x^2 + 5x + 6 && \text{collect like terms} \\ (x^2 + 5x + 6)(x + 5) &= x^3 + 5x^2 + 6x + 5x^2 + 25x + 30 && \text{six products} \\ &= x^3 + 10x^2 + 31x + 30 && \text{collect again} \end{aligned}

Why this step? Once two factors form a quadratic, you are back to the method in the previous part.

Answer. $x^3 + 10x^2 + 31x + 30$ .

Check. At $x = 1$ , the product is $3 \times 4 \times 6 = 72$ , and the expansion is $1 + 10 + 31 + 30 = 72$ .

Change just the sign in the final factor, from $x + 5$ to $x - 5$ . The first quadratic stays exactly the same; the second row changes sign. This is the opening puzzle.

Worked example 4: the squared terms disappear

Question. Expand and simplify $(x + 2)(x + 3)(x - 5)$ .

Plan. Reuse $x^2 + 5x + 6$ , then distribute $x - 5$ .

\begin{aligned} (x + 2)(x + 3)(x - 5) &= (x^2 + 5x + 6)(x - 5) && \text{multiply the first pair} \\ &= x^3 + 5x^2 + 6x - 5x^2 - 25x - 30 && \text{distribute } x \text{ and } -5 \\ &= x^3 - 19x - 30 && \text{the squared terms cancel} \end{aligned}

Why this step? The terms $+5x^2$ and $-5x^2$ have opposite coefficients, so their sum is zero. A cubic need not contain every lower power.

Answer. $x^3 - 19x - 30$ .

Check. At $x = 1$ , the original gives $3 \times 4 \times (-4) = -48$ , and the expansion gives $1 - 19 - 30 = -48$ .

Here is another neat cancellation. The product of three consecutive integers centred on $n$ is $(n - 1)n(n + 1) = n(n^2 - 1) = n^3 - n$ . For example, $9 \times 10 \times 11 = 10^3 - 10 = 990$ . Pairing the two outside factors reveals a shortcut that multiplying left to right hides.

Watch out

Wrong: $(x - 3)^2 = x^2 + 9$ . Write the repeated bracket: $(x - 3)(x - 3) = x^2 - 6x + 9$ . Both mixed products are needed.

For the next product, the outside factors are the easy pair. Changing the final constant again gives a different cancellation to notice.

Combining expanded expressions

When a question contains more than one product, expand each product separately before you combine their like terms.

This keeps three jobs distinct: multiplying within each product, applying the sign or multiplier between products, and collecting the final coefficients. A negative leading coefficient is perfectly acceptable: keep the sign your calculation gives, rather than changing it to fit an expected shape.

Worked example 5: adding a second expression

Question. Expand and simplify fully $(x^2 - x + 2)(2x + 3) + 2(x^3 - 2x^2 + x - 1)$ .

Plan. Expand the product and double the second bracket, then add coefficients of each power.

\begin{aligned} (x^2 - x + 2)(2x + 3) &= 2x^3 - 2x^2 + 4x + 3x^2 - 3x + 6 && \text{six products} \\ &= 2x^3 + x^2 + x + 6 && \text{collect the first product} \\ 2(x^3 - 2x^2 + x - 1) &= 2x^3 - 4x^2 + 2x - 2 && \text{double every term} \\ \text{total} &= 4x^3 - 3x^2 + 3x + 4 && \text{add like coefficients} \end{aligned}

Why this step? A separate line for the second bracket prevents its multiplier from being applied to only its first term.

Answer. $4x^3 - 3x^2 + 3x + 4$ .

Check. At $x = 1$ , the original is $2 \times 5 + 2(-1) = 8$ , and the answer is $4 - 3 + 3 + 4 = 8$ .

Worked example 6 changes only the sign between the two expressions. Use a multiplier of $-2$ for the entire second bracket.

Worked example 6: subtracting a whole expression

Question. Expand and simplify fully $(x^2 - x + 2)(2x + 3) - 2(x^3 - 2x^2 + x - 1)$ .

Plan. Keep the first expansion, distribute $-2$ to the second bracket and collect.

\begin{aligned} (x^2 - x + 2)(2x + 3) &= 2x^3 + x^2 + x + 6 && \text{the first product} \\ -2(x^3 - 2x^2 + x - 1) &= -2x^3 + 4x^2 - 2x + 2 && \text{multiply every term by } -2 \\ \text{total} &= (2 - 2)x^3 + (1 + 4)x^2 && \text{combine equal powers} \\ &\quad + (1 - 2)x + (6 + 2) \\ &= 5x^2 - x + 8 && \text{the cubic terms cancel} \end{aligned}

Why this step? Subtracting $2$ times a negative term produces a positive term, so both $-2x^2$ and $-1$ contribute positively here.

Answer. $5x^2 - x + 8$ .

Check. At $x = 1$ , the original is $2 \times 5 - 2(-1) = 12$ , and the result is $5 - 1 + 8 = 12$ .

The opening puzzle and this example show the same point: cancellation can remove a whole power. Keep the zero coefficient out of the final answer, but keep enough working to show where it went.

Exam tip

Write the individual expansions before the final collected answer so your products and signs can be checked. For “expand and simplify fully”, finish with collected terms rather than leaving a product or several uncollected rows.

The final check varies the expressions while keeping the subtraction of a whole multiple. Attend to the sign of each term in the second bracket.

Quick Recap

Every term in one factor multiplies every term in the other; a three-term factor times a two-term factor gives six products.
For three brackets, choose a useful pair, multiply it, then multiply by the remaining factor.
A squared bracket has mixed terms. A difference of squares loses them by cancellation.
Expand separate products clearly, distribute any outside negative and collect only equal powers.

The opening product is $(x + 2)(x + 3)(x - 5) = x^3 - 19x - 30$ : its $+5x^2$ and $-5x^2$ terms cancel.

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2.6 - Expanding and simplifying brackets

Multiplying a quadratic by a linear expression

Expanding three linear factors

Combining expanded expressions

Quick Recap

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