E2.2 - Several brackets and variables (Extended) | Cambridge IGCSE International Mathematics

Is $(2x+3)^2$ really just $4x^2+9$ ? A small numerical check says otherwise. This Extended E2.2 lesson develops complete expansions with two variables, squared brackets and products of three brackets, keeping every cross-product and collecting powers carefully.

Two-variable products

The opening claim fails already at $x=1$ : $(2+3)^2=25$ , while $4+9=13$ . Missing products are the problem. First make sure every product is included when two different letters appear.

In $(2x+y)(x+3y)$ , treat $2x$ , $y$ , $x$ and $3y$ as whole terms. There are still four products. Use $x\times x=x^2$ , $y\times y=y^2$ and $yx=xy$ , then collect only identical variable parts.

Mixed-variable term

A mixed-variable term contains more than one variable, such as $6xy$ . The terms $6xy$ and $xy$ are like terms; neither is like $x^2$ or $y^2$ .

Changing the number of letters changes how the products look, not the distributive method.

Worked example 1: four products with two letters

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

Plan. Form four products, writing both middle products in $xy$ order.

\begin{aligned} (2x+y)(x+3y)&=2x^2+6xy+xy+3y^2\\ &=2x^2+7xy+3y^2 &&\text{combine the matching middle terms} \end{aligned}

Answer. $2x^2+7xy+3y^2$ .

Check. At $x=1$ , $y=2$ , the original is $4\times7=28$ . The answer gives $2+14+12=28$ .

Now change only the sign of $3y$ . Both products involving it change sign.

Worked example 2: signed mixed-variable terms

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

Plan. Multiply the signed terms, then combine the $xy$ -terms.

\begin{aligned} (2x+y)(x-3y)&=2x^2-6xy+xy-3y^2\\ &=2x^2-5xy-3y^2 &&\text{since }-6+1=-5 \end{aligned}

Answer. $2x^2-5xy-3y^2$ .

Check. At $x=1$ , $y=2$ , the brackets give $4\times(-5)=-20$ . The answer gives $2-10-12=-20$ .

Writing a standard order helps you see what can combine: $x^2$ , then $xy$ , then $y^2$ .

Watch out

$6xy+xy=7xy$ , not $7x^2y^2$ . The powers change during multiplication; collecting the resulting terms adds only their coefficients.

Keep the two middle products visible until they are combined.

Squared brackets

Squaring means multiplying an expression by itself. It does not mean squaring its separate terms. Write the bracket twice so that all four products appear.

Squared binomial

A binomial has two terms. Its square is the product of two identical brackets: $(A+B)^2=(A+B)(A+B)$ .

The cross-products $AB$ and $BA$ are equal, so together they give $2AB$ . This is the term missing from the opening claim.

Worked example 3: a squared sum

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

Plan. Write two identical brackets, multiply all four pairs, then collect.

\begin{aligned} (2x+3)^2&=(2x+3)(2x+3) &&\text{meaning of the square}\\ &=4x^2+6x+6x+9\\ &=4x^2+12x+9 &&\text{two equal cross-products} \end{aligned}

Answer. $4x^2+12x+9$ .

Check. At $x=1$ , the original is $5^2=25$ ; the expansion is $4+12+9=25$ . The missing 12 explains the failed opening claim.

The next bracket changes only the joining sign. Both cross-products become negative, but the final product of two negatives stays positive.

Worked example 4: a squared difference

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

Plan. Multiply two copies of $(2x-3)$ , including both negative cross-products.

\begin{aligned} (2x-3)^2&=(2x-3)(2x-3)\\ &=4x^2-6x-6x+9 &&\text{last product is }(-3)(-3)\\ &=4x^2-12x+9 \end{aligned}

Answer. $4x^2-12x+9$ .

Check. At $x=1$ , the original is $(-1)^2=1$ . The expansion gives $4-12+9=1$ , which is non-negative as a real square must be.

These expansions give two useful identities.

Squaring two terms

\begin{aligned} (A+B)^2&=A^2+2AB+B^2\\ (A-B)^2&=A^2-2AB+B^2 \end{aligned}

They hold for all real $A$ and $B$ . Square each whole term, and include twice their product.

Not on the supplied formula lists: learn them.

You can always return to two brackets rather than relying on a remembered sign pattern.

Watch out

$(2x-3)^2\ne4x^2-9$ . A squared bracket and a conjugate product have different cross-products. Here the two cross-products add; they do not cancel.

Apply the same meaning of “square” to a bracket with two variables.

Three brackets

For three brackets, multiply two first, simplify their product, then multiply that whole result by the remaining bracket. Multiplication can be regrouped, so any pair may be chosen first. Choose the pair whose expansion is easiest to manage.

A quadratic with three terms times a two-term bracket produces six products. The highest power comes from $x^2\times x=x^3$ . Collect cubic, squared, linear and constant terms in that order.

Worked example 5: three linear brackets

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

Plan. Expand the first pair, then distribute both terms of the third bracket over the quadratic.

\begin{aligned} (x+1)(x+2)&=x^2+3x+2 &&\text{expand and collect the first pair}\\ (x^2+3x+2)(x+3)&=x^3+3x^2+3x^2+9x+2x+6\\ &=x^3+6x^2+11x+6 &&\text{collect matching powers} \end{aligned}

Answer. $x^3+6x^2+11x+6$ .

Check. At $x=1$ , the original is $2\times3\times4=24$ , and the expansion gives $1+6+11+6=24$ . At $x=0$ , both forms give 6.

The next example changes only the coefficient of $x$ in the last bracket. The first-pair calculation can be reused, but every term it contains must meet $2x$ .

Worked example 6: a coefficient in the last bracket

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

Plan. Use $x^2+3x+2$ for the first pair and multiply each of its terms by both $2x$ and 3.

\begin{aligned} (x+1)(x+2)(2x+3)&=(x^2+3x+2)(2x+3)\\ &=2x^3+3x^2+6x^2+9x+4x+6\\ &=2x^3+9x^2+13x+6 &&\text{collect each power separately} \end{aligned}

Answer. $2x^3+9x^2+13x+6$ .

Check. At $x=1$ , the original gives $2\times3\times5=30$ ; the expansion gives $2+9+13+6=30$ . The leading term $2x^3$ also agrees with $x\times x\times2x$ .

Clear intermediate working protects the method when there is a later arithmetic slip. Write the first-pair result and the uncollected products before the final line.

Watch out

Do not add the three pairwise products. $(A+B)(C+D)(E+F)$ is one multiplication of three whole factors. Multiplying the first pair leaves that entire result still to be multiplied by the third factor.

Choose a first pair, finish it, then bring in the third bracket.

Quick Recap

Two variables still use one product for every pair of terms.
A squared bracket is two identical brackets; include both cross-products.
For three brackets, finish one pair before multiplying by the remaining factor.
Collect only terms with exactly the same variable powers.

The opening square is $4x^2+12x+9$ : its two cross-products supply the missing $12x$ .

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E2.2 - Several brackets and variables (Extended)

Two-variable products

Squared brackets

Three brackets

Quick Recap

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