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Math Academy SG

Special Algebraic Identities

Secondary 2 G3 Mathematics · Study guide

Understand the three special identities through area, expansion and factorisation, then apply them to numerical shortcuts and algebraic reasoning.

Concept → worked examples → practice14 min read · practice linked below
A square divided into four areasA square of side a plus b is split into an a by a square, two a by b rectangles and a b by b square. The total area is a squared plus two a b plus b squared. Positive lengths a and b are shown in the ratio three to two.ababa²ababb²Whole square: (a + b)²

Recognise a structure before multiplying

An identity is true for every allowed value of its variables. Special algebraic identities are shortcuts for products that repeat the same structure. They also work backwards: recognising a perfect square or a difference of squares helps you factorise.

First be comfortable multiplying brackets and collecting like terms. If that is still unfamiliar, start with expansion and factorisation.

(a+b)2=a2+2ab+b2,(a−b)2=a2−2ab+b2(a+b)^2=a^2+2ab+b^2,\qquad (a-b)^2=a^2-2ab+b^2

(a+b)(a−b)=a2−b2.(a+b)(a-b)=a^2-b^2.

The letters stand for whole terms, including their coefficients. In (3x−2y)2(3x-2y)^2, the two terms are 3x3x and 2y2y. These identities hold for all real values, even though an area model uses positive lengths.

Why the middle term is there

Multiplying (a+b)(a+b)(a+b)(a+b) gives four products: a2a^2, abab, another abab, and b2b^2. The two equal rectangles explain the term 2ab2ab. Squaring the terms separately would leave out both rectangles.

A square divided into four areasA square of side a plus b is split into an a by a square, two a by b rectangles and a b by b square. The total area is a squared plus two a b plus b squared. Positive lengths a and b are shown in the ratio three to two.ababa²ababb²Whole square: (a + b)²
The two rectangles have the same area, even though their orientations differ.

Worked example 1 — a square with a negative middle term

Expand (3x−4)2(3x-4)^2.

  1. Match a=3xa=3x, b=4b=4 with (a−b)2(a-b)^2. The minus sign determines the middle term.
  2. Square each whole term and calculate twice the product: (3x−4)2=(3x)2−2(3x)(4)+42=9x2−24x+16.(3x-4)^2=(3x)^2-2(3x)(4)+4^2=9x^2-24x+16.
  3. The constant is positive because (−4)(−4)=16(-4)(-4)=16. It is the middle term that is negative.

Check: multiplying (3x−4)(3x−4)(3x-4)(3x-4) gives 9x2−12x−12x+169x^2-12x-12x+16, the same result. Substituting x=2x=2 gives 44 on both sides; this catches many errors, but a single substitution alone does not prove an identity.

Opposite middle terms cancel

In (a+b)(a−b)(a+b)(a-b), the cross-products are −ab-ab and +ab+ab, which cancel. The result is a difference of two squares. In contrast, a2+b2a^2+b^2 does not factorise as (a+b)(a−b)(a+b)(a-b).

Worked example 2 — calculate efficiently without a calculator

Find 103×97103\times97.

Both numbers are 3 away from 100, so use the common centre:

103×97=(100+3)(100−3)=1002−32=10000−9=9991.103\times97=(100+3)(100-3)=100^2-3^2=10000-9=9991.

Why this is useful: the distance from the centre must be the same in the two brackets. For 103 and 96, this particular centre would not give a conjugate pair.

Check: 103(100−3)=10300−309=9991103(100-3)=10300-309=9991, agreeing with the identity method.

Worked example 3 — factorise completely

Factorise 18x2−5018x^2-50.

  1. Take out the common numerical factor first: 18x2−50=2(9x2−25)18x^2-50=2(9x^2-25).
  2. Recognise 9x2=(3x)29x^2=(3x)^2 and 25=5225=5^2.
  3. Apply the difference-of-squares identity backwards: 18x2−50=2(3x+5)(3x−5).18x^2-50=2(3x+5)(3x-5).

The factor 22 must remain outside both brackets. This is a factorisation, so there is no equation to solve and no need to set either bracket equal to zero.

Check: 2(3x+5)(3x−5)=2(9x2−25)=18x2−502(3x+5)(3x-5)=2(9x^2-25)=18x^2-50.

Use identities without finding each variable

Sometimes a question gives a sum or product, but not the separate values. An identity can connect the information directly to what is asked.

Worked example 4 — reason from a sum and a product

Given p+q=9p+q=9 and pq=14pq=14, find p2+q2p^2+q^2 and (p−q)2(p-q)^2.

  1. Expand the square of the known sum: (p+q)2=p2+2pq+q2(p+q)^2=p^2+2pq+q^2.
  2. Rearrange, then substitute: p2+q2=(p+q)2−2pq=92−2(14)=53.p^2+q^2=(p+q)^2-2pq=9^2-2(14)=53.
  3. Now use the square of the difference: (p−q)2=p2+q2−2pq=53−28=25.(p-q)^2=p^2+q^2-2pq=53-28=25.

Check: 7 and 2 have sum 9 and product 14; 72+22=537^2+2^2=53 and (7−2)2=25(7-2)^2=25. We cannot conclude p−q=5p-q=5 unless we also know p>qp>q: either 55 or −5-5 has square 25.

Explore the missing rectangles

Change the two positive lengths. Compare the whole square with the sum of only the two small squares. The gap is exactly the area of the two rectangles.

Interactive square area modelFor a equals 3 and b equals 2, areas 9, 6, 6 and 4 fill a square of area 25.9664a = 3b = 232Side 5; area 25
Lengths use a common scale. Labels inside each region give its area in square units.

(3 + 2)² = 25, but 3² + 2² = 13. The missing 12 is 2 × 3 × 2.

Common mistakes and useful checks

  • Missing the coefficient square: (3x)2=9x2(3x)^2=9x^2, not 3x23x^2.
  • Missing the middle term: (a−b)2(a-b)^2 has three terms; a2−b2a^2-b^2 has two.
  • Assuming any three terms form a square: x2+6x+9=(x+3)2x^2+6x+9=(x+3)^2, because 6x=2(x)(3)6x=2(x)(3). The expression x2+6x+4x^2+6x+4 fails this middle-term check.
  • Stopping too soon: extract a common factor, then look for identities inside the remaining expression. Expand your final factors to check.

Practise the identities

Use the expansion and factorisation question bank for identity questions and worked solutions. Identify the whole terms first, choose an identity, and check by expansion.