Three patterns worth recognising on sight — FOIL still works, it is just slower.
Expand
(2x + 5)2
(a + b)² = a² + 2ab + b², so (2x)² + 2(2x)(5) + 5² = 4x² + 20x + 25. D drops the middle term entirely — squaring a sum is not squaring each piece, and that missing 20x costs more marks than any other slip in algebra.
1
Square of a binomial: it is never two terms.
(a + b)² does not equal a² + b². There is always a middle.
(2x + 5)2 = (2x + 5)(2x + 5)
first2x · 2x = 4x²square the front
outer2x · 5 = 10xhalf the middle
inner5 · 2x = 10xthe other half
last5 · 5 = 25square the back
4x² + 20x + 25
The classic error
Writing (2x + 5)² = 4x² + 25 loses the 20x entirely. The outer and inner
products are identical, which is exactly why the middle term is doubled.
2
The shortcut, once you trust it.
(a + b)2 = a2 + 2ab + b2
a = 2x
a² = 4x²
b = 5
b² = 25
2ab
2 · 2x · 5 = 20x
Same answer, three multiplications instead of four.
For a minus, only the middle changes sign: (a − b)² = a² − 2ab + b².
The last term is a square, so it is always positive.
3
Difference of squares: the middle cancels itself.
(3x − 4)(3x + 4)
first3x · 3x = 9x²keep
outer3x · 4 = +12xwatch this
inner−4 · 3x = −12xand this
last−4 · 4 = −16keep
+12x − 12x = 0 — the middle vanishes.
9x² − 16
4
The three patterns, side by side.
Pattern
Expands to
Middle term
(a + b)²
a² + 2ab + b²
doubled, positive
(a − b)²
a² − 2ab + b²
doubled, negative
(a − b)(a + b)
a² − b²
gone
Two brackets that match give a middle. Two that differ only in sign do not.
Read these backwards and you have three factoring rules for free — which is
how you spot 9x² − 16 as (3x − 4)(3x + 4) in one glance.
5
See it as an area.
A square of side (a + b), cut into four pieces.
Two ab rectangles — that is where the 2ab comes from.
The whole square is a² + ab + ab + b², which tidies to a² + 2ab + b².
6
Bigger brackets: treat a block as one letter.
Expand (a + b + 4)(a + b − 4). The brackets differ only in the sign of the 4, so let A = a + b.
(A + 4)(A − 4) = A² − 16
difference of squares
A² = (a + b)² = a² + 2ab + b²
now open the block
= a² + 2ab + b² − 16
keep the 2ab
Square of three terms: every term squared, plus twice every pair. (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac. Expand (x − 2y + 3)².
Piece
Work
Result
squares
x², (−2y)², 3²
x² + 4y² + 9
pair x, −2y
2(x)(−2y)
−4xy
pair −2y, 3
2(−2y)(3)
−12y
pair x, 3
2(x)(3)
6x
x² + 4y² + 9 − 4xy − 12y + 6x
Two squares multiplied: pair the brackets first. (a + 3)²(a − 3)² = [(a + 3)(a − 3)]²
= (a² − 9)² = a⁴ − 18a² + 81 — one difference of squares, then one square of a binomial.
7
Variations: both terms negative, the brackets in the other order, a fraction coefficient.
Expand (−3a − 2)².
(−3a − 2)² = (−1)²(3a + 2)²
take out the −1; its square is 1
9a² + 2(3a)(2) + 4 = 9a² + 12a + 4
a = 1: (−5)² = 25 = 9 + 12 + 4 ✓
Different: two minus signs give a + middle; the square of a bracket never cares about an overall sign.
Expand (2 − 3a)(3a + 2).
(2 − 3a)(2 + 3a)
reorder: the shared term is 2
2² − (3a)² = 4 − 9a²
a = 1: (−1)(5) = −5 = 4 − 9 ✓
Different: the “a” of a² − b² is whichever term has the same sign in both brackets — here the 2, not the 3a.
9a² − 4 is the sign flip
Reading the brackets as (3a − 2)(3a + 2) by habit gives 9a² − 4, the exact negative
of the answer. Find the term whose sign changes; that one is subtracted.
Expand (½x − 4)².
(½x)² = ¼x²
square the ½ too
2(½x)(4) = 4x
the 2 and the ½ cancel
¼x² − 4x + 16
x = 2: (−3)² = 9 = 1 − 8 + 16 ✓
Different: a fraction in front shrinks the square term but not the middle — ¼x² beside 4x looks odd and is right.
8
Use the identity to evaluate.
The exam often hands you pieces of an identity and asks for another piece. You never need x itself.
53² − 47² = (53 − 47)(53 + 47) = 6 × 100 = 600
a² − b² backwards: no squaring
102 × 98 = (100 + 2)(100 − 2) = 10000 − 4 = 9996
a product that straddles a round number
x + 1x = 4 → x² + 2 + 1x² = 16
square it: 2(x)(1x) = 2
x² + 1x² = 16 − 2 = 14
the middle term is a plain 2
x + y = 9, x² + y² = 41 → 81 = 41 + 2xy
(x + y)² = x² + 2xy + y²
2xy = 40 → xy = 20
check: 4 and 5 ✓
Do not forget the 2 — or to halve
Squaring x + 1x and writing x² + 1x² = 16 drops the middle term. With x − 1x
the middle is −2, so you add 2 instead. And when 2xy = 40, the product
is 20, not 40.
On the exam
Special products appear as “find the product” with one dropped middle term among
the choices, and as evaluate-without-solving items like the ones above. Square
what you are given, then subtract what you know. Test a finished expansion with
x = 2, as in the check below.
✓
Check: substitute a number.
Let x = 2 and test both answers.
(2x + 5)²(4 + 5)² = 81the question
4x² + 20x + 2516 + 40 + 25 = 81matches ✓
(3x − 4)(3x + 4)2 · 10 = 20the question
9x² − 1636 − 16 = 20matches ✓
Remember the flow
Same signs · Double the middle · Opposite signs · No middle
When in doubt, FOIL it out.
Now you try
Practice problem
Expand
(5m − 3)2
(a − b)² = a² − 2ab + b², giving 25m² − 2(5m)(3) + 9 = 25m² − 30m + 9. The final term is +9 because (−3)² is positive — B leaves it negative, which no square ever is.
Hint: with a minus inside, only the middle term turns
negative — the last term is a square, so it stays positive. And the middle is
2ab, not ab: double the product.
