Squaring a binomial always makes three terms.
(a + b)² = a² + 2ab + b² · (a − b)² = a² − 2ab + b²
(a + b)² is never a² + b²
The two ab rectangles are real area. Try it with numbers: (3 + 4)² = 49, but 9 + 16 = 25. The missing 24 is exactly 2ab.
- Home
- Grade 7-12 Math Review
- Special Products
Special Products
Patterns worth recognising, because they turn a multiplication into a glance.
Expand
(3x − 5)²
(a − b)² = a² − 2ab + b², so (3x)² − 2(3x)(5) + 5² = 9x² − 30x + 25. A is the single most common algebra error in school mathematics — squaring each term and forgetting the middle. Test it with x = 1: (3 − 5)² = 4, and 9 − 30 + 25 = 4 ✓ while 9 − 25 = −16 ✗. C gets the sign of the middle term wrong; the last term is positive either way, because a negative squared is positive.
Sum times difference: the middle cancels.
(a + b)(a − b) = a² − b²
(5x + 2)(5x − 2)
same terms, opposite signs
= 25x² − 10x + 10x − 4
FOIL, the long way
= 25x² − 4
the middle terms cancel
This is the one case where the answer really does have two terms — and it only happens when the signs are opposite. With matching signs you get the three-term square from step 1.
A common term: add the constants, multiply them.
(x + a)(x + b) = x² + (a + b)x + ab
(x + 3)(x + 8) = x² + 11x + 24
(x − 4)(x + 9) = x² + 5x − 36
(x − 2)(x − 7) = x² − 9x + 14
Watch the middle in the second one: −4 + 9 = 5, and the last term is −36 because a negative times a positive is negative. Signs carry through both slots.
This pattern read backwards is exactly how trinomials are factored, so learning it now does double duty.
Cubes and trinomial squares.
| Pattern | Expands to |
|---|---|
| (a + b)³ | a³ + 3a²b + 3ab² + b³ |
| (a − b)³ | a³ − 3a²b + 3ab² − b³ |
| (a + b + c)² | a² + b² + c² + 2ab + 2ac + 2bc |
(x + 3)³
a = x, b = 3
x³ + 3x²(3) + 3x(9) + 27
substitute
= x³ + 9x² + 27x + 27
tidy
In (a − b)³ the signs alternate, because an odd power of a negative stays negative. The 3s in the middle do not change.
These are mental-arithmetic tricks too.
99 × 101
awkward by hand
= (100 − 1)(100 + 1)
sum and difference
= 10,000 − 1 = 9,999
in your head
51²
(50 + 1)² = 2,500 + 100 + 1
= 2,601
47 × 53
(50 − 3)(50 + 3) = 2,500 − 9
= 2,491
Two more patterns: the cube product, and squaring to find a hidden sum.
(a − b)(a² + ab + b²) = a³ − b³ · (a + b)(a² − ab + b²) = a³ + b³
Multiply (x − 2)(x² + 2x + 4).
(x − 2)(x² + 2x + 4)
a = x, b = 2: the cube pattern
= x³ + 2x² + 4x − 2x² − 4x − 8
the long way, to see why
= x³ − 8
every middle term cancels
Spot it by the bracket: the trinomial is a² ± ab + b², with the middle sign opposite to the binomial's. The product is then just two cubes.
If x − 1x = 3, find x² + 1x².
(x − 1x)² = 3²
square both sides
x² − 2 + 1x² = 9
middle term: 2 · x · (1x) = 2, with a minus
x² + 1x² = 11
add 2 to both sides
The middle term is still there
Squaring each part on its own gives x² + 1x² = 9. But (a − b)² always has a −2ab, and here ab = x · (1x) = 1, so the middle is −2. Moving it across gives 11, not 9.
Combining two squares: (2a + b)² + (2a − b)²
= (4a² + 4ab + b²) + (4a² − 4ab + b²)
expand each with step 1
= 8a² + 2b²
the ab terms cancel
Variations: a negative first term, a fraction, a squared term.
Same patterns, unusual-looking inputs.
(−x − 6)²
= (−1)²(x + 6)²
−x − 6 = −(x + 6)
= x² + 12x + 36
the minus vanishes when squared
A binomial with both terms negative squares to the same answer as the positive one.
(x + 12)²
= x² + 2(x)(12) + 14
a = x, b = 12
= x² + x + 14
2 · 12 = 1
Fractions in the second term: the middle term is 2ab, and here it comes out a whole x.
(x² − 3)(x² + 3)
= (x²)² − 3²
sum and difference, with a = x²
= x⁴ − 9
(x²)² = x⁴
The a in a pattern can be any term, even a power. Square it as a whole: (x²)² = x⁴, not x².
On the exam
Exams ask you to expand quickly, to pick out which pattern fits, or to use one for arithmetic such as 49 × 51 = 2,500 − 1 = 2,499. The harder items hide the pattern, like x + 1x given and x² + 1x² asked — square what you are given. For a 10-second check, put x = 1 into the question and your answer; they must match.
Now you try
Practice problem
Expand
(4a + 3b)(4a − 3b)
The signs are opposite, so this is the sum-and-difference pattern: (4a)² − (3b)² = 16a² − 9b². The middle terms −12ab and +12ab cancel. B keeps a middle term that does not survive. C forgets to square the coefficients — it is 4², not 4.
Hint: check the signs first. Opposite signs mean the middle vanishes and only two terms survive.