What one looks like.
(a + b)² = a² + 2ab + b²
(a − b)² = a² − 2ab + b²
first term
a perfect square
last term
a perfect square, always positive
middle term
exactly 2ab
The last term can never be negative — it is something squared.
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Perfect Square Trinomials
Three terms that came from squaring a bracket — and how to tell at a glance.
Factor
4x² − 12x + 9
4x² = (2x)² and 9 = 3². Check the middle: 2(2x)(3) = 12x, and the sign is minus, so it is (2x − 3)². A squares the 4 instead of recognising 4x² as (2x)² — the coefficient has to come inside the bracket already rooted.
The test, in three moves.
x² + 10x + 25
the candidate
√x² = x, √25 = 5
a = x, b = 5
2ab = 2(x)(5) = 10x
matches the middle ✓
= (x + 5)²
plus, because the middle is plus
Root the ends, double their product, and see whether it lands on the middle term. If not, it is an ordinary trinomial — factor it the usual way.
With coefficients.
4x² − 12x + 9
the candidate
√4x² = 2x, √9 = 3
a = 2x, b = 3
2ab = 2(2x)(3) = 12x
matches, apart from the sign
= (2x − 3)²
minus, from the middle term
Only the middle carries the sign
Both (2x − 3)² and (2x + 3)² end in +9. The middle term is the only clue to which one you have.
Fractions pass the same test: x²9 + 2x3 + 1 has ends x3 and 1, and 2(x3)(1) = 2x3 ✓, so it is (x3 + 1)². And a square with area 4x² − 12x + 9 has side 2x − 3, so its perimeter is 4(2x − 3) = 8x − 12.
Filling in a missing term.
| Given | Missing | How |
|---|---|---|
| x² + 14x + __ | 49 | half of 14, squared |
| x² + __ + 36 | 12x | 2 × 1 × 6 |
| __ − 20x + 25 | 4x² | 20 ÷ (2 × 5) = 2, so 2x |
| 9x² + 30x + __ | 25 | 30 ÷ (2 × 3) = 5 |
“Halve the middle and square it” only works when the first term is a bare x².
Why it matters elsewhere.
completing the square
you are building one on purpose
the discriminant
b² − 4ac = 0 means it is one
repeated roots
(x + 5)² = 0 gives x = −5 twice
x² + 10x + 25 = 0
test the discriminant
100 − 4(1)(25) = 0
exactly zero
one repeated root
the parabola just touches the axis
A zero discriminant and a perfect square trinomial are the same fact seen from two directions.
Variations: an odd middle, a hidden common factor, a fourth power.
Complete x² + 5x + ___ as a perfect square.
half of 5 = 52
odd, so a fraction appears
(52)² = 254
x² + 5x + 254 = (x + 52)²
Different: the missing term need not be a whole number — keep the fraction; do not round to 6.
Factor 3x² − 24x + 48.
3(x² − 8x + 16)
3x² and 48 are not squares — take out 3
2(x)(4) = 8x ✓
now the test works
3(x − 4)²
x = 1: 3 − 24 + 48 = 27 = 3(9) ✓
Different: the square is hiding behind a common factor; the ends only look wrong until it is out.
Factor completely: x⁴ − 18x² + 81.
√x⁴ = x², √81 = 9, 2(x²)(9) = 18x² ✓
the same test, one level up
(x² − 9)²
not finished
(x − 3)²(x + 3)²
x = 1: 64 = (4)(16) ✓
Different: the bracket that comes out is itself a difference of squares, so “completely” needs one more move.
An unknown in the middle: two signs, two answers.
For which values of m is x² + (m + 3)x + 25 a perfect square trinomial? The square could be (x + 5)² or (x − 5)², so the middle may be +10x or −10x.
m + 3 = ±2(1)(5) = ±10
both signs are allowed
m + 3 = 10 → m = 7
gives (x + 5)²
m + 3 = −10 → m = −13
gives (x − 5)²
When the unknown sits in both the middle and the end, use the rule from Step 4 as an equation: with a bare x², (half the middle coefficient)² = last term. For which k is x² + 2(k − 1)x + (k + 5) a perfect square?
half the middle coefficient = k − 1
2(k − 1) ÷ 2
(k − 1)² = k + 5
square it, set equal to the end
k² − 3k − 4 = 0 → (k − 4)(k + 1) = 0
a quadratic in k
k = 4 or k = −1
x² + 6x + 9 and x² − 4x + 4 ✓
Taking only the plus sign
Writing m + 3 = 10 and stopping gives one answer of two. Unless the question says “positive”, a perfect square can have either sign in the middle — check both.
On the exam
The usual item is “for what k is this a perfect square”, sometimes dressed as the area of a square whose side you then need. Root both ends, double the product, and remember the ± — or use b² − 4ac = 0 from Step 5, which gives the same equation.
Check: expand them back.
(x + 5)²
x² + 5x + 5x + 25
= x² + 10x + 25 ✓
(2x − 3)²
4x² − 6x − 6x + 9
= 4x² − 12x + 9 ✓
numeric test
x = 1: 4 − 12 + 9 = 1
(2 − 3)² = 1 ✓
Now you try
Practice problem
Factor
Factor 25x² + 20x + 4
25x² = (5x)², 4 = 2², and the middle term 2(5x)(2) = 20x matches exactly. So (5x + 2)². Always verify that middle term — without it, B and C look every bit as convincing.