How this instrument works
Squaring a binomial is FOIL with both factors set equal to each other: (ax + b)² is shorthand for (ax + b)(ax + b), so the second factor's coefficient and constant are just a and b again. Because the two factors are identical, the Outer product (a times b) and the Inner product (b times a) are not two different numbers that happen to add up — they are the same number counted twice, which is exactly why the middle term collapses to the clean 2ab instead of a sum of two unrelated cross-products.
The pattern has a picture behind it. Draw a square with side length (a + b) and cut it along both edges: the pieces are a square of side a, a square of side b, and two identical rectangles measuring a by b sandwiched between them. Add the four areas — a², b², ab, and ab again — and the total is a² + 2ab + b², the geometric proof Euclid recorded in Book II of the Elements roughly 2,300 years before algebraic notation existed to write it symbolically.
This sheet runs one direction only: from a known binomial to its expansion, which makes it the mirror image of the perfect-square-trinomial calculator elsewhere on this site — that one starts from a finished trinomial ax² + bx + c and tests, via the discriminant b² − 4ac, whether it could have come from squaring a binomial at all. Here there is nothing to test, since squaring (ax + b) always produces a perfect-square trinomial by construction. The trap to watch for is treating the square like a plain distribution over addition and writing (a + b)² as a² + b², silently dropping the 2ab cross term that the identical Outer and Inner products actually contribute.
- Enter the coefficient of x into a, in (ax + b)² — for (3x + 5)² that's a = 3.
- Enter the constant into b, in (ax + b)² — continuing the example, b = 5.
- Read x² coefficient for a², the squared leading term.
- Read x coefficient for 2ab, the doubled cross-product that forms the middle term.
- Read Constant term for b², the squared trailing term — the three together give the full expansion.
Worked example — expanding (3x + 5)²
Set a = 3 and b = 5 for the binomial (3x + 5). x² coefficient comes out to a² = 3² = 9. x coefficient is the doubled cross-product, 2ab = 2 × 3 × 5 = 30. Constant term is b² = 5² = 25. Read the three together and (3x + 5)² expands to 9x² + 30x + 25, with nothing left over and nothing to combine by hand.
A substitution catches any slip without redoing the algebra: at x = 1, the original binomial gives (3 + 5)² = 8² = 64, and the expanded trinomial gives 9 + 30 + 25 = 64, the identical figure reached two different ways. The same check at x = 0 is even quicker — both sides reduce to 25, the constant term standing alone.
Questions
What is the formula for squaring a binomial?
(ax + b)² expands to a²x² + 2abx + b² — the leading term squared, twice the product of the two terms, and the trailing term squared. It follows directly from FOIL: (ax + b)(ax + b) has identical Outer and Inner products, a·b and b·a, which combine into 2ab instead of staying as two separate cross-terms.
Why is the middle term 2ab and not just ab?
Because squaring a binomial multiplies it by itself, the Outer and Inner cross-products from FOIL — ordinarily two different numbers, ad and bc — become the same number twice, ab and ab. Adding an identical pair gives 2ab, which is why the middle coefficient always carries that factor of two whenever a binomial is squared rather than multiplied by a different one.
What's the most common mistake when squaring a binomial?
Writing (a + b)² as a² + b² and skipping the middle term entirely, as though squaring distributed the way plain multiplication does. The cross-product 2ab is real and does not vanish; for (3 + 5)², that shortcut wrongly gives 9 + 25 = 34, while the correct expansion, 9 + 30 + 25 = 64, matches 8² exactly.
How does this relate to the perfect-square-trinomial calculator?
They run in opposite directions. This sheet starts from a binomial (ax + b) and builds its expansion, which is always a perfect-square trinomial by construction. The perfect-square-trinomial calculator starts from a finished trinomial ax² + bx + c and tests — via the discriminant b² − 4ac — whether it could have come from squaring a binomial in the first place.
What happens to the expansion when b is negative?
The constant term b² stays positive regardless of sign, since a negative number squared is positive, but the middle term 2ab flips sign along with b. Squaring (x − 4) gives 1x² − 8x + 16: the −8 carries the minus sign through, while the 16 does not — the pattern that makes a minus-signed perfect-square trinomial recognizable on sight.
Does the pattern still hold when a is not 1?
Yes — a² simply scales the leading term the same way it scales every other piece. (2x + 3)² gives 4x² + 12x + 9, since a² = 4, 2ab = 12, and b² = 9; the pattern first-squared, twice-the-product, last-squared holds for any leading coefficient, not only the case where a equals 1.