Binomial Formulas

The first binomial formula squares a sum: (a + b)² = a² + 2ab $+ b^{2}$. The second squares a difference: (a − b)² = a² − 2ab $+ b^{2}$; only the mixed term changes sign, $b^{2}$ stays positive. The third multiplies sum times difference: $(a + b)(a - b) = a^{2} - b^{2}$; here the mixed terms +ab and −ab cancel each other. All three are special cases of expanding, but so frequent that you must know them by heart. Number check: $(3 + 2)^{2} = 25 = 9 + 12 + 4$ ✓, $(3 - 2)^{2} = 1 = 9 - 12 + 4$ ✓ and $(3 + 2)(3 - 2) = 5 = 9 - 4$ ✓.

Because the square of a sum is a product of two brackets: $(a + b)^{2} = (a + b)(a + b)$. When expanding, every summand of the first bracket meets every one of the second, producing four products: $a^{2}$, ab, ba and $b^{2}$. The two mixed products add up to 2ab, which is completely missing in the incorrect "term-by-term squaring". A number test exposes the error immediately: $(2 + 3)^{2} = 5^{2} = 25$, but $2^{2} + 3^{2} = 4 + 9 = 13$. The difference 12 is exactly 2·2·3 = 2ab. Geometrically 2ab is the area of the two rectangles that lie next to the squares $a^{2}$ and $b^{2}$ inside the big square with side a + b.

Check three patterns. First: two squares with a minus between them, like $9x^{2} - 25$, split by the third formula into (3x + 5)(3x − 5). Second: three terms of the form a² ± 2ab $+ b^{2}$ form a perfect square; verify that the middle term really is twice the product of the roots of the outer ones. For $x^{2} + 10x + 25$ the roots are x and 5, twice their product is 10x ✓, so $(x + 5)^{2}$. If the middle term does not match, as in $x^{2} + 8x + 25$, it is not a perfect square. Third: $a^{2} + b^{2}$ without a mixed term cannot be factored over the reals at all. This pattern recognition is the key to cancelling fractional terms.

They appear everywhere, often hidden. Completing the square, with which you derive the vertex form and the pq formula, is the first binomial formula backwards. When differentiating with the difference quotient you need $(x + h)^{2}$. In integral calculus you simplify integrands like $(x + 1)^{2}$ before integrating. For limits with roots you expand with the third binomial formula to tame $\sqrt{(x + 1)} - \sqrt{x}$. In vector geometry it sits inside |a⃗ + b⃗$|^{2} = |a$⃗$|^{2} + 2\cdot{}a$⃗·b⃗ + |b⃗$|^{2}$. And recognizing $a^{2} - b^{2}$ is standard when cancelling fractional terms. Without secure command of these formulas you lose time in almost every calculus chapter.

Products symmetric around a round number can be computed in a flash: $19\cdot{}21 = (20 - 1)(20 + 1) = 400 - 1 = 399$, and $98\cdot{}102 = (100 - 2)(100 + 2) = 10000 - 4 = 9996$. Squares near round numbers work with the first and second formula: $103^{2} = (100 + 3)^{2} = 10000 + 600 + 9 = 10609$ and $99^{2} = (100 - 1)^{2} = 10000 - 200 + 1 = 9801$. The same trick "rationalizing" helps with roots in denominators: multiply $\frac{1}{(\sqrt{5} - 2)}$ by $(\sqrt{5} + 2)$ and, since $(\sqrt{5})^{2} - 2^{2} = 1$, you simply get $\sqrt{5} + 2$. The pattern is always the same: sum times difference equals square minus square.