How a diamond problem works
A diamond problem has four numbers: two factors a and b on the left and right, their product a × b at the top and their sum a + b at the bottom. Knowing any two is enough to find the rest.
With both factors, multiply and add. With one factor and the product, b = product ÷ a. With one factor and the sum, b = sum − a.
With the product P and the sum S, a and b are the two roots of t² − S·t + P = 0, so t = (S ± √(S² − 4P)) ÷ 2. This is exactly the step used to factor x² + Sx + P into (x + a)(x + b).
Worked examples
Product 12, sum 7: D = 7² − 4 × 12 = 1, so t = (7 ± 1) ÷ 2, giving 4 and 3. Check: 4 × 3 = 12 and 4 + 3 = 7.
Product −24, sum −2: D = (−2)² − 4 × (−24) = 100, so t = (−2 ± 10) ÷ 2, giving 4 and −6.
| Known | Find b | Then |
|---|---|---|
| a and b | — | P = a × b, S = a + b |
| a and product | b = P ÷ a | S = a + b |
| a and sum | b = S − a | P = a × b |
| product and sum | t = (S ± √(S² − 4P)) ÷ 2 | both factors |
When the answer is not a whole number
If S² − 4P is not a perfect square, the factors are irrational and the trinomial does not factor over the integers. If S² − 4P is negative, no real numbers work and the factors are a complex conjugate pair, which the calculator shows as p ± qi.
If a factor is 0, the product is 0 whatever the other factor is, so the product alone cannot recover it; enter the sum instead.