How this instrument works
Multiplying a binomial by a trinomial is the distributive property doing more work than FOIL ever asks of it. FOIL handles two two-term factors and produces four products; here the second factor carries three terms, so distributing the binomial's two terms across all three of the trinomial's terms produces six products — 2 × 3 rather than FOIL's 2 × 2. Nothing new is invented: every one of those six is still an ordinary term-by-term multiplication, there are simply two more of them to keep track of.
Those six products land on only four different degrees, because a binomial of degree 1 times a trinomial of degree 2 always produces a cubic of degree 1 + 2 = 3. The highest-degree term, x³, and the constant term each come from exactly one of the six products — the two leading terms multiplied together, and the two constants multiplied together. The two middle coefficients, on x² and on x, are each the sum of two of the six products that happen to share a degree, which is the one place a term goes missing if the work is done by hand without a system.
The pattern generalises directly: a polynomial with m terms times one with n terms always needs m times n term-by-term products before like terms combine, and always produces a result whose degree equals the sum of the two factors' degrees. FOIL is the m = 2, n = 2 case of that same rule; a binomial times a trinomial is the m = 2, n = 3 case one step further along, six products collapsing into four coefficients instead of four products collapsing into three.
- Enter the binomial's two figures into Binomial: x coefficient, a and Binomial: constant, b — for (2x + 3) that is a = 2 and b = 3.
- Enter the trinomial's three figures into Trinomial: x² coefficient, a, Trinomial: x coefficient, b, and Trinomial: constant, c — for (x² + 4x + 5) that is a = 1, b = 4, c = 5.
- Read x³ coefficient directly; it is the single product of the two leading terms and needs no further combining.
- Read x² coefficient and x coefficient — each already sums the two cross-terms that land on that degree, so no manual addition is needed.
- Read Constant term for the product of the binomial's constant and the trinomial's constant, the last of the six terms.
Worked example — (2x + 3)(x² + 4x + 5)
Enter a = 2 and b = 3 for the binomial (2x + 3), then a = 1, b = 4, and c = 5 for the trinomial (x² + 4x + 5). Six products appear before anything is combined: 2x times each of x², 4x, and 5 gives 2x³, 8x², and 10x; 3 times the same three terms gives 3x², 12x, and 15. Two of those six land on x² and two land on x, so this sheet adds each matching pair before reporting a single coefficient for that degree.
The sheet reports x³ coefficient = 2, x² coefficient = 11 (from 8 + 3), x coefficient = 22 (from 10 + 12), and Constant term = 15, so (2x + 3)(x² + 4x + 5) expands to 2x³ + 11x² + 22x + 15. That is a full cubic built from six term-by-term products, half again as many as FOIL alone would ever produce for two binomials.
Questions
Why does multiplying a binomial by a trinomial produce six terms instead of four?
Because the binomial has two terms and the trinomial has three, so the distributive property produces 2 × 3 = 6 term-by-term products before anything combines — one more pairing per side than FOIL's 2 × 2 = 4 for two binomials. Two of those six land on the same degree, x² and x, so they collapse into single coefficients; the highest and lowest degree terms come from just one product each.
How is this different from FOIL or from multiplying two binomials together?
FOIL and a plain binomial-times-binomial both work on two two-term factors, giving four products that combine into three coefficients of a quadratic. Here the second factor is a trinomial, so the same distributive rule runs across three terms instead of two — six products combining into four coefficients of a full cubic, one degree higher than two binomials can ever reach.
Why do the x³ and constant coefficients need no combining, but x² and x do?
Only one of the six products lands on degree 3, the two leading terms multiplied together, and only one lands on degree 0, the two constants multiplied together, so those coefficients arrive whole. Degree 2 and degree 1 each receive two separate products — one crossing the binomial's leading term against a trinomial term, one crossing the binomial's constant against a different trinomial term — so this sheet adds each pair before reporting the coefficient.
What is the most common mistake when expanding this kind of product by hand?
Missing one of the six products, usually a middle-degree cross-term, since there is no four-letter mnemonic like FOIL to keep six pairings in order. Writing every term of the binomial against every term of the trinomial in a small grid, two rows by three columns, before adding catches the term that a straight left-to-right pass tends to drop.
Does the order of the two factors change the answer?
No — polynomial multiplication is commutative, so (a₁x + b₁)(a₂x² + b₂x + c₂) gives the identical four coefficients as (a₂x² + b₂x + c₂)(a₁x + b₁). This sheet fixes the binomial first and the trinomial second purely to keep the input fields consistent; swapping which factor is entered where changes nothing about the result.
Does this same method extend to two trinomials or to larger polynomials?
Yes — the rule scales with the term counts of both factors: two polynomials with m and n terms always need m times n products before combining. Two trinomials would need 3 × 3 = 9 products collapsing into 5 coefficients of a degree-4 result; this sheet is fixed at the binomial-by-trinomial case, six products collapsing into four coefficients.