How this instrument works
Multiplying two binomials is the distributive property used twice over: (ax + b)(cx + d) opens out to acx² + adx + bcx + bd, four raw pieces before anything gets tidied up. Two of those pieces, adx and bcx, share the same power of x and therefore fold into one another, (ad + bc)x. The FOIL calculator elsewhere on this site names all four raw pieces First, Outer, Inner, and Last and reports them individually; this page carries that same expansion one stage further, folding Outer into Inner before display, so what arrives here is a finished trinomial rather than four pieces still waiting to be added by hand.
The middle coefficient is doing more work than it looks. ad + bc is a sum of two separate products, not one multiplication, and unwinding that sum is exactly the skill needed later to factor a quadratic — hunting for two numbers that multiply to bd and add to ad + bc is the sum-product method run in reverse. Set a = c = 1 and the trinomial simplifies to x² + (b + d)x + bd, precisely the shape every 'find two numbers that multiply and add' textbook exercise begins from, just assembled forwards here instead of pulled apart.
Two extremes show what the formula is doing structurally. Square a binomial by feeding it to itself, (x + 4)(x + 4), and ad + bc doubles rather than cancels — both cross-terms are identical, so the middle coefficient becomes 2ab, the perfect-square trinomial x² + 8x + 16. Send in b = 0 instead, so the first factor collapses to a bare term ax with no constant, and the product bc vanishes with it: the middle coefficient shrinks from a sum of two products down to ad alone. Move past two-term factors altogether — a binomial times a trinomial — and the raw product count jumps from four to six, past what this two-by-two formula was ever built to cover.
- The first factor is (ax + b): its coefficient goes in a, in (ax + b) and its constant goes in b, in (ax + b).
- The second factor is (cx + d): its coefficient goes in c, in (cx + d) and its constant goes in d, in (cx + d).
- x² coefficient reports ac right away — the product of the two leading coefficients, nothing left to add.
- x coefficient (combined) is the finished middle term, ad and bc already summed — where the FOIL calculator would still show those two as separate Outer and Inner figures.
- Constant term gives bd, the product of the two trailing constants and the trinomial's final piece.
Worked example — (2x + 3)(4x + 5) as one trinomial
This trinomial traces back to (2x + 3)(4x + 5), so a = 2, b = 3, c = 4, d = 5. x² coefficient works out to 2 × 4 = 8. x coefficient (combined) adds the two cross-terms directly: 2 × 5 = 10 from one pairing, 3 × 4 = 12 from the other, and 10 + 12 = 22 is the number the field actually shows. Constant term is 3 × 5 = 15. Read the three together and they spell out 8x² + 22x + 15, the whole trinomial delivered in one pass rather than four pieces waiting on a separate addition.
A substitution catches transcription slips without redoing the algebra. At x = 2, the original factors give (2·2 + 3) × (4·2 + 5) = 7 × 13 = 91, and the trinomial gives 8(2²) + 22(2) + 15 = 32 + 44 + 15 = 91 — the same number reached two different ways, which is exactly the agreement worth checking before a result like this feeds into a longer problem.
Questions
What is the formula for multiplying two binomials?
(ax + b)(cx + d) expands to acx² + (ad + bc)x + bd. The x² coefficient is the single product ac, the constant is the single product bd, and the middle x coefficient is a sum of two products, ad plus bc, since those are the only two cross-terms that land on the same power of x.
How is this different from the FOIL calculator?
Both multiply the identical two binomials to the identical answer, but at different stages. FOIL reports First, Outer, Inner, and Last as four separate products, so ad and bc stay visible on their own before anything is added. This sheet takes that same arithmetic one step further and reports ad + bc already combined, alongside ac and bd, as the finished trinomial's three coefficients.
Why do ad and bc combine into one coefficient instead of staying separate?
Because both are ordinary like terms in x — adx and bcx sit on the same power of the variable, so they add exactly the way 3 apples and 4 apples add to 7 apples. Nothing about the distributive property forces them to stay apart once the four raw products exist; combining like terms is the standard next step any polynomial expansion takes.
Where do sign errors usually creep in when multiplying binomials?
In the middle term, once b or d is negative. A negative constant carries its sign into both ad and bc, and the two most frequent slips are dropping that sign during the cross-multiplication or adding a negative product instead of subtracting it — worth a second look whenever either binomial subtracts rather than adds its constant.
Does the order of the two binomials matter?
No. Multiplication is commutative, so (ax + b)(cx + d) and (cx + d)(ax + b) expand to the identical trinomial, term for term. Swapping which binomial is entered first changes nothing about ac, ad + bc, or bd, since each is built from the same four numbers regardless of which pair is labelled a, b and which is labelled c, d.
How does this relate to factoring a quadratic back into two binomials?
It runs the exact same arithmetic in reverse. Factoring x² + (ad + bc)x + bd starts from the finished trinomial and searches for the two numbers that multiply to bd and add to ad + bc — precisely the pair this calculator already has on hand, since it built that trinomial from those numbers in the first place.