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Instrument MI-01-251 · Mathematics

Generic Rectangle Calculator

Picture two binomials as the sides of a rectangle. Split each one into its two terms, multiply out the four resulting sub-areas, and this sheet totals them into the expanded quadratic's three coefficients.

Instrument MI-01-251
Sheet 1 OF 1
Rev A
Verified
Type 05 — Algebra SER. 2026-01251

x coefficient

11.00000000

x² coefficient = ac

2.00000000 x² coefficient
12.00000000 constant term
The working Every figure verified twice
  1. coeffA = 2·1 = 2.00000000
  2. coeffB = 2·4 + 3·1 = 11.00000000
  3. coeffC = 3·4 = 12.00000000
Worksheet log
  1. No entries yet — change an input to log a scenario.

How this instrument works

The generic rectangle, also called the area model, turns multiplying two binomials into finding the area of a rectangle split into four smaller pieces. Draw a rectangle whose width is split into ax and b, and whose height is split into cx and d — four inner boxes appear, one for each pairing: ax times cx, ax times d, b times cx, and b times d. Each box's area is one term of the full expansion, and since area never depends on how a shape gets sliced, adding the four boxes back together must equal the original product exactly.

Two of those four boxes end up describing the same power of x. The box for ax times d gives adx, and the box for b times cx gives bcx — different areas built from different sides, yet both are plain multiples of x, so they merge into one middle coefficient, ad+bc, rather than staying as two separate figures. This sheet performs that merge for you: where a FOIL-style layout on this site reports Outer and Inner as two untouched numbers you would still need to add, the coefficients here already show the combined x coefficient, because the rectangle's two middle regions were always going to sit on the same line once summed.

It also differs from a page built around the sum and product of two plain numbers, which only ever multiplies binomials of the form (x − p)(x − q) with a leading 1 on both sides. Here a, b, c, and d are each free to be anything, including negative values on either binomial, so the same four-box arrangement covers expressions a narrower monic-only tool cannot reach. Whatever the four inputs turn out to be, the rectangle's structure — split, multiply four ways, merge the middle, done — never changes.

Two boundary cases are worth seeing directly. Reduce both binomials to their bare leading terms and three of the four boxes collapse toward zero, leaving a single clean x² result. Flip a sign on one side instead and the middle coefficient can come out negative even while the outer boxes stay positive, since ad+bc simply adds whatever two signed numbers those particular boxes contain.

(ax+b)(cx+d)=acx2+(ad+bc)x+bd(ax+b)(cx+d) = ac\,x^{2} + (ad+bc)x + bd
a, b — the two terms of the first binomial (ax+b) · c, d — the two terms of the second binomial (cx+d) · coeffA — the x² coefficient, ac · coeffB — the merged x coefficient, ad+bc · coeffC — the constant term, bd.
  • Enter the first binomial's numbers into a, from (ax+b), and b.
  • Enter the second binomial's numbers into c, from (cx+d), and d.
  • Read x² coefficient for the product ac — the top-left box of the rectangle.
  • Read x coefficient for the already-merged ad+bc — the two middle boxes added together.
  • Read constant term for the product bd — the last box, at the bottom-right corner.
  • Multiply the two original binomials by hand if you want to double-check that the three coefficients reassemble into the same expanded quadratic.

Worked example — expanding (2x+3)(x+4)

Set a=2, b=3, c=1, d=4. The top-left box multiplies the two leading pieces, a and c: 2 times 1 gives 2, the squared coefficient. The two middle boxes multiply a by d and b by c: 2 times 4 gives 8, and 3 times 1 gives 3; adding those, 8 and 3, gives 11, the merged coefficient. The last box multiplies b by d: 3 times 4 gives 12, the constant. Altogether that reproduces 2, 11, and 12 — exactly the three figures this sheet returns for the squared coefficient, the merged coefficient, and the constant term.

The method holds up away from this tidy pairing too. With a=1, b=0, c=1, d=0, three of the four boxes vanish, leaving just the squared coefficient at 1, with the merged coefficient and the constant both landing on 0. With a=3, b=−2, c=2, d=5, a negative part carries straight into the merge: the leading product becomes 3×2=6, the merge becomes 3×5+(−2)×2=15−4=11, and the last product becomes −2×5=−10 — a squared coefficient of 6, a merged coefficient of 11, and a constant of −10, the same four-box process working unaffected by the negative sign carried through it.

Questions

What exactly is the generic rectangle method?

It is a way of multiplying two binomials by treating them as the two sides of a rectangle, splitting each into its two terms, and filling four inner boxes with the four resulting products. Adding those four areas back together reproduces the full expanded product, since a rectangle's total area never changes no matter how it gets divided into pieces.

How is this different from a FOIL-style calculator?

A FOIL layout names the same four products First, Outer, Inner, and Last, but reports Outer and Inner separately, leaving you to add them into the middle term yourself. This sheet performs that addition already, returning a single x coefficient — ad+bc — because the two middle boxes of the rectangle were always destined to combine once summed.

Why do the two middle boxes combine into one number instead of staying separate?

Because both of them are multiples of x rather than x² or a plain constant. The box ad and the box bc sit on the same power of the variable, so like terms in ordinary algebra, they add directly into one coefficient rather than remaining as two distinct entries in the final expansion.

Does the area model still work with negative coefficients?

Yes. Multiplying (3x−2)(2x+5) carries a negative b straight through every box it touches, giving ad+bc=15−4=11 and bd=−10, for a total of 6x²+11x−10. A negative sign changes the numbers each box produces, not the four-box structure or the rule for combining them.

What happens when b and d are both zero?

Three of the four boxes vanish, since any box multiplied by a zero term contributes nothing, leaving only the ax times cx box standing. Multiplying (x)(x) this way returns x² coefficient 1 with both the x coefficient and the constant sitting at 0.

How does this compare with the box method used for plain numbers?

The box method splits ordinary numbers by place value, such as 20+3 for 23, and keeps all four resulting boxes separate because each sits at a different place value with nothing to merge. Here the four boxes involve a shared variable, x, so two of them land on the same power of x and combine — the rectangles look alike, but only one of the two setups has terms worth adding together.

References