Chapter 2.4 — Exercise 2.3 — Remainder Theorem

Division, Restated for Polynomials

Dividing p(x) = 3x² + 2x − 12 by x − 2 by long division gives quotient 3x + 8 and remainder 4 — so 3x² + 2x − 12 = (x − 2)(3x + 8) + 4. Now evaluate p(2) directly: p(2) = 3(2)² + 2(2) − 12 = 12 + 4 − 12 = 4. The remainder from the division and the value of p at the divisor's zero are exactly the same number — not a coincidence specific to this example, but the pattern the rest of the exercise is built on. It also explains a fact worth keeping in mind whenever a division problem is set up in the first place: the process only ever stops once the remainder's degree drops below the divisor's, which is exactly why dividing by a linear (degree 1) polynomial always ends with a constant (degree 0) left over, never another x term.

The Remainder Theorem Itself

Stated formally: if p(x) has degree one or more, and a is any real number, then dividing p(x) by (x − a) leaves remainder p(a). The proof is short — write p(x) = (x−a)·q(x) + k, where k is a constant (since the remainder's degree must be less than the divisor's degree of 1, meaning degree 0). Substituting x = a makes the (x−a) term vanish entirely, leaving p(a) = k — exactly the remainder, established without ever needing to know q(x) explicitly. Once the remainder is known to be 0, something extra follows for free: p(x) = x² − 5x + 6 divided by g(x) = x − 3 gives remainder p(3) = 9 − 15 + 6 = 0, meaning x − 3 divides p(x) exactly, with absolutely no remainder left over at all. This single observation — a zero remainder signals an exact division — is worth remembering by name, because the entire next exercise is essentially just this one sentence, formalised and given its own theorem.

Finding the Divisor's Zero Before Substituting

The one skill this exercise really tests is identifying the correct zero of the divisor before substituting anything. For x + 1, the zero is x = −1; for 2x − 3, it's x = 3/2; for 5 + 2x, it's x = −5/2 — rearranged first into the ax + b shape if the divisor isn't already written that way, since ax + b = 0 always gives x = −b/a directly, the same general result derived in the previous exercise. Once the zero is found, substitution proceeds as usual, with fractions, negative numbers, and even irrational values like π handled the same way as any other number: x³ + 3x² + 3x + 1 divided by x + π leaves remainder p(−π) = −π³ + 3π² − 3π + 1, an expression rather than a clean number, but a perfectly valid remainder all the same. The Remainder Theorem never actually required the answer to simplify to a nice integer — it only guarantees that whatever comes out of substituting the divisor's zero into p(x) is the correct remainder, clean or not.

Two Polynomials, One Shared Remainder

Some problems give two different polynomials that leave equal remainders when divided by the same linear divisor. If f(x) = 2x³ + ax² + 3x − 5 and g(x) = x³ + x² − 4x + a both leave the same remainder on division by x − 2, then f(2) = g(2) becomes a single equation in a. Working it through: 16 + 4a + 6 − 5 = 8 + 4 − 8 + a, which simplifies to 17 + 4a = 4 + a, giving 3a = −13 and a = −13/3. A near-identical setup — f(x) = x³ + ax² + 5 and g(x) = x³ − 2x² + a, both divided by x + 2 — leads to the same value of a by the same method, despite starting from entirely different polynomials. That repetition is itself informative: the specific coefficients and divisor changed between the two problems, but the underlying equation-building step — evaluate both sides at the divisor's zero, set them equal, solve for the one unknown — stayed identical throughout.

Checking the Theorem Against Actual Long Division

To build confidence that the shortcut and the slow method really agree, some problems ask for verification both ways. f(x) = x⁴ − 3x² + 4 divided by g(x) = x − 2: the theorem gives remainder f(2) = 16 − 12 + 4 = 8 directly, and carrying out the full long division independently also produces remainder 8. A trickier case divides f(x) = x³ − 6x² + 14x − 3 by g(x) = 1 − 2x, where the divisor first needs rewriting as −2x + 1 before its zero (x = 1/2) can be read off — both the theorem and long division agree here too, landing on remainder 21/8. Rewriting a divisor into ax + b form first, before hunting for its zero, is the step that trips learners up most often in this kind of problem — 1 − 2x and −2x + 1 are the same polynomial, but only the second form makes the zero (x = 1/2) readable at a glance.

Two Conditions, Two Unknowns

The hardest problem in this exercise gives two separate remainder conditions on the same polynomial and asks for two unknown coefficients at once. If 2x³ + 3x² + ax + b leaves remainder 2 on division by x − 2, and remainder −2 on division by x + 2, then f(2) = 2 and f(−2) = −2 become a pair of simultaneous equations. The first simplifies to b = −2a − 26; substituting that into the second gives −4a − 30 = −2, so a = −7, and then b = −2(−7) − 26 = −12. The order matters here: solving for a first and only then substituting back to find b is much easier than trying to isolate both unknowns from the two original equations at once, since each equation on its own still has two unknowns tangled together.

Onward to the Factor Theorem

Every calculation in this exercise has quietly been building toward one special case: what happens when the remainder turns out to be exactly 0, rather than some other number. Exercise 2.4 takes that single case and turns it into a theorem of its own — a linear polynomial divides another exactly precisely when this remainder-finding process returns 0 — the same substitution, the same arithmetic, just interpreted as a factor check instead of a division shortcut.