Chapter 2.3 — Exercise 2.2 — Zeroes of a Polynomial

Evaluating a Polynomial at a Point

Question 1 asks for the value of p(x) = 4x² − 5x + 3 at four different inputs — substitute and simplify, watching signs closely. At x = −1: p(−1) = 4(−1)² − 5(−1) + 3 = 4 + 5 + 3 = 12.

Question 2 repeats this for three different polynomials at x = 0, 1, and 2 each — including one written as a quotient, p(t) = (t−1)/(t+1), where the same substitute-and-simplify process still applies, just with a fraction at the end instead of a whole number. For p(y) = 2 + y + 2y² − y³, evaluating at all three points gives p(0) = 2, p(1) = 4, and p(2) = 4 — a reminder that two different inputs can legitimately produce the same output; that alone says nothing about whether either input is a zero, since neither result is 0.

Checking Candidate Zeroes

Question 3 flips the direction: instead of asking for p(x) at some point, it gives a candidate value and asks whether that candidate actually is a zero. The method is identical either way — substitute, simplify, and see whether the result is 0.

• If p(a) = 0, then x = a genuinely is a zero.
• If p(a) ≠ 0, then x = a is not a zero, no matter how close to 0 the result looks — 0.001 is not 0, and the substitution either lands exactly on 0 or it doesn't.
• p(x) = x² − 1 has two zeroes, x = 1 and x = −1 — a quadratic can have as many as two, and this particular pair also happens to be plus-or-minus the same number, since x² treats a value and its negative identically.
• p(x) = (x−1)(x+2) is not a zero at x = −1 (gives (−2)(1) = −2), but is a zero at x = −2 (gives (−3)(0) = 0).

That last case previews an idea the chapter returns to later: a product is zero exactly when at least one of its factors is zero. p(x) = (x−1)(x+2) is zero at x = 1 for the same reason it's zero at x = −2 — either factor collapsing to 0 is enough to send the whole product to 0, regardless of what the other factor evaluates to. Two zeroes from one quadratic-degree product, found without ever multiplying the brackets out first — a shortcut that only gets more valuable once the polynomials involved reach degree three or higher, where multiplying everything out before substituting would be far more error-prone than checking each factor separately.

The Zero of a Linear Polynomial, in General

Question 4 asks for zeroes of several specific linear polynomials, which is the natural point to derive the general rule rather than solve each one separately. For f(x) = ax + b, setting f(x) = 0 and solving gives x = −b/a directly.

Applied instantly: f(x) = x + 2 has zero x = −2; f(x) = 2x + 3 has zero x = −3/2; f(x) = x² has zero x = 0 (a repeated root, since x² = 0 forces x = 0 regardless of the exponent); and f(x) = px, with p ≠ 0, has zero x = 0 as well. The general case f(x) = px + q, with p and q real and p ≠ 0, works exactly the same way: px + q = 0 gives x = −q/p, a single formula that covers every specific case in this question at once rather than needing to be re-derived for each one.

A linear polynomial always has exactly one zero; a constant polynomial (other than the zero polynomial itself) has none at all, since a nonzero constant can never equal 0 for any input, no matter what x is substituted in. The zero polynomial is the one genuine exception to nearly everything said so far — since 0 = 0 for literally every value of x, every real number is simultaneously a zero of it, which is why it's described separately rather than folded into the "constant polynomial" case.

Working Backward From a Known Zero

The last two questions reverse the whole process again: given that a certain value is a zero, find an unknown coefficient that makes that true. If 2 is a zero of p(x) = 2x² − 3x + 7a, then p(2) = 0 gives a solvable equation for a.

The structure of this kind of problem is always the same regardless of which coefficient turns out to be unknown: substitute the given zero, simplify everything that can be simplified numerically, and what's left is a plain linear equation in the one unknown letter — no different, mechanically, from solving 2 + 7a = 0 for a directly.

When two zeroes are given for the same polynomial — say 0 and 1 for p(x) = 2x³ − 3x² + ax + b — both conditions get used together: p(0) = 0 pins down b directly, and p(1) = 0 then becomes a simpler equation for a alone, since b is already known by that point. Substituting one result into the next equation, rather than solving both simultaneously from scratch, is usually the faster route whenever one of the two zeroes happens to be 0. Working through it fully: p(0) = 0 gives 2(0)³ − 3(0)² + a(0) + b = 0, which collapses immediately to b = 0. With b already known, p(1) = 0 becomes 2(1)³ − 3(1)² + a(1) + 0 = 0, i.e. 2 − 3 + a = 0, giving a = 1. Two conditions, applied one after the other rather than together, turned what looked like a pair of simultaneous equations into two much simpler ones solved back to back.

Where the Zero-Finding Skill Leads

Recognising when a substitution gives exactly 0 turns out to be the single mechanical step underneath the rest of this chapter — every remaining technique for handling polynomials, from finding remainders to factorising cubics, is built on repeating this one check at the right moment. Exercise 2.3 reframes this same substitution as a shortcut for polynomial division, and Exercise 2.4 goes one step further, using a zero of exactly 0 to confirm that a linear expression divides another polynomial with nothing left over. Nothing about the underlying check changes across these three exercises — only what the result of "p(a) = 0" is used to conclude.