Chapter 3.2 — Exercise 3.1 — Zeroes of Polynomial

Coefficient, Degree, and Constant — One Example

For p(x) = 5x⁷ − 6x⁵ + 7x − 6, three separate readings are asked for at once:

• Coefficient of x⁵ — the number multiplying x⁵ is −6.
• Degree of p(x) — the highest power present is x⁷, so the degree is 7.
• Constant term — the term with no variable at all is −6 (the same number as the coefficient above, purely by coincidence in this example).

It helps to name every single term in p(x) = 5x⁷ − 6x⁵ + 7x − 6 before moving on, since missing a term is the easiest way to misjudge a polynomial's degree. Written out fully with every power present, this is 5x⁷ + 0x⁶ − 6x⁵ + 0x⁴ + 0x³ + 0x² + 7x − 6 — the powers x⁶, x⁴, x³, and x² simply have coefficient 0 and are usually left out of the written expression entirely. Recognising that "missing" powers are really just zero-coefficient terms, not evidence the polynomial has some smaller degree, is what makes coefficient-and-degree questions reliable rather than guesswork.

Five True-or-False Statements

This exercise also tests whether the definitions above have genuinely sunk in, not just been memorised as words.

Two traps repeat across all five statements. The first is treating "degree" and "number of terms" as related — the last statement exists specifically to break that habit, since a polynomial like x¹⁰⁰ + 1 has only two terms but degree 100. The second is reading a coefficient without tracking its sign — the coefficient of x² in 3x³ − 4x² + 5x + 7 is −4, not 4, precisely because the minus sign belongs to the coefficient itself, not to some separate subtraction happening afterward. A third, quieter trap sits in the denominator statement: 1/(x² − 5x + 6) looks quadratic at first glance simply because a quadratic expression appears somewhere in it, but where that expression sits — numerator or denominator — is what actually decides whether the whole thing is a polynomial at all.

Evaluating a Cubic at Five Points

For p(t) = t³ − 1, substituting five different values of t shows how quickly a cubic's output grows compared to a quadratic's:

The negative inputs are where sign mistakes creep in most easily — cubing a negative number keeps the result negative (unlike squaring, which always turns it positive), so (−2)³ = −8, not 8. Losing track of that single sign flip is the most common error in this kind of substitution.

Notice also that p(1) = 0 in this table — t = 1 is a genuine zero of t³ − 1, sitting right there among the other four values without any special treatment needed to find it. This is worth remembering as a general habit: whenever a table of values is built for any reason, scanning it for a zero afterward costs nothing extra and can save an entire separate substitution step later in a problem.

Confirming Zeroes of a Biquadratic

For p(x) = x⁴ − 16, checking whether −2 and 2 are zeroes means substituting each directly:

• p(−2) = (−2)⁴ − 16 = 16 − 16 = 0
• p(2) = 2⁴ − 16 = 16 − 16 = 0

Both substitutions land on zero, so both −2 and 2 are confirmed zeroes of this degree-4 polynomial. This substitute-simplify-check pattern is the standard verification method used throughout the chapter, and it's also the seed idea behind the Factor Theorem covered later.

A degree-4 polynomial can have up to four zeroes, yet x⁴ − 16 shows only two here. Factorising fully explains why: x⁴ − 16 = (x² − 4)(x² + 4) = (x − 2)(x + 2)(x² + 4), and x² + 4 can never equal zero for any real number x, since a square is never negative and adding 4 only pushes it further from zero. So this particular polynomial simply doesn't use up its full "at most 4" allowance — a reminder that "at most n zeroes" is a ceiling, not a guarantee every polynomial of degree n reaches it.

Confirming Zeroes of a Quadratic

For p(x) = x² − x − 6, the same method checks 3 and −2:

• p(3) = 3² − 3 − 6 = 9 − 9 = 0
• p(−2) = (−2)² − (−2) − 6 = 4 + 2 − 6 = 0

Both results confirm 3 and −2 as the polynomial's zeroes — and, as expected for a quadratic, there are exactly two of them, no more and no fewer.

Why Substitution Never Lies

It's worth being explicit about why this checking method is completely reliable, even though it can feel like guesswork when a question hands you candidate numbers to test. The definition of a zero is purely computational: k is a zero of p(x) exactly when p(k) simplifies to 0, with no other condition attached. There's no need to factorise first, graph anything, or use any formula — substituting and simplifying is both necessary and sufficient. The only way this method can go wrong is through an arithmetic slip during the substitution itself, which is exactly why working through each step carefully, rather than skipping ahead to the answer, matters more here than almost anywhere else in the chapter. It's also the only method available before the tools introduced later in the chapter exist — Exercise 3.2's graphs and Exercise 3.3's coefficient formulas both still rely on substitution as their final check, so the habit built here never actually gets replaced, only supplemented.

Where This Leads

Verifying zeroes by substitution here sets up Exercise 3.2, which shows why these same zeroes are exactly the points where each polynomial's graph touches the x-axis — turning this exercise's algebra into a picture. For the definitions of degree, value, and zero this exercise assumes, revisit the chapter introduction, which works through the same p(x) = x² − 5x − 6 example that reappears indirectly throughout this exercise's own questions.