Zeroes of a polynomial

Evaluating a polynomial. If $p(x)$ is a polynomial in $x$ and $k$ is any real number, replacing every $x$ in $p(x)$ with $k$ gives a single number, written $p(k)$ and called the value of $p(x)$ at $x = k$. This is nothing more than substitution, but it is the operation everything else in this topic is built on.

Definition of a zero. A real number $k$ is called a zero of $p(x)$ if substituting it makes the polynomial vanish:

where: $p(x)$ is the polynomial, $k$ is the candidate value being tested, and $p(k)$ is the number you get after substitution — a zero is any $k$ for which that number is exactly $0$.

Zero of a linear polynomial. For $p(x) = ax + b$ with $a \neq 0$, setting $p(x) = 0$ and solving for $x$ gives

where: $a$ is the coefficient of $x$ (never zero, or the polynomial would not be linear), $b$ is the constant term, and $x = -b/a$ is the unique solution. Every linear polynomial has exactly one zero — no more, no fewer — because this equation has exactly one solution.

Not every polynomial has a real zero at all. Take $p(x) = x^2 + 1$. For any real $k$, $p(k) = k^2 + 1 \geq 1$, which never touches $0$. So "find the zero" can legitimately have the answer "there is no real zero" — a point students often miss, expecting every polynomial to hand them a number.

More than one zero is possible. A quadratic polynomial can have up to two zeroes (verified in the worked example below with $p(x) = x^2 - x - 2$, which has zeroes $2$ and $-1$), a cubic up to three, and so on — a degree-$n$ polynomial has at most $n$ zeroes.

Question: (a) Check whether $x = 2$ and $x = -1$ are zeroes of $p(x) = x^2 - x - 2$. (b) Use the Remainder Theorem to find the remainder when $p(x) = 3x^3 - 4x^2 + x - 6$ is divided by $(x-2)$, and relate it to the zero of $g(x) = 3x - 6$.

Step 1. Test $x = 2$ in $p(x) = x^2 - x - 2$: $p(2) = (2)^2 - (2) - 2 = 4 - 2 - 2 = 0$. Since $p(2) = 0$, $x = 2$ is a zero.

Step 2. Test $x = -1$: $p(-1) = (-1)^2 - (-1) - 2 = 1 + 1 - 2 = 0$. Since $p(-1) = 0$, $x = -1$ is also a zero, confirming this quadratic has two zeroes, $2$ and $-1$, matching the picture above.

Step 3. For the second polynomial, apply the Remainder Theorem directly instead of dividing: dividing $3x^3 - 4x^2 + x - 6$ by $(x - 2)$ leaves remainder $p(2)$. Compute $p(2) = 3(2)^3 - 4(2)^2 + (2) - 6 = 3(8) - 4(4) + 2 - 6 = 24 - 16 + 2 - 6 = 4$. So the remainder is $4$, found without any long division.

Step 4. Find the zero of $g(x) = 3x - 6$: set $3x - 6 = 0$, so $3x = 6$ and $x = 2$. The zero of $g(x)$ is the same number, $2$, that we substituted into $p(x)$ — because $g(x) = 3x - 6$ is a constant multiple of $(x - 2)$, and evaluating any polynomial at the zero of its linear divisor is exactly what the Remainder Theorem's substitution step does. Zeroes: $x = 2$ and $x = -1$ for part (a); remainder $= 4$ for part (b), with the zero of $g(x)$ confirmed as $x = 2$.

Competitive stretch

For a quick check without full computation: if a linear polynomial $g(x) = kx - c$ has zero $x = a$, then $a = c/k$, and dividing any $p(x)$ by $g(x)$ (after normalising it to a monic $(x-a)$ form) still leaves remainder $p(a)$ — the constant multiple does not change where the zero is, only how you write the divisor.

• Confusing the value $p(k)$ (a number) with the zero (the input $x = k$ that makes that number $0$) — a value of $5$ is not a zero, only a value of $0$ is.
• Assuming every polynomial must have a real zero; $p(x) = x^2 + 1$ has none, since $p(k) = k^2 + 1 \geq 1$ for every real $k$.
• Sign slip when solving $ax + b = 0$: the zero is $x = -b/a$, not $b/a$ — forgetting the minus sign is the single most common error.
• In the Remainder Theorem, substituting $x = -a$ instead of $x = a$ when dividing by $(x - a)$ — always match the sign inside the bracket exactly.

Ask the tutor: If checking zeroes by substitution feels slow, ask the AI tutor to walk through a few more practice polynomials and show how the Remainder Theorem shortcuts the arithmetic.