Zeroes, Roots and How Many

Sketch any curve you like that starts low on the left, ends high on the right, and is drawn without lifting your pen. How many times must it cross the horizontal axis? Now try to draw one that crosses zero times. What stops you?

A zero belongs to a function: it is a value $a$ with $P(a) = 0$. A root belongs to an equation: it is a solution of $P(x) = 0$. An $x$-intercept belongs to a graph. All three are the same numbers, named after whichever object is being discussed.

Take $P(x) = x^2 - 5x + 6 = (x-2)(x-3)$.

The numbers $2$ and $3$ are the zeroes of the function $P$, because $P(2) = 0$ and $P(3) = 0$. They are the roots of the equation $P(x) = 0$, because they are its solutions. And they are the $x$-intercepts of the graph $y = P(x)$, or more precisely the $x$-coordinates of the points $(2,0)$ and $(3,0)$ where the curve meets the horizontal axis.

Nothing changes but the object under discussion:

a function has zeroes; an equation has roots; a graph has intercepts.

Exam questions use all three, and switching between them without hesitating is a real skill. "Find the roots of $x^3 = 2x^2 + 3x$" and "find the zeroes of $P(x) = x^3 - 2x^2 - 3x$" and "find where $y = x^3 - 2x^2 - 3x$ meets the $x$-axis" are the same question three times, with answers $-1$, $0$ and $3$.

Claim. A polynomial of degree $n$ has at most $n$ distinct real zeroes.

Why. Suppose $P$ has degree $n$ and $a_1, a_2, \ldots, a_k$ are distinct zeroes. By the factor theorem each $(x - a_i)$ is a factor of $P$, and since the $a_i$ are distinct, these are $k$ different linear factors. So

$$P(x) = (x-a_1)(x-a_2)\cdots(x-a_k)\,S(x)$$

for some polynomial $S$. Taking degrees, and remembering from Lesson 3 that degrees add when polynomials are multiplied:

$$n = k + \deg S \quad \Longrightarrow \quad k \leq n$$

since $\deg S$ cannot be negative. So there are at most $n$ distinct zeroes. The cap is a consequence of degrees adding, nothing more.

Note what the claim does not say. It does not say there are exactly $n$: a quadratic such as $x^2 + 1$ has none at all. It says $n$ is a ceiling, and finding $n$ distinct zeroes tells you that you have found them all and may stop looking.

Two different counts are in circulation, and questions rely on you knowing which is being asked for.

Take $P(x) = (x-1)(x-4)^2$, a cubic.

Distinct zeroes: $1$ and $4$. That is two numbers, and it is the answer to "how many solutions does $P(x) = 0$ have" and to "how many $x$-intercepts does the graph have".

Zeroes counted with multiplicity: $1$ once and $4$ twice, giving three, which equals the degree. The zero $4$ has multiplicity $2$.

Counting with multiplicity is the more useful convention because it makes the count predictable: over the real numbers, a polynomial of degree $n$ has at most $n$ zeroes counted with multiplicity, and the shortfall is always made up by irreducible quadratic factors.

For $x^3 - 1 = (x-1)(x^2+x+1)$, there is one real zero and the irreducible quadratic accounts for the missing two. The degree is never wrong; it is only ever partly invisible over the reals.

The upper bound came from algebra. The lower bound comes from the shape of the graph, and it is a different kind of argument.

For large $|x|$, the leading term dominates every other term, so a polynomial's behaviour far from the origin is decided entirely by $a_nx^n$. When $n$ is odd, $x^n$ has opposite signs for large positive and large negative $x$. So the graph goes to $+\infty$ at one end and to $-\infty$ at the other.

A polynomial graph is also unbroken: it can be drawn without lifting the pen, because a polynomial is defined and finite at every real number, with no asymptotes or gaps. A curve running from below the axis to above it, without lifting the pen, must meet the axis somewhere.

$$\text{odd degree} \quad \Longrightarrow \quad \text{at least one real zero}$$

An even degree has no such guarantee, because both ends go the same way and the curve can stay entirely above the axis, as $x^2 + 1$ does.

Combining the two results for a cubic: at least one real zero, at most three. And in fact only $1$, $2$ or $3$ are possible, never $0$.

Five items on zeroes and their number. Work each one, then reveal the answer.

1. 1 How many distinct real zeroes does $(x-2)^3$ have?

   Show answer
   One bracket, repeated three times, is zero at one value.$1$

2. 2 What is the maximum number of real zeroes a degree-$6$ polynomial can have?

   Show answer
   The degree caps the count.$6$

3. 3 Can a polynomial of degree $5$ have no real zeroes?

   Show answer
   Odd degree forces the two ends in opposite directions.No, it has at least one

4. 4 How many real zeroes does $x^2 + 4$ have?

   Show answer
   No real number squares to $-4$.None

5. 5 $P$ has degree $4$ and you have found four distinct zeroes. Should you keep looking?

   Show answer
   The cap has been reached.No, those are all of them