Polynomial Functions and Complex Zeros

Topic 1.4 found where polynomial functions turn and bend. This topic is about where they are zero: how many zeros there are, how the graph behaves at each one, and what happens when some zeros are not real numbers. It ends with two kinds of symmetry, even and odd.

Zeros from factors, and multiplicity

If $(x - a)$ is a factor of a polynomial $p$, then $p(a) = 0$. When the factor appears to the power $k$, and no higher, $a$ is a zero of multiplicity $k$. In $q(x) = x^2(x - 4)^3$, $0$ has multiplicity $2$ and $4$ has multiplicity $3$. A zero of multiplicity $2$ is also called a double zero.

Multiplicity controls the graph at the zero:

The reason: near $x = a$ every other factor keeps its sign, and $(x - a)^k$ changes sign only when $k$ is odd. An even power is never negative.

Counting zeros: real and non-real

A polynomial of degree $n$ has exactly $n$ complex zeros, counting multiplicity. Some may be real (and show as $x$-intercepts); the rest are non-real numbers $a + bi$. When every coefficient is real, the non-real zeros come in conjugate pairs: if $2 + i$ is a zero, so is $2 - i$.

Counting is a good check. A degree $5$ polynomial with real coefficients whose zeros include $3$ (multiplicity $2$) and $1 + i$ already has $3, 3, 1 + i, 1 - i$: one more zero remains, and it must be real.

Degree from differences

Topic 1.3 showed that a quadratic has constant second differences. The pattern continues: for equally spaced inputs, a polynomial of degree $n$ has constant $n$th differences. For $f(x) = x^3 - 2x^2 + 1$ at $x = 0, \ldots, 5$, the outputs $1, 0, 1, 10, 33, 76$ have first differences $-1, 1, 9, 23, 43$, second differences $2, 8, 14, 20$ and third differences $6, 6, 6$: degree $3$.

Even and odd functions

A function is even if $f(-x) = f(x)$ for every $x$ in its domain. Its graph is symmetric about the $y$-axis: $x^2$, $x^4 - 3x^2$. A function is odd if $f(-x) = -f(x)$. Its graph is symmetric about the origin: $x^3$, $x^3 + x$. Most functions are neither; $x^3 + 1$ is not odd, because $-(x^3 + 1) = -x^3 - 1$, not $-x^3 + 1$.

To test, replace $x$ with $-x$ and simplify. If you get the original back, it is even; if you get its negative, it is odd. For an odd function, $f(2) = 5$ means $f(-2) = -5$.