Polynomial functions and complex zeros

For a polynomial p, three statements are interchangeable: a is a real zero of p, (x − a) is a linear factor of p, and (a, 0) is an x-intercept of the graph. Any one of them buys you the other two. Read p(x) = 2(x − 3)(x + 5) and the zeros fall out: x = 3 makes the first factor zero, x = −5 makes the second factor zero.

Watch the sign carefully. The factor (x + 5) is (x − (−5)): the zero is negative five. Reversing that dictionary, reading the zero of (x + 5) as +5, is the single most common slip in this topic. The leading constant 2 never contributes a zero; multiplying by 2 cannot make anything equal zero.

If the factor (x − a) appears m times, the zero a has multiplicity m, and m controls the graph's behavior at the intercept:

1. m odd: the graph crosses the axis at a. For m = 1 it cuts straight through; for m = 3, 5, ... it flattens momentarily as it crosses, an S-shaped pause.
2. m even: the graph touches and turns. It comes down to the axis, kisses it at a, and leaves on the same side it arrived from, like the vertex of a parabola sitting on the axis.

So (x − 2)²(x + 1) crosses at −1 and only bounces at 2. Counting intercepts to find the degree fails for exactly this reason: that cubic shows two intercepts, not three, because the zero at 2 is doubled.

Some zeros are not real numbers at all. The polynomial x² + 9 is never zero for real x; its zeros are 3i and −3i. Non-real zeros of a polynomial with real coefficients always arrive in conjugate pairs: if a + bi is a zero, a − bi is too. One is never alone.

Two consequences do a lot of exam work. First, a degree-n polynomial has exactly n complex zeros counted with multiplicity, always, whether or not the graph shows them: a quartic with only two x-intercepts still has four zeros, the missing ones either repeated or hiding as a conjugate pair off the real line. Second, because non-real zeros come two at a time, an odd-degree polynomial with real coefficients cannot pair off all its zeros: at least one must be real, so its graph must cross the x-axis at least once.

And to say it plainly: non-real zeros make no mark on the graph. There is no point on the x-axis labeled 3i. The graph of x² + 9 floats above the axis, intercept-free, while its two zeros do their work invisibly.

A function is even when f(−x) = f(x) for every x in the domain: the graph is symmetric across the y-axis. A function is odd when f(−x) = −f(x): the graph is symmetric through the origin, so p(3) = −4 forces p(−3) = +4.

The names tempt a shortcut that does not exist: even degree does not mean even function. The test is about every term, not the leading one. f(x) = x⁴ + x fails the even test because f(−x) = x⁴ − x, which is neither f(x) nor −f(x): the x⁴ term is even-friendly, the x term is not, so the function is neither even nor odd. A polynomial is even exactly when every exponent is even, and odd exactly when every exponent is odd. Run the f(−x) computation; never rule by the degree's name.