Complex numbers and polynomials - University College Cork

Since the coefficients of x in each factor are real numbers, the result follows in thiscase. If exactly two are non-real, they must be each other’s complex conjugate, andso, this time, the roots can be denoted by α, ᾱ, γ, δ, where γ, δ are real numbers. Asbefore,p(x) = a(x 2 − (α + ᾱ)x + |α| 2 )(x 2 − (γ + δ)x + γδ),and, once more, we see that each factor has real coefficients. □5.1 Graphs of quarticsHere, we consider quartics with real coefficients, and say a few words about theirgraphical representation in the cartesian plane.The simplest quartic is p(x) = x 4 = (x 2 ) 2 , whose graph is symmetric about x = 0and convex; it is a large smile. Whereas the graph of every quadratic is symmetricabout some vertical line, the graph of a quartic need not be symmetric about anysuch line. For example, the graph of the quartic p(x) = x 4 + x is not symmetricabout any vertical line. For, if for some real (or complex) h, p(x + h) = p(h − x)for all x, then (?) every number is a root of the cubic 4hx 3 + (4h 3 + 1)x, which isabsurd.Since x 4 is positive for all non-zero x, the sign of the general quartic p(x) = ax 4 +bx 3 + cx 2 + dx + e, a ≠ 0, is the same as that of a, for all large values of x. Inother words, if a > 0, the graph lies in the upper half-plane for all sufficiently largex; whereas, if a < 0, it lies in the lower half-plane for all sufficiently large x. Thus,to describe its graph completely we must examine the behaviour for all x in somesymmetric interval. Thus, we need to determine its turning points, and points ofinflexion, if any. These will provide information about the shape of the graph, anddetermine where it’s convex, concave, increasing and decreasing. To do this, wemust use calculus methods. The turning points, are found by finding the first andsecond derivative p ′ , and solving the equation p ′ (x) = 0; the points of inflexion arefound by determining the second derivative, p ′′ , and solving p ′′ (x) = 0. Since p ′ hasreal coefficients there must be at least one turning point, but there need not be anypoint of inflexion. For instance, this is the case when p(x) = x 4 + 2x 2 + 1, which issymmetric about x = 0, and convex. It has no real root, only one turning point atx = 0, where it has a local minimum, and no point of inflexion.5.2 Exercises1. Let p be an arbitrary quartic. Prove thatp(x + h) = p(h) + p ′ (h)x + p′′ (h)x 22!+ p′′′ (h)x 33!+ p(iv) (h)x 4,4!where p ′ (h), p ′′ (h), p ′′′ (h), p (iv) (h) stand for the first, second, third and fourthderivatives of p evaluated at h.12