Complex numbers and polynomials - University College Cork

2. Let p be a arbitrary quartic. Use the previous exercise to show that we maychoose h so that the coefficient of x 3 in the expansion of p(x + h) is zero.3. Suppose the roots of p(x) = ax 4 + bx 3 + cx 2 + dx + e, a ≠ 0, are in arithmeticprogression. Prove that b 3 − 4abc + 8a 2 d = 0.4. Determine the roots of x 4 − 12x 3 + 49x 2 − 78x + 40. Show that it’s graph issymmetric about x = 3.5. Show that(x − 1)(x − 2)(x − 4)(x − 5) ≥ −9/4, ∀x.6. Prove that the quartic p(x) = ax 4 + bx 3 + cx 2 + dx + e, a ≠ 0 has no pointsof inflexion iff 3b 2 < 8ac.6 Cubic polynomialsWe consider cubics of the formwhere a ≠ 0.p(x) = ax 3 + bx 2 + cx + d,Theorem 6. Suppose a, b, c, d are real numbers. Then p has at least one real root.Proof. We’ve already outlined one proof of this fact based on the fact that ap(x)is positive for large positive x and negative for large negative x, a statement whichextends to cover the case of any polynomial of odd degree having real coefficients.Here’s another approach which uses Theorem 2 due to Gauss. According to this,every cubic (with real or complex coefficients) has at least one root, which may becomplex, and at most three distinct roots. So, p has three roots, which may not allbe different. Since its coefficients are real, by Theorem 3, its complex roots come inpairs. Hence, one of the roots must be real. □As the next example shows, this is the most we can say, in general.Example 3. Find the roots of p(x) = x 3 − 1.Solution. Clearly 1 is a root. Hence x − 1 is a factor of p. In other words, thereis a quadratic q such that p(x) = (x − 1)q(x). It’s straightforward to verify thatq(x) = x 2 + x + 1. Now, the discriminant of the latter is b 2 − 4ac = 1 − 4 = −3,which is negative, and so, by the usual formula, the roots of q are complex, andequal to−1 ± √ 3i.2Thus, p has only one real root. □13