Chapter 3: Polynomial and Rational Functions 3.1 Polynomial Functions A polynomial on degree n is a function of the form P(x) = a n x n + a n 1 x n 1 + + a 1 x 1 + a 0, where n is a nonnegative integer and a n 0. Graphs of polynomials are smooth, unbroken curves with no sharp points. The end behavior is determined by the highest power term. If the highest power is even and a n > 0, then y as x ±. If the highest power is odd and a n > 0, then y as x and y as x. In both cases, if a n < 0 then reflect the end behavior across the x-axis. If y = P(x) is a polynomial and c is a number such that P(c) = 0, then c is a zero of the poly. Thus x = c is a root of P(x) = 0, x c is a factor of P(x), and c is an x-intercept of the graph of y = P(x). The only place y = P(x) can change signs is at an x-intercept. This is a consequence of the Intermediate value theorem for polynomials. If the poly is written in factored form, only factors to odd powers give rise to x-intercepts where the graph crosses the x-axis. If P(x) is a poly of degree n, then the graph of y = P(x) has at most n 1 local extrema. To graph a polynomial equation, write it in factored form. Find the x-intercepts and y-intercept. Determine using test points where the poly is + or. Determine the end behavior and graph. Example: Graph y = x 3 + x x 1. We note there are at most local maxima or minima. The end behavior is like that of y = x 3, with y as x and y as x. In factored form y = x (x + 1) 1(x + 1) = (x 1)(x + 1) = (x + 1) (x 1). This shows there are x-intercepts at 1 and 1; the graph does not cross the axis at x = 1 since that factor is raised to an even power. Letting x = 0 we find y = 1 for the y-intercept. The graph is shown below. 4 3 1 - -1 1-1 - -3 1
3. Dividing Polynomials Study long division and synthetic division of polynomials to prepare for factoring. 3 x 7 Example of synthetic division: x 3 3 1 0 0 7 3 9 +7 1 3 9 0 Quotient is: 1x + 3x + 9 with remainder 0 Thus we see x 3 is a factor of x 3 7. Remainder Theorem: If poly P(x) is divided by x c, then the remainder is P(c). Note above P(3) = 3 3 7 = 0, which was the remainder. Thus if c is a zero of the poly, so that P(c) = 0, x c is a factor of P(x). Factor Theorem: c is a zero of P if and only if x c is a factor of P(x). Example: Find the remainder when 6x 1000 17x 56 + 1x + 6 is divided by x + 1 = x ( 1). Use the remainder theorem! Evaluate P( 1) and this must be the remainder. 6( 1) 1000 17( 1) 56 + 1( 1) + 6 = 6 17 1 + 6 = 3. Thus remainder = 3. Example: Is x 1 a factor of x 567 3x 400 + x 9 +? Use the Factor Theorem! If P(1) = 0, then x 1 is a factor of the poly. (1) 567 3(1) 400 + (1) 9 + = 1 3 + 1 + = 1. Since P(1) 0, x 1 is not a factor of the poly.
3.3 Real Zeros of Polynomials Rational Zeros Theorem: If a poly P(x) = a n x n + a n 1 x n 1 + + a 1 x 1 + a 0 has integer coefficients, then every rational zero of P, and every rational root of P(x) = 0, has the form p / q. Here p is a factor of the constant coefficient a 0 and q is a factor of the leading coefficient a n. Descartes' Rule of Signs: If P is a poly with real coefficients: 1. The number of positive real zeros of P(x) is either equal to the number of variations in sign of P or less than that by an even number.. The number of negative real zeros of P( x) is either equal to the number of variations in sign of P or less than that by an even number. Upper and Lower Bounds' Theorem: If P is a poly with real coefficients: 1. Divide P by x b (b > 0) using synthetic division. If the quotient row has no negative entry, then b is an upper bound for the real zeros of P.. Divide P by x a (a < 0) using synthetic division. If the quotient row has entries alternately nonpositive and nonnegative), then a is an lower bound for the real zeros of P. Example: Find all the real zeros of P(x) = x 3 + 5x + x. Since P has integer coefficients we can use the rational zeros theorem. This assures us that the only possible rational roots are of the form p / q where p is a factor of the constant term, and q is a factor of the leading coefficient. Thus the only possible rational roots are ±, ±1, and ±1/. P(x) has 1 variation in sign so there is only 1 positive real zero. P( x) = ( x) 3 + 5( x) + ( x) = x 3 + 5x x has variations in sign and so either or 0 negative real zeros. We will soon see that every poly of degree n has exactly n zeros, counting complex zeros discussed in the next two sections. Thus P either has 1 positive real and negative real roots, or 1 positive real root and two complex roots. Test the possible rational roots using synthetic division. If one is found, the remaining roots can be found from the reduced quadratic equation which is the quotient. 5 1 4 18 38 9 19 36 since remainder 0, is not a zero. But this bottom row has no negative entry, and so by the upper bounds theorem there is no real root greater than. 1 5 1 1 6 7 1 6 7 5 not a zero, but another upper bound (the lub so far). 1 5 1 1 3 + 6 4 0 thus 1/ is a zero. Now find the zeros of x + 6x + 4. We can factor x + 6x + 4 = (x + 3x + ) = (x + )(x + 1) and P(x) = (x 1/)(x + )(x + 1) Thus the zeros are 1/, 1, and. As Descartes predicted, 1 positive and negative zeros. 3
3.4 Complex Numbers Consider the quadratic equation x 6x + 5 = 0 and the associated P(x) = x 6x + 5. The discriminant of the equation b 4ac = 64 < 0, indicating no real solutions. No real number squared can ever be negative. But Italians trying to solve more complicated equations found they needed to use solutions to equations like our example, in passing, on their way to real number solutions of the equations that interested them. So imagine, they said, that quadratics like the above have solutions, and we will use them for our own purposes. Define an imaginary unit number i such that i = 1 (or equivalently, define i = 1 ). Then let 4 = 4 1=iand so on, numbers like this being called imaginary. Further, consider numbers of the form a + bi, where a and b are real. These are complex numbers. Solve x 6x + 5 = 0 using the quadratic formula. ( 6) ± ( 6) 4(1)(5) 6± 64 6± 8i x = = = = 3± 4i (1) Thus P(x) = x 6x + 5 has complex zeros and factors as P(x) = [x (3 + 4i)][x (3 4i)] The two roots, differing only in the sign of the imaginary term, are called complex conjugates. Any quadratic with real coefficients and discriminant < 0 will always have a pair of complex conjugate solutions. Arithmetic is defined for these numbers as if they were binomials, with of course i = 1. Addition: (3 + 4i) + ( 5 + 1i) = (3 + 5) + (4i + 1i) = + 16i Subtraction: (3 + 4i) ( 5 + 1i) = (3 5) + (4i 1i) = 8 8i Multiplication: (3 + 4i)( 5 + 1i) = (3)( 5) + (3)(1i) + (4i)( 5) + (4i)(1i) = 63 + 16i Division: 3 + 4i 3 + 4i 5 1i 33 56i 33 56 = = = i 5 + 1i 5 + 1i 5 1i 169 169 169 Note in division we multiplied numerator and denominator by the complex conjugate of the denominator. This is just like rationalizing a denominator (which it is, because i is 1 ). Note when we write 4 = iwe are finding the principal square root. (i)(i) = 4 but also ( i)( i) = 4, so there are really two square roots here just as ± are the square roots of 4. Powers of i form a repeating sequence. i 1 = i i 5 = i 4 i 1 = i i = 1 i 6 = i 4 i = 1 i 3 = i i = i i 7 = i 4 i 3 = i i 4 = i i = 1 i 8 = i 4 i 4 = 1 To find i 78 for example, factor out i raised to a power that is the greatest multiple of 4 less than the power of interest. Thus i 78 = i 76 i 4 19 = (i ) i = 1 If z = 3 + 4i, Re(z), the real part of z, is 3. Im(z), the imaginary part of z, is 4. The complex conjugate z= 3 4i(Sometimes this is written z*). 4
Complex numbers can be represented geometrically as points in the complex plane. (Diagram from Wikipedia) The horizontal axis is the real number line, and the vertical axis is the imaginary number line. Numbers like 3 + 4i can be considered as representing a coordinate pair (3, 4). An arrow, or vector, from the origin to (3, 4) has a certain length and makes a certain angle with the positive "x-axis". The modulus or magnitude of z, written z = 3 + 4 = 5. This is the length of the vector. The argument of z is the angle whose tangent is 4/3 or about 53. Thus we can write 3 + 4i or 5 53 or (3, 4). The TI-86 calculator allows either of the last two forms and can perform all the above arithmetic operations. Caution! There is one operation in which complex numbers are unlike radicals. Positive numbers a and b obey the relation a b = ab, but this is not true if both a and b are negative. Always convert the square root of a negative quantity to the form i r before multiplying to avoid this difficulty. Example: 4 9 = i 3i = 6i = 6 5
3.5 Complex Zeros The theorems below help us to factor any poly and tell us that the poly equation y = P(x) has n roots if P is of degree n. Fundamental Theorem of Algebra: Every poly with complex coefficients has at least one complex zero. This is very hard to prove - it was first proven by Gauss. The following theorems have much easier proofs, given in the text. Complete Factorization Theorem: Every poly of nth degree can be factored into n complex linear factors or n complex zeros. Zeros Theorem: Every poly of degree n 1 has n zeros, provided that a zero of multiplicity k is counted k times. Linear and Quadratic Factors Theorem: Every poly with real coefficients can be factored into a product of linear and irreducible quadratic factors. Solve x 3 + x + 4x + 8 = 0. This factors (grouping) into x (x + ) + 4(x + ) = (x + 4)(x + ) = 0. Thus x = or x = ± 4 = ±i. The poly factors into (x + )(x + i)(x i). Solutions may be checked with the TI-86. Use the Poly function, give the order (degree) of the poly, enter the coefficients, and solve. Mathematica also solves algebraic equations. Solve x 3 1 = 0. By inspection, 1 is a root. What are the other two? Factor: x 3 1 = (x 1)(x + x + 1) = 0. Now solve x + x + 1 = 0 with the quadratic formula. 1± 1 4 1 1 1± i 3 1 3 x = = = ± i. 1 Since x 3 1 = 0 has solution x = 3 1, there are apparently three cube roots of unity if we allow complex roots. The reader should show that the roots above, when cubed, do equal one. There are formulas that solve the general cubic and quartic as the quadratic formula solves any quadratic. These were historically important but are complicated and rarely used. This reference (http://mathworld.wolfram.com/cubicformula.html) discusses the formulas. 6
3.6 Rational Functions A rational function is the ratio of two polynomials, assumed to have no factor in common. We illustrate their properties with an example. Graph y = (x 1)(x + ) (x + 1)(x 3) x-intercepts: Look at numerator for values of x that make y = 0. (1, 0) and (, 0) are the two odd x-intercepts. They are called odd because the factors giving rise to them are raised to odd powers. Recall this means the graph crosses the x-axis at these points. y - intercept: Set x = 0 everywhere to find y = /3. Thus (0, /3) is the y-intercept. Rational functions need not have x or y intercepts! Vertical asymptotes: An asymptote is a line that a graph approaches ever more closely. Since division by zero is not allowed, there is no point of the graph of the given rational function at x = 1 and x = 3. These values are found in general by seeing where the denominator is zero. Near those values y or y. Indicate a vertical asymptote with a dashed line. If the factor giving rise to the asymptote is raised to an odd power, the graph skips across the x-axis (from + to ) at that x value. If raised to an even power, the graph stays on the same side of the x-axis. Both of the vertical asymptotes for our example are odd: x = 1 and x = 3 are the odd vertical asymptotes of this example. Horizontal asymptotes: Consider the leading (highest power) terms of numerator and denominator. If the power to which x is raised is larger in the numerator, the end behavior is increasing y as x increases. But if the power is equal or less than the highest power of the denominator, as x increases y will approach a constant value, called a horizontal asymptote. y = x x 1 + + 1+ + x + x = x x x = x x = 1asx x x 3 x x 3 + + x x x 3 1+ + x x. n Ax More easily remembered: if y =, then the following can occur: m Bx 1. if n > m, then there is no horizontal asymptote.. if n = m, then the horizontal asymptote is y = A / B. 3. if n < m, then the horizontal asymptote is y = 0. The graph may cross a horizontal asymptote in the middle of the graph. It is only as x gets large that the graph gets ever closer to this line. The graph never touched a vertical asymptote. Slant asymptotes and other end behavior: If n > m above, there is no horizontal asymptote, but if n = m + 1 there is a slanted line that is approached closely at the far left and right of a graph. Divide, and discard the remainder. y = quotient will be the slant asymptote. If n = m +, a quadratic type curve serves as the end behavior, and so on. 7
Graphing, it helps to start at a known point like the y-intercept, if not zero. 4-4 - 4 6 8 10-8