partial products must be fractions and the last integral coefficient cannot be canceled by a fraction. The rational roots of (3) are therefore x = 2 and 3, and the irrational root is, from the graph, nearly 1.8. The other two roots (Theorem 1) are imaginary. EXERCISES 1. Find all the roots of the equations below, in each case constructing the graph of the polynomial on the left. 2. Find all the rational roots of the equations below, in each case constructing the graph of the polynomial on the left, and estimating the irrational roots, if any, from the graph. (f) 3x2 + 10x3 + 7x2 - 3x − 7 = 0. Using synthetic division, find the irrational root to the nearest tenth of a unit. 3. Show that the graph of a polynomial of degree n cannot have more than n 1 maximum and minimum points, and that it cannot have more than n - 2 points of inflection. 52. Translation of the y-Axis. For the purpose of determining approximately an irrational root of an equation f(x) = 0, where f(x) is a polynomial, a method is used which reduces the roots of an equation by a constant amount c. A translation of the y-axis along the x-axis in the positive direction diminishes the intercepts on the x-axis of the graph of f(x), and hence such a translation represents graphically a diminution of the roots of an equation. We shall consider the method first from a graphical point of view. and find the function whose graph is the same curve referred to a new axis 2 units to the right of the old one. The graph is constructed by the method of Section 49. If f(x) represents the function in (1), then ƒ(x' +2) will represent the function in (2). The intercepts on the x-axis of the graph referred to the new origin are 2 units less than the intercepts referred to the old origin. Hence the roots of the equation f(x' + 2) = 0, or are 2 units less than the roots of the equation f(x) We shall now derive a method of obtaining the coefficients of (3) which is simpler in general than the preceding method. Expanding the terms on the right by the binomial theorem, collecting like powers of x', and representing the coefficients of the powers of x' by For any value of x and x' connected by the relation x x' + c, If f(x' + c) box'n + b1x'n-1 + + bn-2x′2 + bn-12′ + bn be divided by x', as indicated on the right of (9), the remainder is b, and the quotient is If the quotient q1(x') is divided by x', the remainder is b2-1 and the quotient is Q2(x') = box'n¬2 + b1x'n¬3 + + bn-2. Continuing this process it is seen that the successive remainders are the coefficients of (6) in the order bn, bn−1, bn-29 bo. But if the indicated divisions be performed on both sides of (9), if the respective quotients thus obtained be divided by x c and x', and so on, the successive remainders obtained on the left side of (9) will equal those obtained on the right, namely, bn, bn−1, bn-2, .. bo. Hence we have the Theorem. The coefficients of f(x' + c) may be obtained by dividing f(x) by x- c, the quotient by x- c, etc. 1 - 9 +23 15 | 2 1 - 5 · 1* 1* - 3* The computation of the coefficients in the example may be effected by means of this theorem and synthetic division in compact form as indicated. The successive remainders in the divisions, marked with an asterisk, are the coefficients of the function in (3). EXERCISES 1. Construct the graph of each of the functions below, and find the function having the same graph referred to a new y-axis c units to the right. Verify the result by direct substitution of x' + c for x. = 1. x3 - 6x2 + 7x + 4 will right. 2. Show that the second term of the function y be removed if the y-axis is translated 2 units to the Deduce a rule for removing the second term of ax3 + bx2 + cx + d. 3. Determine a translation of the y-axis so that the graph of x3 + 3x2 – 4 in the old system will be the graph of x3 – 3x + 2 in the new system. Plot the graphs of x3 and 3x - 2 on the same axes and from them determine approximately the roots of the equation x3 – 3x + 2 = 0, and hence of the equation x3 + 3x2 − 4 = 0. 4. Plot the graph of x4 2x3 3x2 + 4x + 2 and determine the coordinates of the maximum and minimum points and of the points of inflection. Find the function which has the same graph referred to a y-axis 1 unit to the left. 53. Horner's Method of Solution of Equations. This method enables us to compute irrational roots as accurately as may be desired. EXAMPLE. Find, correct to two decimal places, the real root of the equation f(x) = x3 + x 47 0. (1) Plotting the graph of f(x) we get the curve in the figure, which shows that there is a real root between 3 and 4. As the coefficients of f(x) are integers, and that of x3 is unity, this root is not fractional, and it therefore must be irrational. Now move the y-axis 3 units to the right. -24· -18 -42 (2) -48 FIG. 72. 3 The new equation, omitting the primes on the x's, is ƒ(x +3) = x3 + 9x2 + 28x - 17 = 0. No confusion should arise from omitting the primes if it be remembered that the graph of the polynomial in (2) is the curve in the figure referred to the axes O'X and O'Y'. Equation (2) has a root between 0 and 1, which from the graph appears to be about 0.5. Dividing (2) by x − 0.5, we have As the remainder is negative, the graph lies below the x-axis at x = and hence, from the figure, the root is larger than 0.5. Try x = 0.6. 0.5 The remainder is now positive and hence the graph lies above the x-axis 0.6. Therefore the root of (2) lies between 0.5 and 0.6. at x = The graph of the polynomial on the left is the same curve referred to the axes ("X and O"Y", and it shows that equation (3) has a root between O and 0.1. As the square and cube of a number less than 0.1 are very small, an approximate value of the root may be obtained by neglecting x3 and x2 in (3) and solving the resulting linear equation As the remainder is positive, the graph shows that the root is less than 0.02. Try 0.01. This remainder is negative, and hence the graph shows that the root is greater than 0.01. Hence the root of (3) lies between 0.01 and 0.02. Then the root of equation (2) lies between 0.51 and 0.52, and that of (1) between 3.51 and 3.52. Hence the real root of (1), correct to two decimal places, is x = 3.51. Which is the closer approximation to the root, 3.51 or 3.52? Why? EXERCISES Find all the real roots of the equations following, obtaining irrational roots to two decimal places. |