when 2 is substituted for x). The preceding partial remainders 2, -2, - 4 are the coefficients of the quotient, the first being the same as the first coefficient of the dividend.

The process of synthetic division may be described as follows: To divide a polynomial by xa,

Write the coefficients in the order of descending powers of x, supplying missing terms with zero coefficients. To the right write a.

Bring down the first coefficient of the dividend.

Multiply the first coefficient by a, and add the product to the second coefficient.

Multiply the sum so obtained by a, and add the result to the third coefficient, etc.

The last term will be the remainder, R, and the preceding partial remainders will be the coefficients of the quotient, q(x), which will be of one degree less than the degree of the polynomial.

The remainder theorem, with the aid of synthetic division, affords a simple method of calculating the value of a polynomial for a given value of x.

3

9

4

17-37 4

For example, if f(x) = 3x4 - 9x3- 4x2 - 17x - 37, find f(4). By the remainder theorem, ƒ(4) is the value of the remainder, R, when f(x) is divided by x 4. By synthetic - 4. By synthetic 3 + 3 + 8 + 15 + 23 division we find that f(4) = 23.

+ 12 + 12 + 32 + 60

23. Verify this result by substituting 4 for x in the given polynomial.

If the multiplications and additions are performed mentally, the partial products in the second row of numbers omitted, and the partial remainders alone put down, the computation may be arranged still more compactly in the following form: The second arrangement is more convenient in calculating a table of values of a polynomial for integral values of x, while we shall find that the first is better adapted to finding zeros of the function if the zeros are irrational.

3 9 4 17 - 37 4 3 +3 +8 + 15 +23

49. Graph of a Polynomial. Example. Construct the

graph of

f(x) = x3 3x2

10x + 24.

The table of values is constructed by means of synthetic division and the remainder theorem, the details of the work and the graph being given below.

Hence we find that f(1) = 12, and ƒ(2) = 0.

The calculations may be arranged compactly by writing the coefficients of the polynomial once for all as in the top row of the following table and by operating on this top row with each value of x, performing the multiplications and additions mentally and entering the partial remainders on the same row with the value of x, under the proper coefficient of the dividend.

50. Extent of the Tables. In the example of the preceding section an examination of the synthetic division by x - 6 shows that the successive remainders 1, +3, +8, + 72 are all positive. If a larger value of x be used, by the nature of synthetic division, it is evident that the successive remainders will be positive and larger. Hence as x increases beyond 6 f(x) will also increase and the curve will run up to the right indefinitely. The synthetic division by x + 4 gave remainders 1, -7, 18, 48 which have alternating signs, As x de

creases below - 4, f(x) also decreases and hence the curve runs down indefinitely to the left.

The extreme values of x that should be included in the table are determined by the following rule:

The largest value included in the table is such that in the synthetic division the signs of the partial remainders are all the same. The smallest value (algebraically) included is such that these signs alternate.

EXERCISES

1. Plot the graphs of each of the polynomials below. From the graph find approximate values of the roots of the equation obtained by equating the polynomial to zero.

x pass,

2. The graphs of the functions x3 - x2, 2x3 3x2 + x, 2x3 x2 through the origin and the point (1, 0). Find the slope of the tangent line to each graph at these points, and plot the graphs. How does the slope of the tangent line assist in drawing the graphs?

3. Plot the graph of x4 4x+4x2, and show that it is tangent to the x-axis at two points.

4. A body moves so that its position, s, with reference to a fixed point at any time t, is given by one of the following equations. Find the velocity, v, and the acceleration, a, at any time t. Construct the graphs of s, v, and a on the same set of axes. When and where is the body at rest? What is the value of the acceleration when the velocity is a minimum? Where is the body at this moment and what is its velocity?

Suggestion. To find the acceleration, a, at any time t, find the limit of the average rate of change of v with respect to t.

Thus the rate 8x3. It will be

NOTE. Rate of change of a polynomial. It was seen on page 111 that the rate of change of x" is nx"-1, and it may be shown that the rate of change of ax" is anx-1. of change of 43 is 12x2, and that of 2x4 is show in a later chapter that the rate of change of a polynomial, or the slope of the line tangent to the graph, may be obtained by adding the rates of change of the successive terms, noting that the rate of change of the constant term is zero.

is

For example, the rate of change m of the polynomial

5. Find the slope m of a line tangent at any point to the graph of each of the following equations. Find the rate of change of the slope m with respect to x. Symbolize the rate of change of m with respect to x by F and plot the graphs of y, m and F on the same axes. Find the ordinate of the point on the graph of y, and also of the point on the graph of F, which has the same abscissa as the minimum point of m. Where is the line tangent to the graph of y horizontal?

NOTE. The point on a graph where m ceases to decrease and begins to increase (or vice-versa) is called a point of inflection. The graph is concave downward on one side of this point and concave upward on the other side. For this value of x the value of F is zero.

Hence, to find the abscissas of the points of inflection on the graph of a function, equate F to zero and solve for x. Substitute these

values of x in the equation y = f(x) to find the ordinates.

6. Find the coördinates of the point of inflection of the graph of each of the following equations. Translate the axes to this point as a new origin. Show that the graph is symmetric with respect to the point of inflection, and plot the graph of the equation on the new axes.

(c) y x3+6x2 + 12x + 8.

=

=

11.

6.

7. Find m and F for the function ax + bx2 + cx + d. If the axes are translated to the point of inflection of the graph of this function, show that the equation of the graph referred to the new origin is y' ax'3 + m'x', where m' is the slope of the line tangent to the graph of y at the point of inflection, and that the graph is symmetrical with respect to the point of inflection.

(b) the average of the velocities at t = 2, and t = 4, (c) the velocity at

t

=

3, (d) the value of t at which the velocity is equal to the average velocity for the interval t = 2 to t

=

4.

9. Find the points of inflection of the following functions and plot their graphs.

(a) x4 8x3 + 3x2.

(b) x5

10x2 - 8.

(c) x4 - 4x3.

10. For what positive integral values of n does TM have a point of inflection?

51. Solution of Equations. Rational Roots. Any algebraic equation in one variable may be reduced to the form

where f(x) is a polynomial, by the elementary rules for simplifying an equation involving fractions and radicals. The real roots of the equation (or zeros of the function) are represented graphically by the intercepts on the x-axis of the graph of f(x). (See page 23.)

If any integers are roots of (1), they will appear when the table of values for the graph is being constructed. Thus the integral roots of the equation

obtained by equating to zero the polynomial in the example of Section 49 are 2, 4, and 3, since the table of values shows that

f(2) = 0,

f(4) = 0, and

That these are all the roots follows from

Theorem 1. An equation of degree n has n roots, of which some may be imaginary and some equal to each other.

This theorem is assumed without proof.

Returning to equation (1), any real roots other than integers are either fractions or irrational numbers. In this section we shall see how to find the fractional roots of an equation, which with the integral roots, constitute the rational roots.

=

=

a.

Theorem 2. If x = a is a root of an equation of degree n, f(x) = 0, then f(x) = (x − a)q(x), where q(x) is of degree n · 1. By the definition of a root of an equation f(a) 0. Then by the remainder theorem, if f(x) is divided by x-a, the remainder is zero, so that f(x) is exactly divisible by x Since the degree of the divisor is 1, the quotient q(x) will be a polynomial of degree n 1. 1. Since the dividend is equal to the product of the divisor and quotient, we have the identity f(x) = (x − a)q(x). (2)