curve is on the X axis. synthetically by 2, and in § 161. To find the quotient of our equation we first divide then the quotient by 2, using the principle given Thus the quotient of the polynomial and (x − 2) (x + 2) is x2 — 6x + 4. Solving the equation x2-6x+4= 0, we obtain the two remaining roots, x = 3√5. These remaining roots might also be found by the method of exercise 2. (k) How many imaginary roots can an equation of the 5th degree have? (1) x3 — ax2 + bx − c = 0 has two roots whose sum is zero. What is the third root? What are the two roots whose sum is zero? (m) x3 + x2 + bx + c = 0 has one root the reciprocal of the other. What are the values of the roots? (n) x3 — 4x2 + ax + b2 = 0 has the sum of two roots equal to zero. What must be the values of a and b? (0) x2 - 3 x3 + ax2 + bx + 9 = 0 has the sum of three of its roots equal to zero. What must be the value of b? 175. To multiply the roots by a constant. Suppose we have the equation whose roots are a1, a2,, a. An equation of this type for values of n greater than 2 is usually not solvable by elementary methods. It often happens, however, that by changing its form slightly we may obtain an equation one or more of whose roots we can find. We shall see that if an equation has rational roots we may always find them if we change the form of the equation as indicated on the following page. We seek to form from (1) an equation whose roots are equal to the roots of (1) multiplied by a constant factor, as k. Thus the equation we seek must have the roots ka1, kα2, kaz. We carry out the proof, which is perfectly general, on the equation of the third order f(x)= αx3 +а1x2 + α2x + аz = 0, whose roots are a1, a2, α3. The equation that we seek must have roots ka1, ka, kag. Since now (§ 53) ƒ(x) = 0 is satisfied by a, where a stands for any one of the roots, that is, since ƒ(a) = 0, evidently f =0 is satisfied by ka, that is, 2 Hence we obtain an equation that is satisfied by ka1, ka, ka, RULE. To multiply the roots of an equation by a constant k, multiply the successive coefficients beginning with the coefficient of x-1 by k, k2, ..., k" respectively. In performing this operation the lacking powers of x should be supplied with zero coefficients. EXAMPLE. Multiply the roots of 2 x3 − 3x + 4 = 0 by 2. Simplifying, 2 x3 +2.0 x2 - 4·3x + 8.4 = 0. x36x + 16 = 0. When an equation in form (2), p. 166, has fractional coefficients, an equation may be formed whose roots are a properly chosen multiple of the roots of the original equation and whose coefficients are integers. COROLLARY I. When k is a fraction this method serves to divide the roots of an equation by a given number. COROLLARY II. When k 1 this method serves to form an equation whose roots are equal to the roots of the original equation but opposite in sign. This is equivalent to the statement that f(x)=0 has roots equal but opposite in sign to those of f(x)=0. or EXERCISES 1. Form the equation whose roots are three times the roots of : x4 - 6x3 x + 1 = 0. Solution Supplying the missing term in the equation, we have x4 - 6x3 +0x2 - x + 1 = 0. Since k = 3, we have by the rule x4 - 3.6x3 +9.0x2 - 27. x2 + 81 = 0, x4 18 x3 27 x2+81 0. = 2. Find the equation whose roots are twice the roots of x4+3x3- 2 x + 4 = 0. 3. Find the equation whose roots are one half the roots of 4. Find the equation whose roots are two thirds the roots of x3- 4x6 = 0. 5. By what may the roots of the following equations be multiplied so that in the resulting equation the coefficient of the highest power of x is unity and the remaining coefficients are integers? Form the equations. (a) 3x3- 6x + 2 = 0. Solution: We wish to bring into every term such a factor that all the resulting coefficients are divisible by 3. 6. Form equations whose roots are the negatives of the roots of the following equations. 7. What effect does changing the sign of every term of the member involving x have on the graph of an equation? 8. What is the graphical interpretation of the transformation which changes the signs of the roots of an equation, that is, what relation does the graph of the equation before transformation bear to the graph of the equation after transformation (a) when the degree is an even number, (b) when the degree is an odd number? 9. If 4x4 + 16 x3 + 85 x2+4x+21 = O has as two roots what are the roots of 4x4 16 x3+85 x2 - 4x+21=0? - and - 3, 10. If a root of x3 11 x2+36x360 is 2, what are the roots of x3 + 11 x2 + 36 x + 36 = 0 ? 176. Descartes' rule of signs. A pair of successive like signs in an equation is called a continuation of sign. A pair of successive 'unlike signs is called a change of sign. In the equation 2x4 3x32x2+2x-3=0 (1) are one continuation of sign and three changes of sign. This may be seen more clearly by writing merely the signs, + ++ Let us now inquire what effect if any is noted on the number of changes of sign in an equation if the equation is multiplied by a factor of the form x a when a is positive, that is, when the number of positive roots of the equation is increased by one. Let us multiply equation (1) by x-2. We have then. in which there are four changes of sign, that is, one more change of sign than in (1). If an increase in the number of positive roots always brings about at least an equal increase in the number of changes of sign, there must be at least as many changes of sign in an equation as there are positive roots. This is the fact, as we now prove. DESCARTES' RULE OF SIGNS. An equation f(x)=0 has no more real positive roots than f(x) has changes of sign. ILLUSTRATION. In the equation of degree one x 2 = O there is one change of sign and one real root. In the case of a linear equation there is no possibility of more than one change of sign. In the quadratic equation 2+2x+1= 0 there is no change of sign, and also no positive root since for positive values of x the expression x2 + 2x + 1 is always positive and hence never zero. In the equation x2+2x-3=0 we have one change of sign, and one positive root, +1. We shall prove this general rule by complete induction. First. We have just seen that the rule holds for an equation of degree one. Second. We assume that the rule holds for an equation of degree m, and prove that its validity for an equation of degree m+1 follows. We shall show that if we multiply an equation of degree m by xa, where a is positive, thus forming an equation of degree m +1, the number of changes of sign in the new equation always exceeds the number of changes of sign in the |