16. 27 (cos 75° + i sin 75o). 17. 16 (cos 200° + i sin 200°). 18. Solve the following equations and plot their roots. B z = √1 (cos 0° + i sin 0°) = VI [cos ( (where k takes on the values 0, 1, 2, 3, 4) = 0, COS 0° + i sin 0° 1, when k = cos 720+ i sin 72°, when k = 1, cos 144° + i sin 144°, when k = = 2, cos 216° + i sin 216°, when k = 3, cos 288° + i sin 288°, when k = 4. These numbers we observe lie on a circle of unit radius at the vertices of a regular pentagon. D E CHAPTER XVII THEORY OF EQUATIONS 159. Equation of the nth degree. Any equation in one variable in which the coefficients are rational numbers can be put in the form where a is positive and a,, a, are all integers. (1) The symbol f (x) is read "ƒ of x" and is merely an abbreviation for the righthand member of the equation. Often we wish to replace x in the equation by some constant, as α, 2, or 0. We may symbolize the result of this substitution by f (a), f (-2), or ƒ (0). Thus We symbolize other expressions similarly by (x), Q(x), etc. When we speak of an equation we assume that it is in the form of (1). This equation is also written in the form The 's are integers only when a = 1. ao 160. Remainder theorem. We now prove the following important fact. THEOREM. When f(x) is divided by x-c, the remainder is f(x) with c substituted in place of the variable. Divide the equation (1) by x - C. Let R be the remainder, which must (§ 26) be of lower degree in x than the divisor; that is, in this c is the divisor, R must be a constant and not involve x at all. Let the quotient, which is of degree n 1 in x, be represented by Q(x). since x case, But since this equation is an identity it is always satisfied whatever numerical value x may have (§ 53). COROLLARY. If c is a root of f(x) = 0, then x − c is a factor of the left-hand member. For if c is a root of the left-hand member, it satisfies that member and reduces it to zero when substituted for x. Thus by the previous theorem we have, since 161. Synthetic division. In order to plot by the method of § 103 the equation when the a's are replaced by integers, we should be obliged laboriously to substitute for x successive integers and find corresponding values of y, which for large values of n involves considerable computation. We can make use of the preceding theorem to lighten this labor. The object is to find, with the least possible computation, the remainder when the polynomial f(x) is divided by a factor of form xc, which by the preceding theorem is the value of f(x) when x is replaced by c, that is, the value of y corresponding to x = c. For illustration, let f(x) = 2 x1 3 x3 + x2 - x - 9 and c = = 2. By long division we have x22x-3x2 + x2 - X- 92x3 + x2+3x+5 We can abbreviate this process by observing the following facts. Since x is here only the carrier of the coefficient, we may omit writing it. Also we need not rewrite the first number of the partial product, as it is only a repetition of the number directly above it in full-faced type. Our process now assumes the form 1-22-3+1-1- 92+1 + 3 + 5 1 +1 - 2 +3 - Since the minus sign of the 2 changes every sign in the partial product, if we replace 2 by + 2 we may add the partial product to the number in the dividend instead of subtracting. This is also desirable since the number which we are substituting for x is 2, not 2. Thus, bringing all our figures on one line and placing the number substituted for x at the right hand, we have to We observe that the figures in the lower line, 2, 1, 3, 5, up the remainder are the coefficients of the quotient 2 x3 + x2+3x+5. RULE FOR SYNTHETIC DIVISION. Write the coefficients of the polynomial in order, supplying O when a coefficient is lacking. Multiply the number to be substituted for x by the first coefficient, and add (algebraically) the product to the next coefficient. Multiply this sum by the number to be substituted for x, add to the next coefficient, and proceed until all the coefficients are used. The last sum obtained is the remainder and also the value of the polynomial when the number is substituted for the variable. 162. Proof of the rule for synthetic division. This rule we now prove in general by complete induction. Let the polynomial be аx2 + α1x2-1 + α2x2-2 + + an -2 Let the number to be substituted for x be a. La аo аox + а1 (αα + a1) α + a2 = ·аα2 + α1α + A2. Second. Assume the validity of the that its validity for n = m + 1 follows. carried out on ƒ(x)= α ̧x2 + α1×TM−1 + affords the remainder rule for n = m, and prove Assume then that the rule +a, m Hence the next to the last remainder obtained by applying the rule to this polynomial would be ƒ(a), since the succession of coefficients is the same for both polynomials up to am+1. By the rule the final remainder is obtained by multiplying the expression just obtained, in this case ƒ(a), by a and adding the last coefficient, in this case am +1. This affords the final remainder |