But we know that x - a divides a3 - a3; it will therefore also divide x-a1. And, since x - a divides x2 - a1 it will also divide a". And so on indefinitely. Hence "a" is divisible by x-a, when n is any positive integer. a the 87. Since "a" = x" — a" + 2a" it follows from the last Article that when "+a" is divided by xremainder is 2a”, so that x" + a” is never divisible by x-a. If we change a into a becomes x − (− a) = x+a; also x" - a" becomes x" - (-a)”, and x" − (− a)” is x2 + a” a" according as n is odd or even. Hence, when n is odd or xn a, Ꮖ x"+a" is divisible by x+a, and when n is even x" - a" is divisible by x + a. Thus, n being any positive integer, x"+a" never, xn a" when n is even, x+a" when n is odd. The above results may be written so as to shew the quotients: thus the upper or lower signs being taken on each side of the second formula according as n is odd or even. 88. Theorem. If any rational and integral expression which contains a vanish when a is put for x, then will x-a be a factor of the expression. be Let the expression, arranged according to powers of x, -1 n-2 n-1 = ax" + bx”-1+ cx"2+...... - (ax" + bx"¬1 + ca"-2+...) = a (x” — a”) + b (xn−1 — a”−1) + c (xn−2 − an−2) + ... n-1 n-2 But, by the last Article, "a", x”-1 — a”-1, x2-2 — a”-2, &c. are all divisible by x- α. Hence also a” + bx2-1 + cx22+......... is divisible by x-a. The proposition may also be proved in the following manner. n-2 Divide the expression ax" + bx + cx2+...... by x-a, continuing the process until the remainder, if there be any remainder, does not contain x; and let Q be the quotient and R the remainder. Then, by the nature of division, and this relation is true for all values of x. Now since R does not contain x, no change will be made in R by changing the value of x: put then x = a, and we have = ...... Q (a− a) + R = R. = 0 Hence, if any expression rational and integral in x be divided by x-a, the remainder is equal to the result obtained by putting a in the place of x in the expression. It therefore follows that the necessary and sufficient condition that an expression rational and integral in x may be exactly divisible by x-a is that the expression should vanish when a is substituted for x. Ex. 1. Find the remainder when x3- 4x2 + 2 is divided by x-2. Ex. 2. Find the remainder when x3-2a2x+a3 is divided by x-a. Ex. 3. Shew by substitution that x-1, x-5, x+2 and x+4 are factors of x4-23x2-18x+40. Ex. 4. Shew by substitution that a -b is a factor of a3 (b −c) + b3 (c − a) + c3 (a − b). Put ab and the expression becomes a3 (a−c)+a3 (c − a), which is clearly zero: this proves that a -b is a factor. Ex. 5. Shew that a is a factor of (a+b+c)3-(a+b+c)3 − (a−b+c)3- (a+b-c)3. 89. We have proved that xa is a factor of the expression ax + bxanî + can-2 + ......, provided that the expression vanishes when a is put for a. If the division were actually performed it is clear that the first term of the quotient, which is the term of the highest degree in x, would be ax-1. Hence the given expression is equivalent to -1 a and ax2-1+ ...... a does not vanish ...... must vanish Now suppose that the given expression also vanishes. when x = ẞ; then the product of x will vanish when x=ß; and since x when xß, it follows that ax2-1+ when x = B. Hence x-ẞ is a factor of ax11+ &c.; and, if the division were performed, it is clear that the first term of the quotient would be ax2-2. = Hence the original expression is equivalent to (x − a) (x − B) (αx2-2 + &c.................). Similarly, if the original expression vanishes also for the values y, d, &c. of x, it must be equivalent to (x-α) (x-B) (x − y) (x — 8)......(ax+ &c.......), where is equal to the number of the factors x-a, x-B, &c. If therefore the given expression vanish for n values a, B, y &c. there will be n factors such as x -a, and the remaining factor, ax+ &c. will reduce to a; and hence the given expression is equivalent to 90. Theorem. An expression of the nth degree in x cannot vanish for more than n values of x. For if the expression vanish for the n values a, ẞ, y......, it must be equivalent to a (x − a) (x − B) (x − y)................ If now we substitute any value, k suppose, different from each of the values a, ẞ, y, &c. ; then, since no one of the factors ka, k- B, &c. is zero, their continued product cannot be zero, and therefore the given expression cannot vanish for the value x=k, except a itself is zero. 2-2 But, if a is zero, the original expression reduces to bx2¬1 + cx2-2 +......, and is of the (n-1)th degree; and hence as before it can only vanish for n-1 values of x, except b is zero. And so on. Thus an expression of the nth degree in x cannot vanish for more than n values of x, except the coefficients of all the powers of x are zero; and when all these coefficients are zero, the expression will clearly vanish for all values of x. 91. Theorem. If two expressions of the nth degree in x be equal to one another for more than n values of x, they will be equal for all values of x. and If the two expressions of the nth degree in x be equal to one another for more than n values of x, it follows that their difference, namely the expression n-1 (a− p) x" + (b − q) x2-1 + (c − r) x2-2 + ......, will vanish for more than n values of x. Hence, by Art. 90, the coefficients of all the different powers of a must be zero. Thus a − p = 0, b − q = 0, c — r = 0, &c. that is, a=p, b=q, c=r, &c. Hence, if two expressions of the nth degree in x are equal to one another for more than n values of x, the coefficient of any power of x in one expression is equal to the coefficient of the same power of x in the other expression. When any two expressions, which have a limited number of terms, are equal to one another for all values of the letters involved, the above condition is clearly satisfied, for the number of values must be greater than the index of the highest power of any contained letter. Hence when any two expressions, which have a limited number of terms, are equal to one another for all values of the letters involved in them, we may equate the coefficients of the different powers of any letter. 92. Cyclical order. It is of importance for the student to attend to the way in which expressions are usually arranged. Consider, for example, the arrangement of the expression bc + ca+ab. The term which does - not contain the letter a is put first, and the other terms can be obtained in succession by a cyclical change of the letters, that is by changing a into b, b into c and c into a. In the expression a2 (b − c) + b2 (c − a) + c2 (a - b) the same arrangement is observed; for by making a cyclical change in the letters of a2 (b−c) we obtain b2 (c- a), and another cyclical change will give c2 (a - b). So also the second and third factors of (b−c) (c − a) (a - b) are obtained from the first by cyclical changes. |