Was professor of mathematics in London and wrote papers on the foundations of algebra and on logic. To an English journal, the Athenæum, he contributed amusing articles on circle squarers. These articles were afterwards collected in a book, entitled A Budget of Paradoxes. Observe that the remainder, a3 - 5 a2 + 7 a 2, is the same as the dividend, if in the dividend a is substituted for x. This illustrates the Remainder Theorem: If a rational integral expression in x is divided by x a, the remainder is the same as the original expression with a substituted for x. Now 23 5 x2 + 7 x 2 would be exactly divisible by x = : 0. Hence such a polynomial will be exactly divisible by a binomial of the form a a, if, when a is substituted for x, the polynomial vanishes, i.e. equals zero. This illustrates the Factor Theorem: If a rational integral expression in x becomes zero when a number a is substituted for x, then x a is a factor of the expression. Thus, the expression x + 2 x5 —x−2, for x=1, becomes 1+2—1—2=0, hence x - 1 is a factor of the expression. Again, for x=— 2, this expression becomes 64 − 64 + 2 · x-(-2) or x + 2 is a factor of it. A third factor can be found by dividing x6 + 2 x5. uct of x-1 and x + 2; that is, by x2+x- - 2. x2 + x3 + x2+x+1. Hence we have, х 2 = 0, hence 2 by the prodThe quotient is = 1 and x=- 2. Notice In the example above we took x = that 1 and 2 are both integral factors of the absolute term 2 in the expression x6 +25 х 2. There is a theorem bearing on this point which we shall use along with the Factor Theorem. It relates to rational integral expressions, The theorem, which we give without proof, is as follows: When a rational integral expression has 1 as the coefficient of the highest power of x, and the other coefficients a, b, ......, h, k are all integral numbers (either positive or negative), then, in searching for rational values of x that will make the given expression zero, only integral factors of the absolute term k need be tried. The coefficient of x4 is 1; the other coefficients, 6, 42, - 49 are integers. Hence both conditions are satisfied, and we need to try only integral factors of 49; namely, ± 1, ± 7, ± 49. Hence, by the Factor Theorem, When * x = 1, 1 — 6 — 42 — 490; x + 1 is not a factor. When x = 7, 2401 + 2058 + 294 − 49 ‡ 0, x 7 is not a factor. When x=- 7, 2401 — 2058 — 294 — 49 = 0, x + 7 is a factor. In the same way, x - 49 and x + 49 are found not to be factors. Dividing the expression by (x − 1)(x+7) yields another factor, x2+7. Hence, x2 + 6 x3 + 42 x − 49 = (x − 1)(x + 7)(x2 + 7). * The symbol means "is not equal to." See § 76, footnote. - 1 is 5. What is the remainder when 2x3 3x2 + 5 x divided by x-c? By x-b? By x - 1? 6. What is the remainder when 5x+2x2+3x+6 is divided by x+n? By x+1? 7. Is x + 2 an exact divisor of 4 x 6? 8. If a rational integral expression, with 1 as the coefficient of the highest power of x, and with all other coefficients integers, is exactly divisible by xa, where a is an integer, what relation is a to the last term? 9. Find a factor of x5 204 +203 x2 + x 10. Find a factor of 23 11. Find a factor of 2 x4 9 x2 + 17 x - 6. 1. x32x2 - 3 x 10. Hence, by the Factor Theorem, x- a is a factor of x5 a5. Division of x5 - a5 by x-a yields a second factor. Hence x5 — a5 = (x − a) (x4 + x3a + x2a2 + xa3 + a1).` The quotient may be obtained by the following special method of short division: If -a is substituted for x, then (− a)5 + a5 Hence by the Factor Theorem, x-(-a) or x + a is a factor of x5+ a5. By division, we find a second factor. Hence x5 + a3 = (x + a) (x1 — x3α + x2α2 xa3 + a1). The special rule of division given in Ex. 13 applies to this case, except that the signs in the quotient are alternately + and |