286. If x3-3x2+4x+8 is divided by x 2 and there is a remainder (which does not contain x), then x3-3x2+4x+8= (x − 2) × Quotient + Remainder. Or, substituting Q and R respectively for "Quotient" and "Remainder," and transposing, R = x3-3x2 + 4 x + 8 − (x − 2) Q. As R does not contain x, we could, if Q was known, assign to x any value whatsoever and would always obtain the same answer for R. If, however, we make x = 2, then (x-2)Q = 0, no matter what the value of Q. Hence, even if Q is unknown, we can find the value of R by making x=2. R=23-3.22 +4.2+80=12. Ex. 1. Without actual division, find the remainder obtained by dividing 3x+2x-5 by x-3. Let then R=3x+2x-5-(x-3)Q. x = 3, R = 3.81 +2.3-5-0 = 244. Ex. 2. Without actual division, find the remainder when ax2 + bx3 + cx2+ dx+e is divided by x-m. 287. The Remainder Theorem. If an integral rational expres- m, the remainder is obtained sion involving x is divided by x by substituting in the given expression m in place of x. E.g. The remainder of the division (4 x5 - 4x+11) ÷ (x+3) is 4 (− 3)5 — 4 (− 3) + 11 - 949. The remainder obtained by dividing (x+4)1 − (x + 2) (x −1) + 7 by x-1 is 54 - 3 · 0 + 7 = 632. EXERCISE 109 Without actual division, find the remainder obtained by dividing: 1. x3-3x2-3x+2 by x-4. 2. 210-728+4x-2x-4 by x-1. 4. a1+5a2+2a+1 by a +2. 5. (x+1)3 −3(x − 2) (x+2) + (x + 1)2 by x − 2. - m. 7. 23 - 2x2n +4 xn2 - n3 by x+2 n. 8. (x+4)2+(x+3)3 + (x+2)* by x+1. 9. (a−1)(a − 3) − 4 [(a − 3)(a — 4) — 3] by a −3. 10. b5 by x-b. 11. a5 + b5 by a+b. 12. a20+b20 by a+b. 288. If the remainder is zero, the divisor is a factor of the dividend. The Factor Theorem. If a rational integral expression involving x becomes zero when m is written in place of x, x-m is a factor of the expression. 3x2-2x - 8 is divided by x-4, the remainder equals 2.480, hence (x-4) is a factor of x3-3x2 - 2 x −8. E.g. if x3 43 - 3.42 289. A rational integral expression, ax + bxn-1... ex+f, is divisible by x m only, if m is a factor of f. (§ 131.) Hence, to factor the expression, we substitute for m exact divisors of f, and determine by means of the factor theorem whether x m is a factor or not. Let x=-1, then x3- 7 x2 + 7 x + 15 = 0. Therefore x-(− 1), or x + 1, is a factor. By dividing by x + 1, we obtain x3- 7 x2 + 7 x + 15 = (x + 1)(x2 − 8 x + 15). The substitution x = a makes the expression vanish. Without actual division, show that: 1. 10 29-4x4-13 x2+7 is divisible by x-1. 2. a*-4 a2 - 7 a -24 is divisible by a - 3. 3. a1+a3 — ab3 - b3 is divisible by a-b. 4. x2+3x2-4x-12 is divisible by x-2 and x + 2. 5. (x+1)2(x-2)-4(x-1) (x-3)+ 4 is divisible by x-1. 6. 6 x[4(x+1)(x + 2) — 47] − ∞3 − x2+3x-6 is divisible by x-2. 7. 6x-3x-5 x3+5 x2 -2x-3 is divisible by x+1. 21. 23-(3 a—b)x2 + (2 a2 − 3 ab)x+2 a2b. 22. x3-(a+b+c)x2 +(ab+bc + ca)x — abc. 23. 2-(3a+2 b)x2 +(6 ab +2 a2)x — 4 a2b. 24. Find the H. C. F. of 3 a3+5 a2-a+2 and a3+a2-a+2. 25. Find the H. C. F. of 9 x3 + 18 x2 − x − 10 and 3 x + 13 x2 +2x-8. 26. Find the H. C. F. of x3- x2-5x-3 and 2-4 x2 - 11 x 6. 290. If n is a positive integer, it follows from the Factor Theorem that 1. x2 — y2 is always divisible by x— y. For substituting y for x, x-y"y" —y" = 0. 2. x2+y" is divisible by x+y, if n is odd. For (y)"y"=0, if n is odd. By actual division we obtain the other factors, and have for any positive integral value of n, x2 e.g. x5 — y5 = (x − y) (x2 + x3y + x2y2 + xy3 + y1). x5 + y5 = (x + y) (x± − x3y + x2y2 — xy3 + y1). 291. It can readily be seen that x2+y" is not divisible by either x+y or x- y, if n is even, and that "y" cannot be divided by x+y, if n is odd. EXERCISE 111 State whether the following expressions are prime or not, and factor whenever possible. [For additional examples of this type, see Appendix II.] |