Problems producing Simultaneous Quadratic Equations. 900. The sum of the squares of two numbers is 130, and their product is 63. Find the numbers. 901. The sum of the cubes of two numbers is 126, and the sum of the numbers is 6. Find them. 902. One of two numbers exceeds 30 by as much as the other is less than 30, and their product is 875. Find them. 903. The diagonal of a rectangle is 100 feet, and the longer side exceeds the shorter by 20 feet. Find the area of the rectangle. 904. The sum of the squares of two numbers is 13 times the smaller number, and the sum of the numbers is 10. Find them. 905. The sum of the squares of two numbers is 2 a2 + 2, and the sum of the numbers is 2 a. What are the numbers, and what is the difference of their squares? 906. How many yards of picture molding will be needed for a room whose ceiling area is 1200 square feet, the diagonal of the ceiling being 50 feet? 907. If twice the product of the ages of two children is added to the sum of their ages, the result is 13 years. One child is 3 years older than the other. Find the age of each. 908. If the sum of two numbers is multiplied by the lesser the product is 5. If their difference is multiplied by the greater, the product is 12. Find the numbers. 909. The difference between the numerator and denominator of a certain improper fraction is 2, and if both terms of the fraction are increased by 3, the value of the fraction will be decreased by Find the fraction. 910. A certain floor having an area of 50 square feet can be covered with 360 rectangular tiles of a certain size; but if the masons use a tile 1 inch longer and 1 inch wider, the floor can be covered with 240 tiles. Find the sizes of the different tiles. CHAPTER XXIX SUPPLEMENTARY TOPICS THE REMAINDER THEOREM 468. If any polynomial of the form, C12"+C1x-1 + Czan--2 + ... C1, be divided by x-a, the remainder will be Cа" +С2α-1+ Сçα-2 +C; which expression differs from the given expression in that a takes the place of x. ... Proof: Let Q denote the quotient and R the remainder when C1x2 + C2xn−1 + Cзxn−2 + ... Cn is divided by x α. ... Спо Continue the division until the remainder does not contain x. Application: ... Cn. ... Cn. 1. Without division obtain the remainder when 7x+3 x 469. If any rational and integral expression containing x becomes equal to 0 when a is substituted for x, the expression is exactly divisible by x a. α Proof: Let E be the given expression, and let E be divided by x until the remainder no longer contains x. Let Q denote the quotient obtained and R the remainder. This identity being true for all values of x, we may assume that x = a. By the hypothesis the substitution of a for x makes E equal to 0. Therefore, 0 = E (a − a) + R, 0=0+ R, R = 0. Therefore, the remainder being 0, the given expression is exactly divisible by x — a. Or, x- a is a factor of the given expression, E. Illustrations: 1. Factor 2-12x+16. By trial we find that the expression, x3 −12x+16, equals 0 when x = 2. Therefore, if x = 2, x-ax 2, and x 2 is a factor. Then (x3-12x + 16) ÷ (x − 2) = x2 + 2 x − 8. The factors of x2 + 2 x Therefore, x3- 12 x + 16 = (x − 2)(x + 4)(x − 2). Result. 2. Factor 2+9x2+23x+15. (In an expression whose signs are all plus, it is evident that no positive number can be found the substitution of which will make the expressior equal to 0.) By trial we find that if x = - 1, the expression becomes 0. Hence, if x = 1, x − a = [x-(-1)]= x + 1, a factor required. Then, (x2 + 9 x2 + 23 x + 15) ÷ (x + 1) = x2 + 8 x + 15. Furthermore, x2 + 8x + 15 = (x+3)(x + 5). Therefore, x+9x2 + 23 x + 15 = (x + 1)(x+3)(x+5). Result. Exercise 138 Without division show that 1. a3 +3a2+3 a +2 is divisible by a +2. 2. 24-8x3+24x2-32x+16 is divisible by x-2. 3. cc3-8c2+9c-9 is divisible by c+3. 4. 27 a3+9a2-3 a-10 is divisible by 3a-2. 5. m3 -5 m1 +9 m3 — 6 m2 — m +2 is divisible by m-2 THE THEORY OF DIVISORS OF BINOMIALS 470. The following proofs depend directly upon the prin ciples established in Arts. 468 and 469. If n is a positive integer, we may establish as follows the divisibility of the binomials, "y" and "+y", by the binomials, x ·y and x+y. I. x-y" is always divisible by x-y. For, if y is substituted for x in x2 Therefore, x པ་ y", we have xn · yn = yn — yn = 0. y is always a divisor of x" — yo. II. xy" is divisible by x+y if n is even. xn — yn = (− y)n — yn — yn — yn = 0. Therefore, x + y is a divisor of x" — y" when ʼn is even. For, if y is substituted for x in x" + y", we have x2 + yn = yn + yn = 2 yn, which result does not reduce to 0 by the substitution. y is never a divisor of x2 + yn. Therefore, x IV. x2+y" is divisible by x + y if n is odd. For, ify is substituted for x in x2 + y2, we have Therefore, x + y is a divisor of x + y when n is odd. Then, (x2)5+(y2)5 is divisible by x2 + y2. (Art. 470, IV.) Whence, (x2) 5 + (y2) 5 = (x2) 4 — (x2)3 (y2) + (x2)2(y2)2 − (x2) (y2)8 + (y2)4 = x8 — x3y2 + x1y1 — x2y® + y3. Result. 2. Divide 64 x12 — y6 by 2 x2 - y. We may write Then, (2 x2)6 — yo 2x2-y = -- = (2 x2)5 + (2 x2)1y + (2 x2)3y2 + (2 x2)2y8 + (2 x2) y1 + y5 =32x10+ 16 x3y + 8 xôy2 + 4 x1y3 + 2 x2y2 + y5. Result. 5. What are the exact binomial divisors of 81 - a1? |