FACTORING. PREPARATORY STEPS. 174. STEP I.—Find by inspection all the exact divisors of each of the following numbers, and write them in order on your slate, thus : 6 = 3 x 2, 10 = 5 x 2. 1. 6 10 14 15 21 22 26 5. 3 11 115 119 123 129 141 Refer to the results on your slate and observe (1.) Each prime exact divisor is called a prime factor of the number of which it is a divisor. (2.) Each number is equal to the product of its prime factors. STEP II.—The same prime factor may enter into a number two or more times. Thus, 18 = 2 X 3 X 3. Hence the prime factor, 3, enters twice into 18. Resolve the following numbers into their prime factors, and name how many times each factor enters into a number. 1. 4 8 16 32 64 9 27 DEFINITIONS. 175. A Factor is one of the equal parts of a number, or one of its exact divisors. Thus, 15 is composed of 3 fives or 5 threes ; hence, 5 and 3 are factors of 15. 176. A Prime Factor is a prime number which is a factor of a given number. Thus, 5 is a prime factor of 30. 177. A Composite Factor is a composite number. which is a factor of a given number. Thus, 6 is a composite factor of 24. 178. Factoring is the process of resolving a composite number into its factors. 179. An Exponent is a small figure placed at the right of a number and a little above, to show how many times the number is used as a factor. Thus, 35 = 3 x 3 x 3 x 3 x 3. The 5 at the right of 3 denotes that the 3 is used 5 times as a factor. 180. A Common Factor is a number that is a factor of each of two or more numbers. Thus, 3 is a factor of 6, 9, 12, and 15; hence is a common factor. 181. The Greatest Common Factor is the greatest number that is a factor of each of two or more numbers. Thus, 4 is the greatest number that is a factor of 8 and also of 12. Hence 4 is the greatest common factor of 8 and 12. ILLUSTRATION OF PROCESS. 182. Find the prime factors of 462. 2 ) 462 EXPLANATION.-1. We observe that the number 462 is divisible by 2, the smallest prime number. Hence 3 ) 231 we divide by 2. . g) 77 2. We observe that the first quotient, 231, is divisible 11 by 3, which is a prime number. · Hence we divide by 3. 3. We observe that the second quotient, 77, is divisi. ble by 7, which is a prime number. Hence we divide by 7. 4. The third quotient, 11, is a prime number. Hence the prime factors of 462 are 2, 3, 7, and 11 ; that is, 462 = 2 x 3 x 7 x 11. Any composite number may be factored in the same manner. Hence the following 183. RULE.—Divide the given number by any prime number that is an exact divisor, and the resulting quotient by another, and so continue the division until the quotient is a prime number. The several divisors and the last quotient are the required prime factors. EXAMPLES FOR PRACTICE. 184. Find the prime factors of the following numbers : 23. 5184. 24. 9160. 25. 8030. 26. 4165. 27. 62500. 28. 81000. %. 3234. 18. 1000. 29. 64000. 8. 4361. 19. 4515. 30. 45500. 9. 30030. 20. 17854. 31. 16875. 10. 11025. 21. 2310. 32. 18590. 11. 14600. 22. 5450. 33. 16380. CANCELLATION. PREPARATORY PRO POSITIONS. 185. Study carefully the following propositions : PROP. I.—Rejecting a factor from a number divides the number by that factor. Thus, 72 = 24 x 3. Hence, rejecting the factor 3 from 72, we have 24, the quotient of 72 divided by 3. Prop. II.-Dividing both dividend and divisor by the same number does not change the quotient. Thus, 60 = 12 = 20 threes ; 4 threes = õ. Observe that the unit three, in 20 threes • 4 threes, does not in any way affect the size of the quotient; therefore, it may be rejected and the quotient will not be changed. Hence, dividing both the dividend 60 and the divisor 12 by 3 does not change the quotient. ILLUSTRATION OF PROCESS. 186. Ex. 1. Divide 462 by 42. 6 ) 462 typy EXPLANATION.–We divide both the 6 / 42 = m = ll. divisor and dividend by 6. According to Prop. II, the quotient is not changed. 13 8 11 EXPLANATION.-1. We divide any factor in the dividend by any number that will divide a factor in the divisor. Thus, 65 in the dividend and 15 in the divisor are divided each by 5. In the same manner, 55 and 35, 13 and 39, 24 and 3 are divided. The remaining factors, 8 and 11, in the dividend are prime to each of the remaining factors in the divisor. Hence, no further division can be performed. 2. We divide the product of 8 and 11, the remaining factors in the dividend, by the product of 3 and 7, the remaining factors in the divisor, and find as a quotient 4,4, which, according to (185-II), is equal to the quotient of 65 x 24 x 55 divided by 39 x 15 x 35. All similar cases may be treated in the same manner; hence, the following 187. RULE.—1. Cancel all the factors that are common to the dividend and divisor. II. Divide the product of the remaining factors of the dividend by the product of the remaining factors of the divisor. The result will be the quotient required. WRITTEN EXAMPLES. 188. 1. Divide 9009 by 6006. Ans. 11. 2. Divide 8475 by 525. Ans. 161. 3. Divide 3328 by 216. Ans. 1511. 4. Divide 8 x 15 x 40 by 10 x 24. Ans. 20. 5. Divide 49 x 25 x 12 by 16 x 36 x 5. Ans. 545 6. Divide 12500 by 175. Ans. 1667. 7. Divide 64 x 81 x 25 by 24 x 27. Ans. 200. 8. Divide 12 x 49 x 27 by 42 x 14. Ans. 27. 9. What is the quotient of 16 times 5 times 4 divided by 8 times 20 ? Ans. 2. 10. Multiply 8 times 66 by 5 times 18, and divide the product by 33 times 72. Ans. 20. 11. If 10, 12, 84, and 42 are the factors of the dividend, and 12, 5, 24, and 7 are the factors of the divisor, what is the quotient ? Ans. 42. 12. How many barrels of flour, at 12 dollars a barrel, are worth as much as 16 cords of wood, at 3 dollars a cord ? 13. At 18 dollars a week, how many weeks must a man work to pay 3 debts of 180 dollars each ? Ans. 30 weeks. 14. When a laborer can buy 36 bushels of potatoes, at 4 shillings a bushel, with the earnings of 24 days, how many shillings does he earn a day? Ans. 6 shillings. 15. How many loads of potatoes, each containing 15 bushels, at 42 cents a bushel, will pay for 12 rolls of carpeting, each containing 56 yards, at 75 cents a yard ? Ans. 80 loads. 16. A man exchanged 75 bushels of onions, at 90 cents a bushel, for a number of boxes of tea, containing 25 pounds each, at 54 cents a pound; how many boxes did he receive? 17. How many pounds of tea, at 72 cents a pound, would pay for 3 hogsheads of sugar, each weighing 1464 pounds, at 15 cents a pound ? Ans. 915 pounds. |