94. When the Coefficient of the Highest Power is not Unity. Again, referring to Chap. IV. Art. 51, we may write the following results: (3x+2)(x+4)= 3 x2 + 14 x + 8 (1), (2), (3), (4). The converse problem presents more difficulty than the cases we have yet considered. Consider the result 3x2. 14x+8 = (3 x ·2) (x-4). The first term 3x2 is the product of 3 x and x. The third term +8 is the product of 2 and -4. The middle term - 14 x is the result of adding together the two products 3 x x -4 and ≈ × — 2. Again, consider the result 3 2−10 x−8=(3x+2)(x−4). The first term 3x2 is the product of 3 x and x. The third term 8 is the product of +2 and The middle term 10x is the result of adding together the two products 3 x x -4 and x x 2; and its sign is negative because the greater of these two products is negative. `Considering in a similar manner results (1) and (4), we see that: 1. If the third term of the trinomial is positive, then the second terms of its factors have both the same sign, and this sign is the same as that of the middle term of the trinomial. 2. If the third term of the trinomial is negative, then the second terms of its factors have opposite signs. 95. The beginner will frequently find that it is not easy to select the proper factors at the first trial. Practice alone will enable him to detect at a glance whether any pair he has chosen will combine so as to give the correct coefficients of the expression to be resolved. Ex. Resolve into factors 7x2 - 19 x 6. Write (7x 3) (x 2) for a first trial, noticing that 3 and 2 must have opposite signs. These factors give 7 x2 and - 6 for the first and third terms. But since 7 x 2-3 × 1 = 11, the combination fails to give the correct coefficient of the middle term. Next try (7x 2) (x 3). Since 7 x 3 2 x 119, these factors will be correct if we insert the signs so that the negative shall predominate. Thus 7 x2 – 19 x − 6 = (7 x + 2) (x − 3). [Verify by mental multiplication.] 96. In actual work it will not be necessary to put down all these steps at length. The student will soon find that the different cases may be rapidly reviewed, and the unsuitable combinations rejected at once. Ex. 1. Resolve into factors 14 x2 + 29 x − 15. (1), (2). In each case we may write (7 x 3) (2 x 5) as a first trial, noticing that 3 and 5 must have opposite signs. And since 7 x 53 x 2 = 29, we have only now to insert the proper In (1) we notice that the factors which give 6 are both positive. In (2) we notice that the factors which give 6 are both negative. And therefore for (1) we may write (5x+)(x+ ). (2) we may write (5x)(x). And, since 5 × 3 + 1 x 2 = 17, we see that 5 x2 + 17 x + 6 = (5 x + 2) (x + 3). In each expression the third term 6 also admits of factors 6 and 1, but this is one of the cases referred to above which the student would reject at once as unsuitable. Ex. 3. 9x2 - 48 xy + 64 y2 = (3 x − 8 y) (3 x − 8y) Ex. 4. =(3x-8y)2. 6+7x-5x2 = (3 + 5 x) (2 − x). NOTE. In Chapter xxvi. a method of obtaining the factors of any trinomial in the form ax2 + bx + c is given. 97. We add an exercise containing miscellaneous examples on the preceding cases. Resolve into factors: 1. x2 + 13x + 42. 2. 143-24 ax + a2x2. EXAMPLES X. f. 3. 2 x2+7x + 6. 4. a2b2-3 abc - 10 c2. 98. By multiplying a +b by a-b we obtain the identity (a + b) (a − b) = a2 — b2, a result which may be verbally expressed as follows: The product of the sum and the difference of any two quantities is equal to the difference of their squares. Conversely, the difference of the squares of any two quantities is equal to the product of the sum and the difference of the two quantities. Ex. 1. Resolve into factors 25 x2 - 16 y2. 25x216 y2 (5x)2- (4y)2. = Therefore the first factor is the sum of 5x and 4 y, and the second factor is the difference of 5x and 4 y. .. 25x216 y2 (5x+4y) (5x-4y). = The intermediate steps may usually be omitted. Ex. 2. 1 - 49 c6 = (1 + 7 c3) (1 − 7 c3). The difference of the squares of two numerical quantities is sometimes conveniently found by the aid of the formula a2b2 = (a+b) (a - b). 47. (121)2(120)2. 50. (753)2(253)2. 53. (1639)2(739)2. 48. (750)2-(250)2. 51. (101)2(99)2. 54. (1811)2(689)2. 99. When One or Both of the Squares is a Compound Expression. We employ the method of the preceding articles, as is shown in the following examples: Ex. 1. Resolve into factors (a + 2b)2 – 16 x2. The sum of a + 2b and 4x is a + 2b + 4x, and their difference is a+2b-4x. ...(a+2b)216x2 = (a + 2 b+4x) (a + 2b-4x). |