WHEN THE TERMS CAN BE GROUPED SO AS TO CONTAIN A COMMON FACTOR. 80. Example 1. Resolve into factors x2-ax+bx-ab. Noticing that the first two terms contain a factor x, and the last two terms a factor b, we enclose the first two terms in one bracket, and the last two in another. Thus, x2- ax + bx-ab= (x2 - ax) + (bx − ab) =x : (x − a) + b (x − a).. since each bracket of (1) contains the same factor x-a. Example 2. Resolve into factors 6x2-9ax +4bx – 6ab. 6x2-9ax +4bx-6ab= (6x2 - 9ax)+(4bx − 6ab) Example 3. Resolve into factors 12a2-4ab - 3ax2 + bx2. 12a2-4ab - 3ax2+ bx2=(12a2 - 4ab) - (3ax2 - bx2) --- ..(1) NOTE. In the first line of work it is usually sufficient to see that each pair contains some common factor. Any suitably chosen pairs will bring out the same result. Thus, in the last example, by a different arrangement, we have 12a2-4ab-3ax2+ bx2= (12a2 - 3ax2) - (4ab – bx2) The same result as before, for the order in which the factors of a product are written is of course immaterial. 27. 2ax2+3axy-2bxy-3by2. 28. amx2+bmxy — anxy-bny2. TRINOMIAL EXPRESSIONS. 81. Before proceeding to the next case of resolution into factors the student is advised to refer to Chap. v. Art. 44. Attention has there been drawn to the way in which, in forming the product of two binomials, the coefficients of the different terms combine so as to give a trinomial result. Thus, by Art. 44, We now propose to consider the converse problem : namely, the resolution of a trinomial expression, similar to those which occur on the right-hand side of the above identities, into its component binomial factors. By examining the above results, we notice that: 1. The first term of both the factors is x. 2. The product of the second terms of the two factors is equal to the third term of the trinomial; thus in (2) above we see that 15 is the product of -5 and -3; while in (3) — 15 is the product of +5 and -3. 3. The algebraic sum of the second terms of the two factors is equal to the coefficient of x in the trinomial; thus in (4) the sum of -5 and +3 gives 2, the coefficient of x in the trinomial. The application of these laws will be easily understood from the following examples. Example 1. Resolve into factors x2+11x+ 24. The second terms of the factors must be such that their product is +24, and their sum +11. It is clear that they must be +8 and +3. .. x2+11x+24=(x+8)(x+3). Example 2. Resolve into factors x2 - 10x + 24. The second terms of the factors must be such that their product is +24, and their sum - 10. Hence they must both be negative, and it is easy to see that they must be 6 and -4. Example 3. Example 4. .. x2-10x +24=(x − 6) (x − 4). x2-18x+81= (x − 9) (x −9) = (x-9)2. x2+10x2+25=(x2+5)(x2+5) =(x2+5)2. Example 5. Resolve into factors x2 - 11ax + 10a2. -a. The second terms of the factors must be such that their product is +10a2, and their sum - 11a. Hence they must be - 10a and .. x2-11ax+10a2= (x − 10a) (x − a). NOTE. In examples of this kind the student should always verify his results, by forming the product (mentally, as explained in Chap. v.) of the factors he has chosen. EXAMPLES XI. c. Resolve into factors: 1. a2+3a+2. 4. a2-7a+12. 7. x2-19x+90. 2. a2+2a+1. 3. a2+7a+ 12. 5. x2-11x+30. 6. 8. x2+13x+42. 9. x2-21x+110. 10. x2-21x+108. 11. x2-21x+80. 12. x2+21x+90. 82. Next consider a case where the third term of the trinomial is negative. Example 1. Resolve into factors x2+2x-35. The second terms of the factors must be such that their product is - 35, and their algebraical sum +2. Hence they must have opposite signs, and the greater of them must be positive in order to give its sign to their sum. The required terms are therefore +7 and -5. .. x2+2x−35=(x+7) (x−5). Example 2. Resolve into factors x2-3x-54. The second terms of the factors must be such that their product is - 54, and their algebraical sum - 3. Hence they must have opposite signs, and the greater of them must be negative in order to give its sign to their sum. The required terms are therefore - 9 and +6. .. x2-3x-54=(x−9)(x+6). Remembering that in these cases the numerical quantities must have opposite signs, if preferred, the following method may be adopted. Example 3. Resolve into factors x2y2+23xy – 420. Find two numbers whose product is 420, and whose difference is 23. These are 35 and 12; hence inserting the signs so that the positive may predominate, we have x2y2+23xy - 420 = (xy+35) (xy - 12). 83. We proceed now to the resolution into factors of trinomial expressions when the coefficient of the highest power is not unity. Again, referring to Chap. v. Art. 44, we may write down the following results: (3x+2)(x+4)=3x2+14x+8..... (3x-2) (x-4)=3x2-14x+8... (3x+2) (x-4)=3x2-10x-8..... .(1), .(2), ..(3), .(4). The converse problem presents more difficulty than the cases we have yet considered. Before endeavouring to give a general method of procedure it will be worth while to examine in detail two of the identities given above. Consider the result 3x2 - 14x+8=(3x − 2) (x − 4). The third term +8.. - 2 and - 4. The middle term 14.c is the result of adding together the two products 3x × 4 and x x -2. Again, consider the result 3x2-10x-8=(3x+2) (x − 4). The third term -8..... +2 and -4. The middle term - 10x is the result of adding together the two products 3x × −4 and x × 2; and its sign is negative because the greater of these two products is negative. 84. 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. |