35. 4 c2 36 cde81 d2e2. 47. 48. 49. (a + b)2 + 2 (a + b)(x + y) + (x + y)2. 50. (ab)2 - 2 (a − b) (x − y) + (x − y)2. 51. (b+c)28 (b + c) (x + y) + 16 (x + y)2. 53. 49 (d + k)2 — 70 (d + k)(m + u) + 25 (m + w)2. 21. If an expression has the form of a trinomial square, a2 ± 2 ab + b2, we shall call the "middle term," represented by 2 ab, the finder term. The finder term, ± 2 ab, of an integral rational trinomial square a2 ± 2 ab + b2, contains as factors the terms a and b of the binomial ab of which the trinomial is the square. We may often, by using some particular term as a finder term, select from a set of four or more terms a group of three terms, which taken together form a trinomial square. Ex. 1. Factor x2 + y2 - 2 x + 2 xy + 1 − 2 y. - Since there are three separate squares, x2, y2 and 1, and three terms, 2 xy, - 2 x and 2 y, each having a coefficient 2, we are led to suspect the exist ence of one or more trinomial squares among the terms of the given expression. We may group x2, y2, and 2 xy, since the product of the square roots x and y of x2 and y2 is doubled to form the "finder term " 2 xy. We may write, x2 + y2 - 2x + 2xy + 1 − 2 y = x2 + 2 xy + y2 — 2x - 2y+1 Since (x + y)2 and 1 are squares, we may look for twice the product of their square roots, (x + y) and 1, that is, for 2(x + y). This product is obtained by writing - 2 x 2 y in the form — 2(x + y). Accordingly (x + y)2 − 2 x − 2 y +1 may be written as (x + y)2 — 2 (x + y) + 1, which is the square of x + y − 1. We may obtain the same result by grouping either x2,- 2x and 1, or y2,- 2y and 1. The student should carry out the solutions as suggested above. Ex. 2. Factor m2 + 2 mn + n2 + 2 mz + 2 nz + z2. The terms may be grouped in any one of the ways (1), (2), or (3) as suggested below: (1) m2 + 2mn+n2 with the cor- ( (1) + 2(m + n)z leaving ( (1)+z2. (2) m2 + 2 mz+z2 responding (3) n2 + 2 nz + z2 groups ་ (2)+n2. (2) + 2(m + z)n the Of these three possible arrangements, we will use the first, (1) m2 + 2 mn + n2 + 2(m + n)z + z2 = (m + n)2 + 2 (m + n)z + z2 The student should show that the adoption of either of the arrangements of the terms, (2) or (3), leads to the same result. EXERCISE XII. 5 Obtain factors of the following expressions, checking all results numerically: 1. a2+2ab+b2 + 2a + 26 +1. 2. m2 + 2 mn + n2 + 10 m + 10 n + 25. 10. a(a + 2k) + w(w + 2 a) + k(k + 2 w). 11. b(b + 2 c) + c(c + 2 d) + d(d + 2b). 12. a(a — 2b) + b(b + 2 c) + c(c − 2 a). 13. (x2+2yz) + (y2 + 2 xy) + (~2 + 2 xz). 14. a2 + b2 + c2 2(abac + be). 15. 4x2+ y2+9z2 + 2(2 xy-6xz3yz). 22. From the converse of the identity in Chap. VII. § 20, we have Or, the difference of the squares of two numbers may be written as the product of the sum and difference of the numbers. Since 64 x2 is the square of 8 x, and 9 is the square of 3, we may write as factors of 64 x2 9 the product of the sum 8x+3 and the difference 8 x of these same numbers. That is, 64 x29 (8x+3)(8x-3). 3 Check. Let x = 2. 256919. 13 247 = 247. The terms of the binomial quotient 4a4b2c6 - d2, obtained by dividing the terms of the given expression by the common numerical factor 25, are squares. Hence, the binomial difference, 4ab2c6 - d2, may be expressed as the product of the sum 2 a2bc + d, multiplied by the difference 2 a2be3 — d. Hence, we have 100 a1b2c - 25 d2 = 25 [4 a+b2c® — d2] = 25 (2 a2bc3 + d) (2 a2bc3 — d). Check. Let a = 3, b = 2, c = 1, d= 4. 32000 32000. 23. The difference of the squares of either binomials or polynomials may be factored by applying the method under consideration. Ex. 1. Factor (a + b)2 - (c + d)2. We have (a+b)2 - (c+d)2 = [(a + b) + (c + d)] [(a + b) - (c+d)] 20 = 20. (x − y)2 - (mn)2 = [(x − y) + (m − n)][(x − y) — (m − n)] [xy+mn][x − y — m +n]. Check. Let x = 4, y = 2, m = 3, n = 2. EXERCISE XII. 7 22 123.1 3 = 3. Obtain factors of the following expressions, checking all results 5. (a + b)2 - (x − y + z)2. 6. (ab)2 (x + y − z)2. 7. (xy-2)2 — (a + b)2. 8. (abc)29 (xy)2. 9. 9 (x + y)2-16 (z+w)2. y)2 10. 9 (ab)2 49 (cd)2. 11. 4x2-49 (a+b+c)2. 12. 64 (g+h)2 - 121 (k+r)2. 13. 36 (g-x)2 - 169 (h — y)2. 14. 225 (a+b)2 - 225 (c + d)2. 15. 49 (a2 + b)2 — 144 (a + b2)2. 16. 16 (a2 + b2)2 — 121 (a + b)2. 17. 64 (a + b + c)2 — 169 d2. 18. 121 (ab+c)2 - 225 d2. 19. 25(a-b+c)-81(x-y-2). 20. 100 y 196 a2 (b + c)2. 21. 121 a2 (b+c)2 81. 22. 169 b2 (c + d)2 - 196 e2. 23. 1962 (y-2)2 — 225 w2. 24. 36 c2 (d—e)2— 121 x2 (y + z)2. 25. 1442 (a+b+c)2 - 225 ď2. 26. 64x2 (mn)2 — 225 k22. 27. 81a2(m+n)2—256b2(x+y—z)2. 24. Many polynomial expressions may be so transformed, by suitably grouping the terms, as to appear as the difference of two squares. Ex. 1. Factor a2 + 2 ab + b2 cả +2 cả - d2. The terms may be so grouped that the expression will appear as the difference of two trinomial squares, as follows; |