Using Rule 2 from Section 1.3 and the techniques developed in Problems 1.5 to 1.11, factor the following polynomials:

(a) x2 + 11xy + 28y2

(1) c · d = 28 [1, 28; 2, 14; 4, 7]

(2) c + d = 11 [4 + 7 = 11]

(b) x2 – 19xy + 60y2 (1) c · d = 60 [−1, −60; −2, −30; −3, −20; −4, −15; −5, −12; −6, −10], for reasons analogous to those in Problem 1.7.
(2) c + d = −19 [−4 + (−15) = −19]

(c) x2 – 13xy − 48y2 (1) c · d = −48 [1, −48; 2, −24; 3, −16; 4, −12; 6, −8], for reasons analogous to those in Problem 1.9.

(2) c + d = −13 [3 + (−16) = −13]

(d) x2 + 11xy – 42y2 (1) c · d = −42 [−1, 42; −2, 21; −3, 14; −6, 7], for reasons similar to those in Problem 1.8.
(2) c + d = 11 [−3 + 14 = 11]

(e) 3x2 + 29xy + 18y2 (1) a · b = 3 [3, 1]
(2) c · d = 18 [1, 18; 18, 1; 2, 9; 9,2; 3,6; 6,3], as in Problems 1.11

(a) and b).
(3) ad + bc = 29 [(3 · 9) + (1 · 2) = 29] Then rearranging the factors for proper multiplication, we obtain

(f) 7x2 – 36xy + 45y2 (1) a · b = 7 [7, 1]
(2) c · d = 45 [−1, −45; −45, −1; −3, −15; −15, −3; −5, −9; −9, −5], as in Problem 1.11 (c).
(3) ad + bc = −36 [(7 · −3) + (1 · −15) = −36](g) 5x2 + 12xy – 44y2 (1) a · b = 5 [5, 1]
(2) c · d = −44 [1, 44; 2, 22; 4, 11; each combination of which must be considered in both orders and with alternating signs as in Problem 1.11 (e)]
(3) ad + bc = 12 [(5 · −2) + (1 · 22) = 12(h) 8x2 + 46xy + 45y2 (1) a · b = 8 [1, 8; 2, 4; 4, 2; 1, 8], as in Problem 1.12 (2) c · d = 45 [1, 45; 3, 15; 5, 9; 9, 5; 15, 3; 45, 1]
(3) ad + bc = 46 [(2 · 5) + (4 · 9) = 46]

(i) 4x2 – 25y2 (1) a · b = 4 [1, 4; 2, 2]
(2) c · d = −25 [1, −25; −25, 1; 5, −5]
(3) ad + bc = 0 [(2 · 5) + (2 · −5) = 0]

Transcribed Image Text:

x + 11xy + 28y = (x+4y)(x + 7y)