Use the techniques and procedures developed in Problems 1.5 to 1.10 to factor the following expressions in which the coefficient of the x2 term is no longer limited to 1.

Of all the possible combinations of factors above, only (5 · 9) + (1 · 2) = 47. Carefully arranging the factors, therefore, to ensure that 5 multiplies 9 and 1 multiplies 2, we have

(b) 3x2 + 22x + 24 (1) a · b = 3. Factors are [3,1], giving (3x + ?)(x + ?).
(2) c · d = 24 [1, 24; 24, 1; 2, 12; 12, 2; 3, 8; 8, 3; 4, 6; 6, 4]
(3) ad + bc = 22 [(3 · 6) + (1 · 4) = 22]. Then arranging the factors to ensure that 3 multiplies 6 and 1 multiplies 4, we have

(c) 3x2 – 35x + 22 (1) a · b = 3 [3,1]

(2) c · d = 22 [−1, −22; −22, −1; −2, −11; −11, −2], as in Problem 1.7.
(3) ad + bc = —35 [(3 · −11) + (1 · −2) = −35]. Here rearranging the factors so 3 multiplies −11 and 1 multiplies −2, we obtain

(d) 7x2 – 32x + 16 (1) a · b = 7 [7,1]
(2) c · d = 16 [−1, −16; −16, −1; −2, −8; −8, −2; −4, −4]
(3) ad + bc = −32 [(7 · −4) + (1 · −4) = −32]

(e) 5x2 + 7x – 52 (1) a · b = 5 [5,1]
(2) c · d = −52 [1, 52; 2, 26; 4, 13; each combination of which must be considered in both orders and with alternating signs]
(3) ad + bc = 7 [(5 · 4) + (1 · −13) = 7]

(f) 3x2 – 13x – 56 (1) a · b = 3 [3,1]
(2) c · d = −56 [1, 56; 2, 28; 4, 14; 7, 8; considered as in (e)]
(3) ad + bc = −13 [(3 · −7) + (1 · 8) = −13]

(g) 11x2 + 12x – 20 (1) a · b = 11 [11,1]
(2) c · d = −20 [1, 20; 2, 10; 4, 5; considered as above]
(3) ad + bc = 12 [(11 · 2) + (1 · −10) = 12]

(h) 7x2 – 39x – 18 (1) a · b = 7 [7, 1]
(2) c · d = −18 [1, 18; 2, 9; 3, 6; considered as above]
(3) ad + bc = −39 [(7 · −6) + (1 · 3) = −39]

Transcribed Image Text:

(a) 5x + 47x+18 (1) a b=5. (2) c d = 18 Factors are [5, 1], giving (5x +?)(x+?). [1, 18; 2,9; 3, 6] (3) ad+bc 47. Here all the pairs of possible factors from step (2) must be tried in both orders in step (1), that is, [5,1] with [1, 18; 18, 1; 2, 9; 9, 2; 3, 6; 6, 3]