29. (16 a1-81 b1) ÷ (2a+3b) 30. (xo + 64 y®) ÷ (x2 +4 y2) 33. (x3 y1 — x1 y3) ÷ (x2 y — x y3) 34. (a3 2:15 + b3 yo) ÷ (a x5 + b y2) 39. (x1o — 32 y2o) ÷ (x2 — 2 y1) Factoring. 1. Definitions and Principles. 92. The quantities multiplied together to produce a given quantity are the Factors of the quantity. 93. The prime quantities multiplied together to produce a given quantity are the Prime Factors of the quantity. 94. A composite quantity may have two or more sets of factors, but it can have only one set of prime factors. Thus, a ba2 × b2 = a2 b × b = a b2 × a = ab xab= a xa x b2 = a2 × b × b = ab×a× b = a × a × b × b. The last is the only set of prime factors. 95. The process of finding the factors of a quantity is Factoring. 96. ab ab; but, a and b are the factors of ab; therefore, Prin. 46.—A divisor of a quantity is one of the two factors of the quantity, and the quotient is the other. 97. Since a divisor of every term of a quantity is a divisor of the quantity [P. 15], and a divisor of a quantity is a factor of the quantity [P. 46], it follows that, Prin. 47.-A factor of every term of a quantity is a factor of the quantity. 2. Problems. 1. To factor a polynomial having a common factor in its terms. Illustration.-Factor ac2 a c2 + a2 c. Form. (a2 c2 — a c2 + a2 c) = a c (a c − c + a) Solution: We see by inspection that a c is a factor of each term of the polynomial; it is therefore a factor of the polynomial [P. 47]. Dividing by a c, the quotient a c-c+a is the other factor [P. 46]. Therefore, (a2 c2 — a c2 + a2 c) = a c (a c c + a). 3. x2+axy 4. 23+3x2-2x EXERCISE 43. +36 a1 b2 c Illustrations.-Factor a2 - b2, x1 — y1, and xo y2 — xa y1. Solutions: 1. a2 — b2 = (a + b) (a — b) [P. 39]. 2. x4 — y1 = (x2 + y2) (x2 — y2) [P. 39] = (x2 + y2) (x + y) (xy) [P. 39]. 3. x y2 - x4y+= x2 y2 (x2 — y2) [P. 47 and 46] = 24y2 3. To factor the sum or difference of the equal odd powers of two quantities. Illustration.-Factor a3 — b3, a3 + b3, and ao — bo. Solutions: 1. a3 — b3 = (a — b) (a2 + ab + b2) [P. 44 and 45]. 2. a3 + b3 = (a + b) (a2 3. ao — bε = (a3 + b3) (a3 (a2 a b + b2) (a ab + b2) [P. 43 and 45]. — b3) [P. 39] = (a + b) 4. To factor a trinomial that is a perfect square. Illustrations. 1. a2+2ab+b2 = (a + b) (a + b), since (a+b)2 = a2 + 2 a b + b2 [P. 31]. 2. a2 - 2ab+ b2 = (a — b) (a — b), since (a — b)2 = a2-2ab+b2 [P. 32]. 3. 4a2+12 ab + 9 b2 = (2 a + 3 b) (2 a + 3 b), since (2a+3b)2 = 4a2 + 12 a b + 9 b2 [P. 31]. 4. 4a-12ab+ 9 b2 = (2 a − 3 b) (2 a 3 b), since ab9b (2a-3b)2 = 4a2 - 12 a b + 9 b2 [P. 32]. 98. A trinomial is a perfect square when two of its terms are perfect squares, and the other term is the square root of their product. twice 5. To factor a trinomial that is the product of two Illustrations. 1. a2 + 3 a +2 = (a + 2) (a + 1), since (a + 2) ( a + 1) = a2+3a+2 [P. 40]. 2. a2 5a+6= (a− 2) (a− 3), since (a− 2) (a-3) = a2 — 5 a + 6 [P. 40]. - 3. a2 + 2 a − 8 = (a + 4) (a − 2), since (a + 4) (a — 2) = a2+2a-8 [P. 40]. 4. (4a2 - 4a - 15) = (2 a + 3) (2 a — 5), since (2a+3) (2a-5)=4a-4a-15 [P. 40]. 99. A trinomial is the product of two binomials having a like term, when the first term is a square, and the last term is the algebraic product of two factors whose sum, |