VIII. FACTORING. 102. Factoring is the process of resolving a quantity into its factors. (Art. 11.) 103. The factoring of monomials may be performed by inspection; thus, 12 a3b2c = 2.2.3 aaabbc. A polynomial is not always factorable; but there are certain forms which can always be factored, the more important of which will be considered in the succeeding articles. CASE I. 104. When the terms of the polynomial have a common monomial factor. 1. Factor a3 +3a. Each term contains the monomial factor a. Dividing the expression by a, we have a2+3. Hence, a3+3a=a (a2+3), Ans. 2. Factor 14 xy1 — 35 x3y2. 14xy-35x3y 7xy2 (2y2-5x2), Ans. = 13. Factor the sum of 54 a1b3, — 72a3c2, and — 90 a2ă. 14. Factor the sum of 96 c'd3, 120c2d', and — 144 c3d*. CASE II. 105. When the polynomial consists of four terms, of which the first two and the last two have a common binomial factor. Factoring the first two and last two terms as in Case I, we have m(a - b)+n(a —b). Each term now contains the binomial factor a - b. Dividing the expression by ab, we obtain m +n. Hence, Note. If the third term is negative, as in Ex. 2, it is convenient, before factoring, to enclose the last two terms in a parenthesis preceded by a sign. 106. If a quantity can be resolved into two equal factors, it is said to be a perfect square, and one of the equal factors is called its square root. Thus, since 9 a1b2 equals 3 a2b × 3 a2b, it is a perfect square, and 3 a2b is its square root. Note. 9a4b2 also equals -3a2bX-3a2b, so that its square root is either 3 a2b or -3 ab. In the examples in this chapter we shall consider the positive square root only. 107. The following rule for extracting the square root of a monomial is evident from Art. 106: Extract the square root of the coefficient, and divide the exponent of each letter by 2. For example, the square root of 25xyz2 is 5x3yz. 108. It follows from Art. 95 that a trinomial is a perfect square when its first and last terms are perfect squares and positive, and the second term is twice the product of their square roots. Thus, 4x2-12xy +9 y2 is a perfect square. 109. To find the square root of a perfect trinomial square, we take the converse of the rules of Art. 95: Extract the square roots of the first and last terms, and connect the results by the sign of the second term. Thus, let it be required to find the square root of 4x2-12xy +9y2. The square root of the first term is 2x, and of the last term 3y; and the sign of the second term is the required square root is Hence 2x-3y. CASE III. 110. When a trinomial is a perfect square (Art. 108). 1. Factor a2 + 2 ab2 + b1. By Art. 109, the square root of the expression is a + b2. Hence, a2+2ab2+b= (a + b2) (a + b2), or (a+b2)2, Ans. 2. Factor 4x2-12xy +9 y. 4x2-12xy +9 y2 = (2 x − 3 y) (2 x — 3y) =(2x-3y), Ans. Note. The given expression may be written 92-12xy + 4x2; whence, 9 y2-12xy + 4x2 = (3 y − 2 x) (3 y − 2 x) = (3 y − 2 x )2; which is another form of the answer. 6. a2 10a + 25. 7. y2+2y+1. 8. m2 - 2 m + 1. 9. +12 +36. 10. n® – 20n3 +100. 11. x2y2+16xy +64. 12. 1-10 ab + 25 a2b2. 13. 16m2-8am + a2. 14. a1+2 a3 + a2. 15. -4x+4x2. 26 6+8+16x1. 19. a2b1+18 ab2c + 81 c2. CASE IV. 111. When an expression is the difference of two perfect squares. Comparing with the third case of Art. 95, we see that such an expression is the product of the sum and difference of two quantities. Therefore, to obtain the factors, we take the converse of the rule of Art. 95: Extract the square root of the first term and of the last term; add the results for one factor, and subtract the second result from the first for the other. 1. Factor 36a2 — 49 y2. The square root of the first term is 6x, and of the last term 7y. Hence, by the rule, 36-49y=(6x+7y) (6x-7y), Ans. 2. Factor (2x - 3 y)2 — (x − y)2. (2x-3y)2 - (x − y)2 = [(2 x − 3 y) + (x − y)][(2x−3y) — (x—y)] |