24. a3-3 a*y* + y3. 25. a3 +a2+a+1. 26. a3 +9 a2 + 16 a +4. 27. 2x2+x3y+2 x2y2 + xy3. 28. m3 +m1a + m3a2 + m2a3 +maa + a3. 29. (x-2)3 — (y — z)3. 30. a+b+2 ab(a1 — a2b2 + b1). 31. x3y3 +x1y1z + x3y3z2 + x2y2z3 + xyz1+z5. 32. 8a3+6 ab (2 a-3b)-27 b3. 33. a (23+y3) — ax (x2 — y2) — y2(x+y). 34. a3-b3+3b2c3 bc2+c3. 35. a+2ab-2 ab2c-b2c2. 36. a*+2 a3b + a2b2 — a1b2 — 2 a2b2c — b2c2. SOLUTION OF EQUATIONS BY FACTORING 94. Many equations of higher degree than the first may be solved by factoring. (See §§ 144-146, E. C.) Factoring the left member of the equation, we have (x − 2)(x + 1) (2 x + 1) = 0. (1) (2) A value of x which makes one factor zero makes the whole left member zero and so satisfies the equation. Hence x = 2, x = xare roots of the equation. – 1, To solve an equation by this method first reduce it to the form A = 0, and then factor the left member. Put each factor equal to zero and solve for x. The results thus obtained are roots of the original equation. Transposing and factoring, (x-4)(x2 - 8x + 3) = 0. Hence the roots of (1) are the roots of x 4 1 0 and x2 From x 4 = 0, x = 4. The quadratic expression x2 not be resolved into rational factors. See § 155. (1) (2) 8x+3=0. 8x + 3 can EXERCISES Solve each of the following equations by factoring, obtaining all roots which can be found by means of rational factors. 1. x33x228x. 2. 6x+8x+5=19 x2. 5. 2-4x=4x+5. 6. 2x+3x=9x2-14. 7. 5232-14x+8=0. 8. 2x+x=14x-3. 9. 12x+14x3+1=3x2+4x. 10. x-4x4-40x3+6—x=58x2. COMMON FACTORS AND MULTIPLES 95. If each of two or more expressions is resolved into prime factors, then their Highest Common Factor (H. C. F.) is at once evident as in the following example. See § 182, E. C. x* — y* = (x2 + y2)(x + y)(x − y), Given (1) In case only one of the given expressions can be factored by inspection, it is usually possible to select those of its factors, if any, which will divide the other expressions and so to determine the H. C. F. Ex. Find the H. C. F. of 6 x3 + 4 x2 − 3 x − 2, The other expression cannot readily be factored by any of the methods thus far studied. However, if there is a common factor, it must be either 2 x2 - 1 or 3 x + 2. We see at once that it cannot be 3x+2. (Why?) By actual division 2 x2 - 1 is found to be a factor of 2x4 + 2 x3 + x2 Xx 1. Hence 2 x2 . 1 is the H. C. F. 96. The Lowest Common Multiple (L. C. M.) of two or more expressions is readily found if these are resolved into prime factors. See § 185, E. C. Ex. 1. Given 6 abx-6 aby=2·3 ab (x − y), 8 a2x+8 a2y=23a2 (x+y), 36 b3(x2 — y2)(x + y) = 2232b3 (x − y) (x+y)2: (1) (2) (3) The L. C. M. is 28. 32 a2b3 (x − y)(x + y)2, since this contains all the factors of (1), all the factors of (2) not found in (1), and all the factors of (3) not found in (1) and (2), with no factors to spare. In case only one of the given expressions can be factored by inspection, it may be found by actual division whether or not any of these factors will divide the other expressions. Ex. 2. Find the L. C. M. of 6x3x2+4x+3, and 6x+3x2-10x-5. (1) (2) (Why?) (1) is not readily factored. Grouping by twos, the factors of (2) are 3 x2 5 and 2x + 1. Now 3 x2 - 5 is not a factor of (1). Dividing (1) by 2x + 1 the quotient is 3x2 - 2x + 3. Hence 6x8x2+4x+3= (2x+1)(3x22x+3), Hence the L. C. M. is (2x+1)(3x2 - 2x+3) (3 x2 - 5). Ex. 3. Find the L. C. M. of a3 +2 a2-a-2, (1) (2) By means of the factor theorem, a 1, a + 1, and a + 2 are found to be factors of (1), but none of the numbers, 1, − 1, stituted for a in (2) will reduce it to zero. Hence (1) and (2) have no factors in common. The L. C. M. is therefore the product of the two expressions; viz. (a + 1)(a − 1)(a + 2) (10 a3 - 3 a2 + 4 a + 1). 97. The H. C. F. of three expressions may be obtained by finding the H. C. F. of two, and then the H. C. F. of this result and the third expression. Similarly the L. C. M. of three expressions may be obtained by finding the L. C. M. of two of them, and then the L. C. M. of this result and the third expression. This may be extended to any number of expressions. EXERCISES Find the H. C. F. and also the L. C. M. in each of the following: 1. x2+ y2, x2+yo. 2. x2 + xy + y2, x3 — y3. 3. x2-5x-6, x2-2x-3, x2+19 x+18. 4. x2-6x2+1, 3+x2-3x+1, x3+3x2+x-1. 5. 162 a3b +252 a2b2 +9 ab3, 54 a3 +42 a2b. 6. 232-8x+3, x2+2x-1. 7. 3 p3 + 5 p2 − 7r−1, 3r2+8r+1. 8. a3-3a2+4, ax-ab-2x+2b. 9. a6+2ab+b6-2 a1b-2 ab, a3-2 ab+b3. 10. 8 a3-36 a2b+54 ab2 - 27 b3, 4 a2 — 9 b2. 11. 36 a1-9a2-24 a-16, 12 a3-6a2-8 a. 12. 2y2+4 by + 3 cy +6 bc, y2-3 by -1062 13. 16-y16, x8 — y3, x1 — y1. 14. m3 +8m2+7m, m3 +3 m2-m-3, m3 — 7 m - 6. ̧ 98. An important principle relating to common factors is illustrated by the following example: We observe that x + 2, which is a common factor of (1) and (2), is also a factor of their sum (3), and of their difference (4). This example is a special case of the following: 99. Theorem 1. A common factor of two expressions is also a factor of the sum or difference of any multiples of those expressions. Proof. Let A and B be any two expressions having the common factor f. Then if k and I are the remaining factors of A and B respectively, Afk and B = fl. Also let mA and nB be any multiples of A and B. Then mA = mfk and nB = nfl, from which we have: mA ±nB = mfk ± nfl = f(mk ± nl). Hence ƒ is a factor of mA ± nB. 100. Theorem 2. If f is a factor of mA±nB and also of A, then f is a factor of B, provided n has no factor in common with A. Proof. Let f be a factor of mA±nB and also of A, where mA and nB are integral multiples of the expressions A and B. and ma expression by cancellation. Now that "B n f mA may each be reduced to an integral mA±nB MA nB f f is also integral. That is, f is a factor of nB. But ƒ is not a factor of n since it is a factor of A, and by hypothesis n and A have no factor in common. Hence ƒ is a factor of B. 101. By successive applications of the above theorems it is possible to find the H. C. F. of any two integral expressions. Ex. 1. Find the H. C. F. of 9 x-x2+2x-1, and 27x58x2-3x+1.. Multiplying (1) by 3x and subtracting from (2) we have (1) (2) (3) By theorem 1, any common factor of (1) and (2) is a factor of (3). |
