14. (x − 2) (x − 1)(x + 1)(x + 2) = 0. 15. (3x-1) (4 x + 1) (5 x − 2) (2 x + 7) = 0. 17. x2 -- x = 0. 18. x22x = 0. 19. x2 + 3x = 0. 20. 2 x2-3x=0. 21. 2x25x = 0. 22. 3x2 + x = 0. 23. x = 5. 24. 2x2 = x. 29. 6x2 = x2 + 5. 30. 5 (x2+5)=3(x2 + 25). 33. 5 (x2+3)-(x − 5)(x+5)=76. CHAPTER XI. HIGHEST COMMON FACTORS. 131. A Common Factor of two or more algebraic expressions is an expression which will exactly divide each of them. Thus a is a common factor of ab and ac. The Highest Common Factor of two or more algebraic expressions is the expression of highest dimensions which will exactly divide each of them. Thus a2 is the highest common factor of a2b and a3c. Instead of Highest Common Factor it is usual to write H. C. F. We proceed to show how to find the H. C. F. of given expressions. 132. H. O. F. of Monomial Expressions. The highest common factor of two or more monomial expressions can be seen by inspection. Take, for example, a3b2c and a2b3c4, The first expression is clearly divisible by a, or by a2, or by a3, but by no higher power of a; and the second expression is divisible by a, or by a2, but by no higher power. Hence a2 will divide both expressions, and it is the highest power of a which will divide them both. Also b2, but no higher power of b, will divide both expressions; and c, but no power of c above the first, will divide both expressions. Hence the H.C.F. of a3b2c and a2b3c4 is a2b2c. Again, to find the H. C. F. of ab3c3d2 and a3c5d3. The highest power of a which divides both expressions is a; b will not divide both expressions; the highest power of c which divides both is c3; and the highest power of d which divides both is d2. Hence the H. C. F. is ac3d2. Also a3bc1, a2b3c3d, and a1bc1d2 are all divisible by a2, by b, and by ; and therefore the H. C. F. of the three expressions is a2bc1. From the above examples it will be seen that the H. C. F. of two or more monomial expressions is obtained by taking each letter which is common to all the expressions to the lowest power in which it occurs. If the expressions have numerical coefficients, the G. C. M. of these can be found by arithmetic, and prefixed as a coefficient to the algebraical H. C. F. 133. H. C. F. of Multinomial Expressions whose factors are known. When the factors of two or more multinomial expressions are known, their H. C. F. can be at once written down. The H. C. F. will be the product obtained by taking each factor which is common to all the expressions to the lowest power in which it occurs. Consider, for example, the expressions (x − a)3 (x + b)3 and (x − a)2 (x + b)4. It is clear that both expressions are divisible by (x − a)2, but by no higher power of (x − a). Also both expressions are divisible by (x + b)3, but by no higher power of x + b. Hence the H. C. F. is (x − a)2(x + b)3. Again, a2b (a − b)2 (a + b)3 and ab2 (a — b)3 (a + b)1 are both divisible by a, by b, by (a - b)3, and by (a + b)3. Hence the H. C. F. is ab (a − b)2 (a + b)3. In the following examples the factors can be seen by inspection, and hence the H. C. F. can be written down. Ex. 1. Find the H. C. F. of a1b2 — a2b1 and a1b3 + a3b1. Since a1b2 — a2b1 = a2b2 (a2 — b2) = a2b2 (a − b)(a + b), and a+b3 + a3b1 = a3b3 (a + b), we see that the H. C. F. is a2b2 (a + b). Ex. 2. Find the H. C. F. of a3 +3a2b+2 ab2 and a1 + 4 a3b + 3 a2b2. a3 +3a2b + 2 ab2 = a (a2 + 3 ab + 2 b2) = a (a + b)(a + 2b), and a2 + 4a3b + 3 a2b2 : = a2 (a2 + 4 ab + 3 b2) = a2 (a + b) (a + 3b); hence the H. C. F. is a (a + b). Ex. 3. Find the H. C. F. of 3 a3 + 2 a2. a and 5 at + 3 a3 — 2 a2. 3 a3 +2 a2 − a = a (3 a2 + 2 a − 1) = a (3 a − 1) (a + 1), a2 (5 a2 + 3 a − 2) = a2 (5 a − 2) (a + 1) ; |