Using rule IV, 89, write out by inspection: = 38. (x-2) (x — 4) = 59. (a+b) (b+c)−(c+ d) (d+a)-(a+c) (bd). 60. (a+b+c+d)2+(—a−b+c+d)2+(a−b−c+d)2+(a+b—c−d)2. 61. (a2+b2+c2)2 — (a+b+c) (a+b-c) (a+c—b) (b + c − a). 62. (a+b) (a - b)3. 63. (ab)2 (a+b)3. CHAPTER VIII FACTORING AND SOLUTION OF EQUATIONS BY FACTORING FACTORING 91. If two factors are given, their product can be found by multiplication. If one of two factors and their product are given the second factor can be found by division. It is often important to determine the factors of a given product. There are many useful examples wherein factors can easily be found by means of the theorems proved in the previous chapters I-V; thus: 92. Case I.-To Factor a Polynomial, Every Term of which has a Common Factor. For example, 1. x2+xy = x(x + y). 2. 3a2 6ab9a2b2=3 a (a−2b+3 ab2). 3. 72ay 84.xy2+60x2y2 = 12 xy (6x-7y+5xy). The rule for factoring this case is an immediate consequence of Law V of multiplication, namely, that RULE. a (b + c) = ab + ac a (b + c + d) = a (b + c) + ad = ab + ac+ad, etc. Divide each term of the polynomial by the product of the factors common to all of the terms of the polynomial, the divisor and the quotient will be the factors required. EXERCISE XV Factor the following expressions: 3. 92-7xy. 5. 7x-35x3y. 7. 49-84x. 9. 8ab2+2 a2bx2 +6 abx31. 6. 16 y2+64 xy3, 8. 422-84 x*y* + 203 x°y®. 10. 90.3y — 130 €3y2 + 270 ayʻ. 11. 21x-1ym+1 — 28,2n−3⁄4y?m+3+63 x3n−5y3n+5 +49¿11ym ̧ 12. ax1уbx2-1yp+1 + c.x2−2yp+2+dxn¬3yp+3 — €xn−4yp+4. 93. Case II. When the Trinomial is a Perfect Square. 1. Find the factors of x2 + 14x + 49. Since this is a trinomial with plus terms and is to be a perfect square, it must have the form (x + a)2 = (x + a) (x + a) = x2 + 2 ax + a2 and [286, I] x2+2 ax + a2 must be the same as x2+14x + 49 Hence the second term a of the binomial whose square is a2+14x+49 must be a number such that two times a is 14 and a squared is 49. The only number whose double is 14 and whose square is 49, x2+14x+49= (x+7) (x+7) = (x+7)°. Since the sign of the middle term of the trinomial is trinomial is to be a perfect square, it must have the form is 7. and the 18x+81. 18x+81, x2-2ax+a2 must be the same as x2 Hence the second term of the binomial whose square is must be a number (-a) such that The only number which multiplied by 2 is 18 and whose square is 81, is 9. 94. Case III.-The Factoring of the Difference of Two Squares. Definition. If a number can be resolved into the product of two equal factors, one of them is the square root of the number. By definition, 4 is the square root of 16, and is indicated 164 or *1′(4)2= 4; and in general V (x)2= x. That is, the operation indicated by the sign | is the inverse of the operation indicated by the sign (), and undoes the work of squaring a number. Further, the v4 a2x® = 1 (2 ax3)2 = 2ax3. An expression in the form of two squares which have a negative sign between them, is the product of two factors which can be determined as follows: Take the square root of the first number, and the square root of the second number. The sum of these square roots will be the first factor and their difference will form the second factor. 3. (x − y)2 — (≈ — w)2 = ¦ (x − y) + (z—w)} {(x − y) — (zw)} =(xy+w) (x y z +w). 95. The terms of an expression can often be arranged so that it is equivalent to two squares with the sign between them, and the expression can then be resolved into factors, thus x2 + y2 — z2-de-2xy-2 dz= x2-2xy + y2 — (z2 + 2dz + d) 96. The difference of two squares may often be resolved into several factors, 1. x16 — y16 — (x8+ y3) (x8 — y8) = (xx8+ y3) (x2+y1) (x1 — y1) = (xx8+ y3) (x*+y1) (x2+y3) (x2 — y3) = (x8+ y3) (x1+ y1) (x2+ y2) (x + y) (x − y). *Owing to the frequent use of the operation of square root the symbol is used instead of V. = 2. 4(xy + zw) — (x2 + y2 — z2 — w2)? = {2(xy+zw)+(x2+y2—z2—w2)} {2(xy+zw)—(x2 + y2 — z2 —w2) } = {(x2+2xy + y2)-(z2-2zw+w2)} {(z2+2zw+w3)-(x2-2xy+y)} = (x+y)2 — (z—w)2} {(z+w)2—(x-y)2} = {(x+y)+(zw)} {(x+y)—(z—w)} {(z+w)+(x−y)} {(z+w)-(x-y)} =(x+y+z-w) (x+y−z+w) (x−y+z+w) (−x+y+z+w). 97. Case IV.-The Factoring of the Sum and the Difference of Two Cubes. Definition. The cube root of a quantity is one of the three equal factors into which it may be resolved. Thus, the cube root of 64 is 4, and is written 64 = 4. The operation of finding the cube root is the inverse of the operation of finding the cube. Thus, 125 Further 5.5.5= 15.5·5 (5)= 5; and in general (c)3 = x. To find the cube root of a quantity A3 is to undo the work accomplished by finding the cube of 4. |