When stated concisely, all such cases of multiplication come under the following formula : Formula IV. (x+m)(x+n) = x2+(m+n)x+mn. It is to be remembered here that m and n represent any numbers, positive or negative. ORAL EXERCISES State the answer by inspection in each of the following 11. (x-12)(x-2). 12. (y+1)(y-8). 13. (y-1)(y+8). 14. (y+8) (y-1). 15. (+11)(2-4). 16. (A+3)(A+7). 17. (n-4)(n+8). 18. (t+5)(t-8). 19. (R+1)(R-6). 26. (c2-2) (c2+3). 27. (a3+4)(a3+6). 28. (a2b2-1)(a2b2+3). [HINT. The result, by Formula IV, is (2 a)2-4(2 a)–21. This may be written 4 a2-8a-21. Ans.] 30. (3x-4) (3x+2). 31. (4x+3)(4 x+1). 32. (6 x−5)(6x+7). 33. (8r+5) (8 r−6). 34. (2 S+6) (2 S+10). 35. (6 V-3)(6 V−6). 36. (H+2)(H — 19). 37. (a-10 b) (a+4 b). [HINT. The result, by Formula IV, is a2+(-10b+4 b)a+(−10 b) (4 b). This reduces to a2-6 ab-40 b2. 38. (r+7 s) (r−6 s). Ans.] 48. 18. 16 39. (x−2r)(x+3 r). 40. (t+10 z) (t−z). 41. (2 2+3 z) (2 ť2+4 z). 49. 28 33 [HINT. Write as (30-2) (30+3).] 50. 31. 28 For further exercises on this topic, see the review exercises, p. 114, and Appendix, p. 299. 58. The Factoring of a Trinomial. In § 57 and in the exercises that follow it we started every time with two given binomials, such as x+2 and x+3, and we saw how to get their product very easily; in fact, we learned to do it mentally. In every case the product turned out to be a trinomial beginning with the square of the common letter. Thus, in multiplying x+2 by x+3 the result is the trinomial x2+5x+6, which begins with x2. Suppose now we try to turn this idea around. Let us start with a given trinomial, such as x2+7x+6, and inquire what two binomials give this when multiplied together. In other words, let us try to find the two binomial factors of x2+7x+6. This can be done as follows. We think of the two numbers whose sum is 7 and whose product is 6. The numbers that do this are 6 and 1. Then x+6 and x+1 must be the factors we want, for when we multiply (x+6) by (x+1), using Formula IV, we get (as we should) our trinomial, x2+7x+6. Similarly, in order to find the factors of x2+2x-8 we think of the two numbers whose sum is 2 and whose product is -8. The numbers are seen to be 4 and −2. Therefore, the desired factors are x+4 and x-2. Ans. NOTE. It is to be observed that in all such problems, the first term of each of the desired factors is simply x; that is, it is the square root of the first term of the given trinomial. WRITTEN EXERCISES Find the two binomial factors of each of the following expressions. 1. x2+7x+12. [HINT. Here we need the two numbers whose sum is 7 and whose product is 12.] For further exercises on this topic, see the Appendix, p. 299. 59. The Square of the Sum of Two Numbers. Suppose we have any two numbers, a and b, and let us form their sum, a+b. Then (a+b)2 means (a+b) (a+b) and when multiplied out it looks as follows: Expressed in words, this says that the square of the sum of any two numbers is equal to the square of the first number plus twice the product of the two plus the square of the second number. For example, (r+6)2= r2+2(r· 6)+62= r2+12r+36. (2a+3b)2= (2 a)2+2(2 a 3 b)+(3 b)2=4 a2+12 ab+9 b2. Check each of these examples by multiplying the long way, as in § 53. NOTE. Formula V is really a special case of Formula IV, for if in that formula we use a for x, b for m, and b for n, we obtain (a+b)·(a+b)=a2+(b+b)a+b2, which is the same as (a+b)2= a2+2 ab+b2. 60. The Square of the Difference of two Numbers. Suppose we have any two numbers, a and b, and that we form their difference, a-b. Then (a-b)2 means (a-b) (a-b) and when multiplied out this looks as follows: a-b a a22-ab - ab+b2 a2-2 ab+b2 Thus we have the following formula: Formula VI. (a - b)2=a2-2 ab+b2. Expressed in words, this says that the square of the difference of any two numbers is equal to the square of the first number minus twice the product of the two plus the square of the second number. For example, (r−6)2= r2 — 2(r · 6)+62= r2−12 r+36. (3x-2b)2=(3x)2-2(3 x 2 b) + (2 b)2=9 x2-12 bx+4 b2. NOTE. Formula VI is a special case of Formula IV, for if we use a for x, -b for m, and -b for n in Formula IV we obtain (a−b) • (a−b) =a2+(−b—b)a+b2 which is the same as (a−b)2= a2—2 ab+b2. 61. Applications of Formulas V and VI. The value of Formulas V and VI lies in the fact that they enable us to read by inspection the square of any binomial in algebra. For example, (x+3)2 is immediately read by Formula V to be x2+2(3x)+32, which reduces to x2+6x+9. Likewise, (2x-3)2 is read by Formula VI to be (2x)2-2(2 x · 3)+32, which reduces to x2-12 x+9. ORAL EXERCISES Give by inspection the expanded values of each of the following expressions. |
