The way in which this principle may be used in solving equations is illustrated below : EXAMPLE 1. Solve the equation x2 – 8 x+15=0. SOLUTION. Since the trinomial 22–8 x+15 may be factored (see § 58) into (2-3)(x-5), the given equation may be written in the form (x-3)(3-5)=0. Thus, we are to have the product of two numbers equal to zero, and it will be, according to the Principle above, whenever either of the factors is equal to zero. That is, when either 2-3=0, or x-5=0. But these last two equations give us x=3 and x=5. Therefore, either x=3 or x=5 is a solution (root) of our equation. Ans. CHECK. When x=3, the left side of the given equation becomes 32–8. 3+15, which reduces to 9-24+15, and this reduces (as the equation demands) to 0. When x=5, the left side of the given equation becomes 52–8.5+15=25–40+15=0. EXAMPLE 2. Solve the equation x2 —4 x–21=0. SOLUTION. Factoring, we have (x-7)(x+3)=0. Whence, by the Principle above, we have either x–7=0, or x+3=0. Therefore, the roots are 7 and -3. Ans. CнECк. 72–4:7–21=49–28–21=0; (-3)2–4(-3)–21=9+12–21=0. NOTE. Each of the equations just considered has two roots. This is true of every equation which contains the second (but no higher) power of the unknown letter. Such equations are called quadratic equations and will be more fully considered in Chapter XVI. WRITTEN EXERCISES Solve the following equations by factoring. Check the first ten. 1. x2 – 7 x+10=0. 3. x2+8 x+15=0. 2. x2–5 x+6=0. 4. X2+7 x-30=0. 5. x2+4 x=0. 17. 52–5–30=0. 6. x2 – 5 x=0. 18. x2 — 3 — 56=0. 7. 22+5 x—6=0. 19. 22—3 2-28=0. 8. x2 — 2 — 6=0. 20. x2 – 6 x-40=0. 9. n2—2 n-35=0. 21. (2+9c+20=0. 10. m2 +8 m+15=0. 22. x2 – 5 x— 14=0. 11. p2=6+r. 23. 4 x2 – 8 x+4=0. (Hint. Transpose all terms to left side.) (Hint. Factor by Formula VI.] 12. n2-32= -4n. 24. 9 x2 – 12 x+4=0. 13. 5 x2 – 4 x=0. 25. { x2 — x+1=0. 14. 3+5 x=2 x2 ở c” - c+ =0. 15. a? — 25=0. 27. s r2 +}r=0. 16. p2 — 36=0. 28. x2 - 4x+1=0. For further exercises on this topic, see the review exercises below, and Appendix, p. 301. ng Formula the following EXERCISES – REVIEW OF CHAPTER VI 1. Using Formulas I and II ($$ 48 and 50) state (orally) a different form for each of the following expressions. (a) 2 x2x2. (c) ma’a3+n(ab)3. (6) (2 xy)? (d) 4(abc) ab – 5(xyz)3x2y3. (e) 4x( —xy)2—5 y( - y2)3+6 z(xy). (f) (3 ab)2(xy)3+4(2 cd)?(yz)2+3 x_x(zw)4. (g) (ab)2(ab)3+(cd)?(cd)3+x2x(x • x)4. (h) (x2y2)2 – 5(abcd)3a262 +4(p?q2r292)2. [Hint. (22)2=x2 • x2=x4.] (i) (x2x3y223)3. (j) (-3 h2k32)3 – (-2 g2h2)?g2h2. 2. Using Formula III (851) state (orally) a different form for each of the following expressions. (a) 3(x+y). (d) m2n2(a2+b2+c). (6) m(a+b). (e) ab(c+d) – pq(r+s). (c) mn(a−b). (f) xy(x+y)+yz(z2+y). 3. Factor each of the following expressions. (See 88 52 and 54.) (a) 5 x*y–10 x3y2. (c) 3q5–12 q3p2 +6 qr4. (6) 5 m4–10 m3-5 m2. (d) 3 xöy3+3 x2y2+12 xy. (e) 60 mon372 — 45 møn2p3 +90 men3r2. (f) 12 a2b2c3 — 16 a2b2c2 — 20 a3b3c3. (g) aq-ar+qx — rx. (m) m3+ m2–3 m-3. (h) gh-gx - rh+rx. (n) 3 a26–9 ab+qa-3 qb. (i) x2 - xy-5x+5 y. (0) 16 ax+12 ay-8 bx — 6 by. (j) x2+xy-ax - ay. (p) mn+m-3 n2-3 n-4 n–4. (k) 1-m+n-mn. (9) px — qx ——py+ay+y. (1) ax—xy — ab+by. (r) ma+mn+mn+na+m+n. 4. By use of Formula IV (857) state (orally) the result of each of the following multiplications. (a) (n-8)(n-12). (h) (2 x-7)(2 x+6). (b) (p2-5)(p2–3). (i) (3 y-2)(3 y+3). (c) (23-7)(x3+6). (j) (4x2+1)(4x2 —3). (d) (a" - 5)(a" +4). (k) (ab – 3)(ab+4). (e) (x*—a)(x—b). (1) (m?n? –a)(m2n2 – 2 a). (f) (m—2 a)(m+36). (m). (2 xy+y)(y2 — xy). (g) (2 x+1)(2 x—3). (n) (abc+d)(abc+1). 5. Factor the following expressions. (See § 58.) (a) x2 +7 x+12. (f) x2–2 3–15. (k) x2 +15 ax+56 a?. (6) x2–7 x+12. (g) x2—X—30. (1) 22–11 ax+30 a?. (c) m2+8 m+12. (h) y2–6 ay+5 a. (m) – 52+96+62. (d) a+a-12. (i) x2–3 nx — 28 na. (n) 100—25x+x?. (e) 62-5 6-14. () x2+18 bx+80 62. 5. If, in the figure at the left, the radius of the large circle is R and the radius of each of the smaller circles is r, show that the shaded area, A, is given by the formula A=+(R+2 r)(R—2r). ameter is D and whose inner diameter is d. If the length is l, show that the volume V occupied by the iron is expressed by the formula · v_ (D+d/D-d. Fig. 37. 7. The result in Ex. 6 may be expressed in words as follows: “To find the volume of a circular tube, multiply a times the length by half the sum of the two diameters by ——.” Finish the sentence. Does this give us a convenient rule for determining any such volume ? Explain. 8. One side of a right triangle measures 12 feet, while the hypotenuse is 2 feet longer than the other side. Find the length of the other side. For further exercises on this Chapter, see Appendix, pp. 298–301. CHAPTER VII DIVISION AND FACTORING | 69. Division. The process of finding one factor when the product and other factor are given is called division. The dividend is the given product, the divisor is the given factor, and the quotient is the required factor. Thus, to divide 6 ab by 3 a means to find the number which when multiplied by 3 a gives 6 ab. Here the dividend is 6 ab, the divisor is 3 a, and the answer (or quotient) is evidently 2 b. 70. Law of Signs for Division. In § 27 we saw that the sign of a quotient is + whenever the dividend and divisor have like signs, and is – whenever they have unlike signs. This may be stated in a table as follows: +ab:(+6)= ta tab=(-6)= -a 71. Law of Exponents for Division. In dividing x5 by x3 we may proceed, as in arithmetic, by removing equal factors from dividend and divisor, the work appearing as below. 205 t ot.*. * • X = 2(5–8) = x2. Ans. 203 $b Similarly, we see that x6 = x2 is equal to x(6–2), or x4. In general, we have the following rule. † 8 27 should be read again at this point. |