Ex. 1. Determine whether 6x2+9x+2 has rational factors.

Hence the roots are irrational, and the expression has no rational factors.

Ex. 2. Factor 3x2-19 x -14.

Solving the equation 3x2 - 19 x 14 = 0 by the formula,

x=

19 ± √192 + 4 .3 . 14

6

Ex. 3. Factor 2x2-2x+1.

Solving the equation by means of the formula, we find the roots

Ex. 4. Factor x2 + xy-2xy2 - 2 y2+8 y3 — 8y'.

The expression is quadratic in respect to x, hence we solve the equation x2+x(y-2 y2)-2 y2+8y8-8y = 0.

NOTE. A quadratic equation cannot have three roots. For if we write the equation ax2 + bx + c = 0 in the form a(x − r1)(x − r2) = 0, no other value rs, not equal to either r1 or 2, can satisfy the equation, as a(rз — r1) (rз — r2) cannot equal zero.

26. 8 x2-16xy — 8 xz + 6 y2 — 8 yz — 30 z2.

27. 12 p2 - 18 pq + 28 pr — 12 q2 + 19 qr − 5 r2.
28. x2-3x-3.

CHAPTER XXI

PROGRESSIONS

357. A series is a succession of numbers formed according to some fixed law.

The terms of a series are its successive numbers.

ARITHMETIC PROGRESSION

358. An arithmetic progression (A. P.) is a series, each term of which, except the first, is derived from the preceding by the addition of a constant number.

The common difference is the number which added to each term produces the next term.

Thus each of the following series is an A. P.:

The common differences are respectively 4, -7, and d.
The first is an ascending, the second a descending, progression.

359. To find the nth term of an A. P., the first term a and the common difference d being given.

The progression is a, a+d, a +2 d, a +3 d.

Since d is added to each term to obtain the next one,

2 d must be added to a, to produce the 3d term,

3 d must be added to a, to produce the 4th term,

(n-1) d must be added to a, to produce the nth term.

Hence

l=a+ (n-1) d.

(I)

Thus the 12th term of the series 9, 12, 15 is 9+11 · 3 or 42.

360. To find the sum s of the first n terms of an A.P., the first term a, the last term l, and the common difference d being given.

s = a + (a + d) + (a + 2 d) ... (l − d) +l.

Reversing the order,

s = l + (l−d) + (1 − 2 d) ... (a + d) + a. Adding, 2s = (a + 1) + (a + 1) + (a + 1) ··· (a + 1) + ( a +1).

(II)

... we have

Thus to find the sum of the first 50 odd numbers, 1, 3, 5

from (I),

Hence

7=1+49.2= 39.

8 = 50 (199) = 2500.

361. In most problems relating to A. P., five quantities are involved; hence if any three of them are given, the other two may be found by the solution of the simultaneous equations :

NOTE. It is possible to find general formulæ expressing any two quantities in terms of any three others. The formule, however, have little value, since all examples can be solved without them.

Ex. 1. The first term of an A. P. is 12, the last term 144, and the sum of all terms 1014. Find the series.

The series is, 12, 23, 34, 45, 56, 67, 78, 89, 100, 111, 122, 133,

144.

But evidently n cannot be fractional, hence n = 6.

EXERCISE 134

1. Find the 11th term of the series 11, 22, 33,

2. Find the 18th term of the series 94, 87, 80,

3. Find the 7th term of the series 12, 151, 19.
4. Find the 15th term of the series 8, 10, 127.
5. Find the 11th term of the series −1, −31, −6.
6. Find the 12th term of the series —7, −1, +5.
7. Find the 13th term of the series

· 8.5, — 10.9, — 13.3

Find the last term and the sum of the following series :