1. Resolve into factors x3 + 3 x2 – 13 x ·

The exact divisors of 15 are 1, − 1, 3, 3, 5,

— 5, 15, — 15.

15, the expression does not

If we put 1 for x in x3 + 3x2 - 13 x
vanish. If we put - 1 for x, the expression vanishes.
Therefore, x-(-1), that is, x + 1, is a factor.
Divide the expression by x + 1, and we have

x8+3x2-13 x 15 = (x + 1) (x2 + 2 x

15)

= (x + 1)(x − 3) (x+5).

An expression can sometimes be resolved into three or

NOTE. more factors.

2. Resolve into factors x8

26 x 5.

By trial we find that the only exact divisor of expression vanish is 5.

Therefore, divide by x + 5, and we have

5 that makes the

x3- 26 x 5 = (x + 5) (x2 — 5 x − 1).

- 1, the exact divisors of

As neither + 1 nor -1, will make x2 - 5x — 1 vanish, this expression cannot be resolved into factors.

136. A compound expression involving x and y is divisible by xy if the expression vanishes when+y is put for x; and is divisible by x+y if the expression vanishes when y is put for x.

If n is a positive integer, prove by the Factor Theorem:

Put y for x in xn+yn; then x2 + yn yn + yn = 2 yn.
As 2 yn is not zero, x2 + yn is not divisible by x

2. xn

Put

y" is divisible by x + y, if n is even.

y.

y for x in xy", then " - yn = (— y)n — yn. If n is even, (— y)n = yn, and (— y)n — yn = yn — yn.

As yn yn = 0, xn yn is divisible by x+y, if n is even.

3. x2 + y2 is divisible by x + y, if n is odd.

+ yn, then x2 + yn = (− y)n + yn.
- yn, and (— y)n + yn :

n=

=

Put y for x in x
If n is odd, (- y)n :
- yn+yn.
As - yn + yn = 0, xn + yn is divisible by x + y, if n is odd.

From § 134 and these three cases, we have

1. For all positive integral values of n,

-1

x1 — yn = (x − y) (xn−1 + xn−2y + x2¬3y2 +

.....

2. For all positive even integral values of n,

x2 — y2 = (x + y) (xn−1 — xn−2y + x2¬3y2 −

3. For all positive odd integral values of n,

.....

x2 + yn = (x + y) (xn−1 — x2-2y + x2-3y2 — ..... +yn-1).

4. x+y" is never divisible by x by x+y, if n is even.

NOTE.

In applying the preceding rules for resolving an expression into factors, if the terms have a common monomial factor, this factor should be removed first.

When an expression can be expressed as the difference of two perfect squares, the method of § 126 should be employed in preference to any other.

9 c2 + 12 bc.

69. 4x2+9y2x2-12xy. 70. (a+b)2 - (c — d)2.

71. a2+a+3b-9b2.

72. 5 ct 15 c3 - 90 c2.
73. a2x - c2x + a2y - c2y.
74. x+16 a2x2 + 256 a*.
75. (x + y) + (2 x − y)3.

76. a2+b2- c2 + 2 ab.
77. y2 a2c2 2 ac.
78. (x + 5a)2 - 25 a2.

81. a2 - 2 ab + b2 + 12xy - 4x2 — 9 y2.