118. Sum and Difference of any Two Like Powers. By performing the division, we find that

We find by trial that

a2 + b2, a2 + b2, a + bo, and so on

are not divisible by a + b or by a - b. Hence,

When n is a positive integer, it is proved in chap. vii,

1. an + bn is divisible by a + b if n is odd, and by neither

a + b nor a - b if n is even.

2. an - bn is divisible by a a + b and a - b if n is even.

b if n is odd, and by both

NOTE. It is important to notice in the above examples that the terms of the quotient are all positive when the divisor is a b, and alternately positive and negative when the divisor is a+b; also, that the quotient is homogeneous, the exponent of a decreasing and of b increasing by 1 for each successive term.

CHAPTER VII.

FACTORS.

119. Rational Expressions. An expression is rational if none of its terms contain square or other roots.

120. Factors of Rational and Integral Expressions. By factors of a given integral number in Arithmetic we mean integral numbers that will exactly divide the given number. Likewise, by factors of a rational and integral expression in Algebra we mean rational and integral expressions that will exactly divide the given expression.

121. Factors of Monomials. The factors of a monomial may be found by inspection. Thus, the factors of 14 a2b are 7, 2, a, a, and b.

122. Factors of Polynomials. The form of a polynomial that can be resolved into factors often suggests the process of finding the factors.

123. When the terms have a common monomial factor.

Resolve into factors 2x2 + 6 ху.

Since 2 and x are factors of each term, we have

.. 2 x2 + 6 xy = 2x (x + 3y).

Hence, the required factors are 2x and x + 3y.

124. When the terms can be grouped so as to show a common compound factor.

1. Resolve into factors ac + ad + bc + bd.

Since one factor is seen in (2) to be c + d, dividing by c + d we obtain the other factor, a + b.

2. Resolve into factors 3x2 + 6 ax + bx + 2ab.

3x2 + 6 ax + bx + 2 ab = (3x2+6ax) + (bx + 2ab)
= 3 x (x + 2 a) + b (x + 2 α)
= (3x + b) (x + 2 α).

3. Find the factors of ac + ad - bc - bd.

ac + ad - bc - bd = (ac + ad) - (bc + bd)

= a (c + d) - b (c + d)
= (a - b) (c + d).

NOTE. Here the last two terms, - bc bd, being put within a parenthesis preceded by the sign -, have their signs changed to +.

4. Resolve into factors 3 x3 - 5 x2 - 6x + 10.

3 x3 5 x2 6 x + 10 = (3 x3 — 5 x2) — (6 х — 10)
= x2 (3x - 5) - 2 (3 х - 5)
= (x2 – 2) (3x - 5).

5. Resolve into factors 5 х3 — 15 ах2 - x + 3a.

5 x3

15 ax2 - x + 3 a = (5 x3 — 15 ах2) — (x - 3а)
= 5 x2 (x - 3 а) – 1 (x - 3)
= (5 x2 - 1) (x – 3 а).

6. Resolve into factors 6 y

27 x2y - 10 x + 45 x3.

6 y 27 x2y 10 x + 45 x3 = 6 y - 10 x - 27 x2y + 45 x3

= (6 у - 10 x) — (27 х2у — 45 x3)

= 2 (Зу - 5x) — 9 x2 (3 у — 5x)
= (2 - 9 x2) (3у — 5х).

5. x2 + ху

11.

x2 + 4x2 + 3x + 12.

125. When a trinomial is a perfect square. A trinomial is a perfect square if its first and last terms are perfect squares and positive, and its middle term is twice the product of the square roots of the first and last terms.

Thus, 16 a2 24 ab +962 is a perfect square.

The rule for extracting the square root of a perfect trinomial square is as follows:

Extract the square roots of the first and last terms, and connect these square roots by the sign of the middle term.

Thus, if we wish to find the square root of

16 a2 - 24 ab + 962,

we take the square roots of 16 a2 and 962, which are 4a and 36, respectively, and connect these square roots by the sign of the middle term. The square root is therefore