92. The Factors of a quantity are those quantities which, being multiplied together, will produce the given quantity.

93. A Prime Factor is one which cannot be produced by the multiplication of two or more factors; it is therefore divisible only by itself and unity.

94. An algebraic expression may be factored by inspection, by trial, or by its law of formation.

To express the prime factors of a monomial, we have only to factor the coefficient, and repeat each letter as many times as there are units in its exponent. Thus,

95. The following remarks will aid in factoring polynomials: 1st. If all the terms of a polynomial have a common factor, the quantity may be factored by writing the other factors of each term within a parenthesis, and the common factor without. Thus, 2a2x3-6a2x2+4a2x-10a3 = 2a2 (x3-3x2+2x-5a).

2d. If two of the terms of a trinomial are perfect squares, and the other term is twice the product of the square roots of the squares, the trinomial will be the square of the sum or difference of these roots (70, I and II), and may be factored accordingly. Thus, in the trinomial, 4a4-20ab+252, the two terms, 4a1 and 2562, are the squares of 2a2 and 5b respectively, and the other term, 20ab, is equal to 2 × 2a2 × 5b; hence,

4a4-20ab25b2 = (2a2 - 5b) (2a2-5b).

3d. If a binomial consists of two squares connected by the minus sign, it must be equal to the product of the sum and difference of the square roots of the terms (70, III). Thus,

9x2y=(3x + y) (3x − y).

4th. Quantities in the form of am±b may be factored by reference to the principles and formulas relating to these quantities. Thus,

`a3 + b3 = (a + b) (a2 — ab + b2).

NOTE.-It may happen that when there is no factor common to all the terms, a portion of the polynomial may be factored.

4. Factor a2 + c2x + cmx.

Ans. a2+c (c + m) x.

5. Factor 23 — x3y + xy2 — y3. ? + b2 (a + b) (a + b).

6. Factor ab2+2a3b®+a2ba.

Ans.

7. Arrange (22- x) a + (x2 + x) (3bc) -q according to the powers of x. Ans. (a+3b-c) x2 (a-3b+c) x―q.

8. Factor a5m - 9am3.

9. Factor Sa3 — x3.

x2 Ans. am (a2-3m) (a2 + 3m).

Ans. (4a2+2ax + x2) (2α-x).

10. Factor y+243. Ans. (y1—3y3+9y2—27y+81) (y+3).

11. Find the factors of 26-yo.

Ans. (x2 + xy + y2) (x2 — xy + y2) (x + y) (x − y).

12. Find the factors of a3 ab2+2abc ac2.

Ans. a (a+b- c) (a − b + c).

SUBSTITUTION.

96. Substitution, in Algebra, is the process of putting one quantity for another, in any given expression.

x3 + x2 — 5x — 3 — y3 — 2y2 — 4y + 2, Ans.

=

Hence, for substitution we have the following

RULE.-Perform the same operations upon the substituted quantity as the expression requires to be performed upon the quantity for which the substitution is made.

Ans. a2-ab+b2.

Ans. x2+2x+1.

1. Substitute a-b for a in a2+ab+b2.

2. Substitute x+2 for a in a2-2a+1.

3. Substitute x + 3 for y in y1 — Qy3 + y2 — 6.

Ans. x4+10x3 + 3x2 + 60x + 30.

4. Substitute 8+ r for x, in x2 + ax + b, and arrange the result according to the descending powers of r.

for

Ans. r2 + (2s + a) r + s2 + as + b.

5. What will a + a3b + a2b2 + ab + b4 become, when ba? Ans. 5a1. 6. What will x2 + ax2 + a2x + as become, when m + 1 is put 1 for a? Ans. 4m (m2 + 1).

X,

and m

7. What will x + y become, when a + b is put for x, and a - b for y? Ans. 2 (a1 + 6a2b2 + b1).

8. What is the value of (x + a + b + c)5 + (x − a − b — c)3, Ans. 2(x+10x32+5x84).

when a+b+c=s ?

9. In 2-7x-6 substitute y-2 for x. Ans. y3-by2+5y. 10. In 25- 2x4 + 3x3

·7x2 + 8x·

3 substitute y + 1 for x. Ans. y5 + 3y1 + 5y3.

11. If a − b = x, b − c=y, and c- - a=z, prove that 2(a—b)2 (b−c)2+2(a−b)2 (c—a)2 + 2 (b−c)2 (c—a)2 = xa+y1+z1.

THE GREATEST COMMON DIVISOR.

97. A Common Divisor of two or more quantities is a quantity which will exactly divide each of them.

98. The Greatest Common Divisor of two or more quantities is the greatest quantity that will exactly divide each of them; it is composed of all the common prime factors of the quantities.

The term, greatest, in this connection, is used in a qualified sense, and has reference to the degree of a quantity, or of its leading term, not to its algebraic or its arithmetical value. Thus, if x 3 and x2 + 4x + 2 are the prime factors common to two or more quantities, then according to the above definition, (x2 + 4x + 2) (x − 3) = x3 + x2-10x6, is the greatest common divisor. But this product is not necessarily greater in value than one of the prime factors. For, if x = 4, then we have

x2 + 4x + 2 = 34, and x3 + x2 - 10x - 6 = 34.

99. Several quantities are said to be prime to each other when they have no common factor.

CASE I.

100. When the given quantities can be factored by inspection.

It is evident (81, 2) that no factor of the greatest common divisor can have an exponent greater than the least with which it enters the given quantities. Hence the following obvious

RULE.-I. Find by inspection, or otherwise, all the different prime factors that are common to the given quantities, and affect each with the least exponent which it has in any of the quantities. II. Multiply together the factors thus obtained, and the product will be the greatest common divisor required.

1. Find the greatest common divisor of a5 — 2a3x2 + ax1, and - 2α3x + a2x2.

Factoring, we have

a3 — 2a3x2 + axa1 = a (a1 — 2a2x2 + x4) = a2 - 2α3x + a2x2= a2 (a2 - 2ax + x2) =

a (a — x)2 (a + x)2, a2 (a — x)2.

The lowest powers of the common factors are a and (a — x)2; and we have

a (ax)2= a3 — 2a2x+ax2

the greatest common divisor required.

2. Find the greatest common divisor of 2a2bc3, 6ab2c5, and 10a3bc2. Ans. 2abc2.

3. Find the greatest common divisor of 5a2y2z2, 6x3yz2, and 12x2yz3. Ans. xyz. 4. Find the greatest common divisor of x2-y2 and 2 — 2xy + y2. ・y.

Ans. x

5. What is the greatest common divisor of a3m b2m and 2ac2m 2c2bm?

6. What is the greatest common divisor of

and 3axz2 ax272 - az2?

Ans. m (a — b).

Ans.

a2x3· 3a2x2 + a2x a (x2-3x + 1).

4x2,

7. What is the greatest common divisor of 16x2 1, x and 1 8x + 16x2?