If, however, the expression be placed in minus parentheses and the terms be arranged according to descending powers of x, we have

7 ax-3a2 - 2x2 = − (2 x2 - 7 ax + 3a2)

Check.

Let x = 2, a = 3.

7=7.

We shall find that the factors first obtained, - (3 ax) (a — 2x), differ from the factors

binomials.

(2 x − a)(x − 3 a) only in the signs of the terms of the

We may show that the factors of the second set are identical with those of the first set, as follows:

-(2x-a)(x -3a)

= ·

=

(2x- a)(x − 3 a)(−1)(− 1)

(2x-a)(− 1)(x − 3 a)(− 1)

=(-2x+a)(− x + 3 a)

三 = (a 2x)(3α-x).

45. Certain expressions may be factored by applying any one of several different methods.

First Method. On examination, we find that two of the terms 16 a1 and 25 are squares. A trinomial square having these same two terms would have as a middle term twice the product of the square roots of these terms, - that is, 40 ac2. The sign of this term will be plus or minus according as the trinomial is the square of a sum or the square of a difference.

Solution I. Assuming first that the middle term is + 40 a2c2, we shall obtain the factors as follows:

The difference between 40 a2c2 and 41 a2c2 is 81 a2c2.

Accordingly we have,

16 at 41 ac2 + 25 c4

Solution II.

16 a1 41 a2c2 + 25 c4 +81 a2c2-81 a2c2
16a+ 40 a2c2 + 25 c4 - 81 a2c2
= (4 a2 + 5 c2)2 — (9 ac)2

[4a2+5c2+9 ac] [4 a2 + 5 c2 - 9 ac]
=(4a+5c)(a + c) (+ a − 5 c) (a — c).

Assuming secondly that the middle term is same result, as follows:

40 a2c2, we obtain the

16a41 ac2 + 25 c1 = 16 a1 — 41 a2c2 + 25 c1 + a2c2 a2c2

Second Method. We may apply the methods of §§ 33-39 to the expression as given, and thus obtain the same factors, as follows:

16 a* - 41 a2c2 + 25 c1 = (16 a2 — 25 c2) (a2 — c2)

-

46. Any homogeneous function of two letters may be factored, provided that it is possible to factor the non-homogeneous expression resulting from giving the value unity to one of the letters. Ex. 7. Factor x3- 9 x2y + 26 xy2 — 24 y3.

By assigning the value 1 to y, the expression 3 - 9 x2y + 26 xy2 — 24 y3, which is homogeneous with reference to x and y, reduces to the nonhomogeneous expression 23 9 x2 + 26 x 24.

The expression x3 9 x2 + 26 x 24 may be factored as follows:

From this identity we may obtain the factors of the given expression by introducing such powers of y as are necessary to make the members homogeneous expressions with reference to a and y.

Hence we have,

x3-9 x2y + 26 xy2 - 24 y3 = (x − 2 y) (x − 3 y) (x — 4 y).

27162312 — 192 =

Check
Let x=3, y = 2.

(− 1)(— 3)(— 5) - 15 = − 15.

126. 4 (a2 + c2) (a2 — c2) + 3 b3 (4 a2 — 4 c2).
127. a3 — b3 — a(a2 — b2) + b(a - b)2.

149. ax + b2y + b2x + a2y + 2 (abx + aby).

150. (a + b)2 + (b + d )2 − (c + d)2 — (c + a)2.

Application of the Principles of Factoring to the
Solution of Equations

47. To solve an equation containing one unknown is to find such

a value, or values, for the unknown as will, when substituted for the unknown, make the two members of the equation identical.

Ex. 1. Solve x2 + 15 = 8x.

By transposing the term 8x to the first member we have,

(1)

Factoring,

x2-8x+ 15 = 0.

(x-5)(x-3)=0.

(2)

By §§ 27, 25, Chap. X., the derived equation (2) is equivalent to the original equation (1).

If either of the factors

5 or x 3 becomes zero for any particular value of x, the other remaining finite, the product of the two factors will become zero. Hence for such a value of x the first member of the equa tion will take the same value as the second, that is, it will become zero. If x be given either of the values 5 or 3, one of the factors will become zero, and the other a finite number.

Accordingly these values are solutions of equation (2).

It

may

be seen that, by placing the factor x 5 of the first member of equation (2) equal to zero and solving the equation thus formed, we shall obtain x = 5, which is one of the solutions of equation (2).