or we may write the expression in the form

1- x2+ax - ax3= 1 − x2+ax (1 − x2),

and the factors 1 - x2, 1+ax are obvious.

For the best arrangement or grouping no general rule can be given: the following cases are however of frequent occurrence and of great importance.

I. When one of the letters occurs only in the first power, the factors often become obvious when the expression is arranged according to powers of that letter.

Ex. 1. To find the factors of ab+be+cd + da.

Arranged according to powers of a we have a (b+d)+bc+cd, which is at once seen to be a (b+d) + c (b + d) = (a+c) (b+d).

Ex. 2. To find the factors of x2+(a+b+c)x+ab+ac.

The expression =a (x+b+c)+x2 + bx + cx=(a+x) (x+b+c).

Ex. 3. To find the factors of ax3+x+a+1.

ax3+x+a+1=a (x23 + 1) + x + 1 = (x+1) { α (x2 -- x+1)+1}.

Ex. 4. To find the factors of a2+2ab2ac - 3b2+2bc.

The given expression is of the first degree in c; we therefore write it in the form a2+2aò - 3b2-2c (a - b)

= (a − b) (a + 3b) — 2c (a - b) = (a - b) (a + 3b − 2c).

II. When the expression is of the second degree with respect to any one of the letters; factors, which are rational and integral in that letter, can be found as in Art. 81.

Ex. 1. Find the factors of a2+3b2 - c2+2bc-4ab.

Arranging according to powers of a, we have

a2-4ab+3b2 - c2+2bca2 - 4ab + 4b2 – 4b2 + 3b2 – c2 + 2bc

= (a − 2b)2 − (b − c)2 = {(a − 2b) + (b − c)} {(a − 2b) − (b − c)} = (a − b − c) (a – 3b+c).

Ex. 2. Find the factors of a2 - b2 - c2 + d2 – 2 (ad – bc).

The expression

=a2-2ad-b2 – c2 + d2 + 2bc

Ex. 3. Find the factors of a2+2ab - ac - 3b2+5bc - 2c2.

Ex. 4. Find the factors of x4+x2 - 2ax+1− a2.

Arranging according to powers of a, we have

:= − {a2+2ax + x2 − 1 − 2x2 - x4}

- {a2+2ax − 1 − x2 − x1 } =

− { (a + x)2 − (1 + x2)2 } = − (a+x+1+x2) (a+x − 1 − x2).

=

III. When the expression contains only two powers of a particular letter and one of those powers is the square of the other, the method of Art. 81 is applicable.

Ex. 1. To find the factors of x4 - 10x2+9.

x4-10x2+9=x1-10x2+25-25+9= (x2-5)2 - 16

=(x2-5+4) (x2-5-4)=(x2-9) (x2 − 1) = (x+3) (x − 3) (x + 1) (x − 1),

or x2-10x2+9= (x2+3)2 – 16x2

=(x2+3+4x) (x2 + 3 − 4x) = (x+3) (x+1) (x − 3) (x − 1).

Ex. 2. To find the factors of x2+ x2+1.

Two real quadratic factors can be found as follows:

x1+x2+1= (x2+1)2 − x2 = (x2+1+x) (x2+1−x).

Ex. 3. To find the factors of x6 - 28x3 +27.

x6-28x3+27=x6-28x3 +142-142+27= (x3-14)2 – 132

= (1⁄23 − 1) (1⁄23 – 27) = (x − 1) (x − 3) (x2+x+1) (x2 + 3x+9).

In this case, and also in Ex. 1, two factors can be seen by inspection, as in Art. 79.

Ex. 4. To find the factors of a+b4+c4 - 2b2c2 - 2c2a2 - 2a2b2.

Arranging according to powers of a, we have

a4 - 2a2 (b2+c2) + b2+c4 − 2b2c2

=a1- 2a2 (b2+c2) + (b2 + c2)2 − (b2 + c2)2 + b4 + ca – 2b2c2

= { a2 − (b2 + c2)} 2 — 4b2c2 = (a2 – b2 – c2 – 2bc) (a2 – b2 – c2 + 2bc)

= {a2 - (b+c)2} {a2 - (b-c)}

=(a+b+c) (a-b-c) (a−b+c) (a+b-c).

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IV. Two factors of aP2+bP+c, where P is any expression which contains a, can always be found by the method of Art. 81; for we have

Ex. 1. To find the factors of (x2+x)2+4 (x2+x) − 12.

= (x2+ x − 2) (x2+x+6)
=(x+2) (x-1) (x2+x+6),

the factors of x2+x+6 being imaginary [see Art. 83, Note].

Ex. 2. To find the factors of (x2+x+4)2+8x (x2+x+4)+15x2.
The given expression = {(x2+x+4)+3x} {(x2+x+4)+5x}
=(x2+4x+4) (x2+6x+4)
=(x+2)2(x2+6x+4).

Since

2(x2+6x+1)2+5 (x2+6x+1)(x2+1)+2(x2+1)2.

2P2+5PQ+2Q2=(P+2Q) (2P+Q),

the given expression

={(x2+6x+1)+2(x2+1)} {2 (x2+6x+1)+x2+1}
=(3x2+6x+3) (3x2+12x+3)

=

=9(x+1)2(x2+4x+1).

Ex. 4. To find the factors of (x2+x+1)(x2+x+2) − 12.

The given expression = (x2 + x)2 + 3 (x2+x) − 10

=

=(x2+x-2) (x2+x+5)

=(x+2) (x-1) (x2+x+5).

24. 2y2 - 5xy + 2x2 - ay — ax — a2.

25.

26.

a2 - 36o - 3c2 + 10bc - 2ca - 2ab.

2a2-7ab - 22b2 - 5a - 356 + 3.

27. 1+ (b-a)
1 + (b − a2) x2 — abx3.

28. 1-2ax-(c — a3) x2 + acx3.

29. a (b-c)+b2 (ca) + c2 (a - b).

30. b°c + bc2 + c2a + ca2 + a2b + ab2 + 2abc.
31. a2b-ab2 + a2c - ac2 - 2abc + b2c + bc2.

32. x3 (a + 1) − xy (x − y) (a − b) + y3 (b + 1).

33.

ax (y3 + b3) + by (bx2 + a3y).

34. 2x3-4x3y - x2z + 2xy + 2xyz - y3z.
35. xyz (x3 + y3 + z3) — y3z3 — z3x3 — x3y3.

36. x1- 2x2a2 - 2x2b2 + a* + b* — 2a2b2.

37. (x+x)-14 (x2 + x)+24.

38. (x2+4x+8)2 + 3x (x2 + 4x+8) + 2x2.
39. (x+1)(x+2) (x+3) (x+4)- 24.

40. (x + 1)(x + 3) (x + 5) (x + 7) + 15.

86. Theorem. The expression x" - a" is divisible by x-a, for all positive integral values of n.

It is known that x-a, x2-a2 and x3

- a3 are all

· x2-1 (x − a) + a (x2−1 — a1-1).

Now if x a divides xn-1 a"- it will also divide x2-1(x-α) + a (a"-- a"-1), that is, it will divide a" - a". Hence, if x- a divides 07-1 an- it will also divide

xn-an.

S. A.

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