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IN MEMORIAM
FLOR1AN CAJORI

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eath's Mathematical Monographs

Issued under the general editorship of

Webster Wells, S. B.

ir of Mathematics in the Massachusetts Institute of Technology

FACTORING

BY

WEBSTER WELLS, S. B.

Professor of Mathematics in the
Massachusetts Institute of Technology

D. C, Heath & Co., Publishers

Boston New York Chicago

Number 7

Price, Ten Cents

FACTORING

AS PRESENTED IN

WELLS' ESSENTIALS ^/ALGEBRA

I. Advanced processes are not presented too early.
II. Large amount of practice work.

No other algebra has so complete and well-graded development of
this important subject, presenting the more difficult principles at those
stages of the student's progress when his past work has fully prepared
him for their perfect comprehension.

The Chapter on Factoring contains these simple processes :
CASE I When the terms of the expression have a common monomid factor —
thirteen examples. ' #

II. When tin expression is the sum of two binomials which have a commoa

binomial factor — twenty examples.
Ill When the expression is a perfect trinomial square — twenty-six examples.
iv! When the expression is the difference of two perfect squares— fifty-nve

examples. , ,

V. When the expression is a trinomial of the rorm x "Tax-TO

— sixty-six examples,
VI. When the expression is the sum or difference of two perfect cubes-— twenty
examples. c

VII. Wlien the expression is the sum or difference of two equal odd powers or
two quantities — thbteen examples.

Ninety-three Miscellaneous and Review Examples.
Further practice in the application of these principles is given in the two following
chapters — Highest Common Factor and Lowest Common Multiple.

In the discussion of Quadratic Equations, Solution of Equations by
Factoring is made a special feature.
Equations of the forms ^-5^-24 = 0, 2X*—X^O, *3+^a~*--4

= 0, and ** — 1 = o are discussed and illustrated by thirty examples.
The factoring of trinomials of the form ax*+bx+c and ax*+bx*+c,

which involves so large a use of radicals, is reserved until Chapter

XXV, where it receives full and lucid treatment.

The treatment of factoring is but one of the many features of
superiority in Wells' Essentials of Algebra.

Half Leather, 33S ftges. Trice $1. JO.

D C HEATH & CO., Publishers

BOSTON NEW YORK CHICAGO

HEATH'S MATHEMATICAL MONOGRAPHS
Number 7

FACTORING

BY

WEBSTER (WELLS, S.B.

PROFESSOR OF MATHEMATICS IN THE MASSACHUSETTS
INSTITUTE OF TECHNOLOGY

BOSTON, U.S.A.

D. C. HEATH & CO., PUBLISHERS

1902

CAJORI

PREFACE.

The present work contains a more varied assortment of
methods and examples in factoring than are to be found
in most high school texts.

Two methods are given for factoring expressions of the
forms x2 + ax + b and ax2 + bx -f c ; the first by inspection,
and the second by the solution of a quadratic equation ; the
latter is applicable to every case.

In factoring expressions of the form xA + axPy2 -f- y\ ex-
amples are given involving radicals, this special case being
of importance in higher mathematics.

References made to sections in the author's "Essentials
of Algebra" give the authority for the statements in
question.

As the examples are different from those in the author's
"'Academic Algebra" and "Essentials of Algebra," they
may be used to supplement the classroom work in those
texts. .

WEBSTER WELLS.

COPYKIGUT, 1902, BY WkBSTKU WlCLLg.

(pA/U

FACTORING.

DEFINITIONS.

1. To Factor an algebraic expression is to find two or
more expressions which, when multiplied together, will pro-
duce the given expression.

In the present treatise, we consider only the factoring of integral
expressions.

2. A Common Factor of two or more expressions is an
expression which will exactly divide each of them.

FACTORING.

3. It is not always possible to factor a rational and inte-
gral polynomial.

There are, however, certain expressions which can always
be factored ; these will be considered in the present treatise.

4. Case I. When the terms of the expression have a com-
mon factor.

1. Factor 14 ab* - 35 aW.

Each term contains the monomial factor 7 ab2.
Dividing the expression by 7 ab2, we have 2 62 — 5 a2.
Then, 14 ab* - 35 a*b2 = 7 ab\2 b2-& a2).

2. Factor (2 ra -f 3)a? + (2 m + S)f.

The terms have the common binomial factor 2 m -+ 3.
Dividing the expression by 2 m -f- 3, we have a2 -f- j/3.
Then, (2m + 3y + (2m + 3)y» = (2 m + 3) (a? + ^).

w

2 Factoring,

3. Factor (a — b) m 4- (5 — a)n.

We have b — a m — (a — 6). (JSJss. u%., § 41.)

Then, (a — 5)m + (b — a)n = (a — 6)m — (a — 6)n

= (a — b) (m — n).

We may also solve Ex. 3 by changing the first term into the form

— (6 — a)m.

Thus, (a — 6)m + (b — a)n = (b — a) n — (b — a)m = (6 — a) (n— ra) .

We may thus have more than one form for the factors of an
expression.

4. Factor 5 a(x — y) — 3 a(x + y).

5a(x — y) — 3 a(x + y)=a [5(x — y) — 3(x -f #)]
= a(5 # — 52/ — 3x — 3y)
= a(2 x-8y)
= 2 a(x — 4 y).

EXERCISES I.

Factor the following :

5. a6-5a5-2a4 + 3a3. 8. (a - 2)b* - (a - 2)cW.

6. m5n2 + msn4 — mne. 9. (3 x + 5)m-l- (3a-f5).

7. 24 arty— 40 afy2+ 56 oty. 10. (m—n)x— (n— m)(y+z).

11. a(a2-2)+3(2-a2).

12. (a + 2/)(m + tt) + (a? + 2/)(™-*0-

13. a(6 + c) - a(6 - c).

14. 3^(a-l)--(l--a;).

15. (a + m)2 — 3(a + m).

16. ar2(52/-2«)~^(22/ + ^).

17. (m — n)3 -f 2 m(m — n)2.

18. 4:x(a — b — c) — 5 2/(&4-c — a).

19. (a - 6) (m2 -f a») - (a - b) (m2 - i/z).

20. (m-7i)4-2m(m-7i)3 + m2(m-7i)2.

Case I — Common Factor. 3

5. The terms of a polynomial may sometimes be so ar-
ranged as to show a common polynomial factor; and the
expression can then be factored as in Case L

1. Factor ab — ay + bx — xy.

By Sec. 4, (ab - ay) + (bx — xy) = a(b - y) -f x(b - y).
The terms now have the common factor b — y.
Whence, ab — ay -f- bx — xy = (a + x) (b —y).

2. Factor a3 + 2 a2 - 3 a - 6.

In this case, it is convenient to enclose the last two terms
in parentheses preceded by a — sign.

Thus, a? + 2 a2 - 3 a - 6 = (a3 + 2 a2) - (3 a + 6)
= a2(a + 2)-3(a + 2)
= (a2 - 3) (a + 2).

EXERCISES II.
Factor the following :

3. ac + ad + be + &o\ 5. mx + my — nx — ny.

4. ay — 3 a; + 2 y — 6. 6. a& — a — 5 6 + 5.

7. 8 ay + 12 ay + 10 &c + 15 a&.

8. m4 + 6m3-7m-42.

9. 6 -10 a + 27 a2 -45 a3.

10. 20 db -28 ad- 5bc + 7 cd.

11. m3 — m2n -f ran2 — n3.

12. aW - aWd3 - aWd2 + cW.

13. 63+36x2 + 56z3 + 32s5.

14. 48 xy + 18 na - 88 my - 33 mn.

15. mx-\- my -{-nx + wy +j%c +jpy.

16. ax — ay + as — 6a; + by — 6z.

17. 3 am — 6 an + 4 6m — 8 bn -f- cm — 2 en.

18. a# -f ay — az — bx — by + bz + ex + cy — cz.

4 Factoring.

6. If an expression when raised to the ?ith power, n being
a positive integer, is equal to another expression, the first
expression is said to be an nth Boot of the second.

Thus, if an = b, a is an nth root of b.

7. The Radical Sign, -y/, when placed before an expres-
sion, indicates some root of the expression.

Thus, Va indicates a second, or square root of a ;
Va indicates a third, or cube root of a ;
Va indicates a fourth root of a ; and so on.

The index of a root is the number written over the radical
sign to indicate what root of the expression is taken.

If no index is expressed, the index 2 is understood.

An even root is one whose index is an even number ; an
odd root is one whose index is an odd number.

8. A rational and integral expression {Ess. Alg., § 108) is
said to be a perfect square, a perfect cube, or, in general, a
perfect nth power, when it has, respectively, a rational and
integral square, cube, or nth root.

9. Since (2 a2b)3 = 8 aebs, a cube root of 8 a66s is 2 a?b.
Again, since (m2)4 = m8, a fourth root of m8 is m2.

We also have (— m2)4 = m8 ; so that another fourth root
of m8 is — m2.

It is evident from this that every positive term which is
a perfect nth power, has a positive nth root ; and in addition,
if n is even, a negative nth root of the same absolute value.

We shall call the positive nth root the principal nth root.

It will be understood throughout the remainder of the work, unless
the oontrary is specified, that when we speak of the nth. root of a term,
we mean the principal nth root.

nth Root. 5

10. Any Boot of a Power.

Required the value of -tya™1, where m and n are any-
positive integers.
We have (am)n = amn.

Then by Sec. 6, VcF" = am.

11. Any Boot of a Product.

Let n be a positive integer, and a, b, c, •••, numbers which
are perfect nth powers.

By Sec. 6, (-Vo6c~^)n = abc~ ..

Also, (Vax Vb x Vc x •••)"= (Va)* x (\/&)nX (V^)nX ...

= abc-", by Sec. 6.

Then, Vafo^ = Va x a/6 x Vc x ••• ;

for each of these expressions is the nth root of abc • • • ; and
we assume as an axiom that, if two perfect nth powers are
equal, their principal nth roots are equal.

12. Any Boot of a Monomial.

From Sees. 9, 10, and 11 we have the following rule for
finding the principal root of a rational, integral, and positive
monomial, which is a perfect power of the same degree as
the index of the required root :

Extract the required root of the numerical coefficient, if
any, and divide the exponent of each letter by the index of
the required root.

Ex. Required the fifth root of 32 a1 W.

By the rule, a/32 a10b5c15 = 2 a2bcs.

13. A rational and integral trinomial is a perfect square
when its first and last terms are perfect squares, and positive,
and its second term plus or minus twice the product of their
square roots (Ess. Alg., §§ 78, 79).

Thus, 4 x2 -f 12 xy2 + 9 yA is a perfect square.

6 Factoring.

To find the square root of a perfect trinomial square, we
have the following rule :

Extract the square roots (Sec. 12) of the first and third terms,
and connect the results by the sign of the second term.

1. Find the square root of 4 x2 + 12 xy2 + 9 y\

By the rule, V4a;2 + 12xf + 9?/4 = 2a; + 3y\
The expression may also be written in the form

(_2a02 + 2(-2aD(-320 + (-32/)2;

which shows that ( — 2 x) + ( — 3 y), or — 2 x — 3 y, is also a square
root.

But the first form is simpler, and will be used in the examples of
the present treatise.

2. Find the square root of m2 — 2 mn -f n*.

By the rule, Vra2 — 2 mn + n2 = m — n.

We may write the expression in the form n2 — 2 mn -f wi2 ; in which
case, by the rule, the square root is n — m.

In the present treatise, when finding the square roots of expressions
of the form m2 — 2 mn + w2, we shall always take the square root of the
first term as given, minus the square root of the third term as given.

14. Case II. When the expression is a trinomial perfect
square.

1. Factor25a2 + 40a&3 + 1666.

By Sec. 13, the square root of the expression is 5 a -f- 4 68.
Then, 25 a2 + 40 ab* -f 16 b6 = (5 a + 4 63)2.

2. Factor x2 - 2 x(y - z) + {y- z)2.

a? _2% - z) + (y - z)2 =0- (y -z)f

« (* — y + *)2.

3. Factor — m4 + 4 m2n2 — 4 n4.

— m4 H- 4 mW — 4 n4 = — (m4— 4 m2n2 -f- 4 n4)
==_(m2_2n2)2.

Case II — Trinomial Perfect Square. 7

EXERCISES III.

Factor the following :

4. ar* + 8a + 16. 8. x*y2 + 14 xy + 49.

5. 9- 6a + a2. 9. 36 a2 - 132 ab + 121 b\

6. m2 + 10 mn + 25 n2. 10. - 16 a2 + 24 ax - 9 a2.

7. 4a6-4a3&c2 + 62c4. 11. 81 m2 + 144 mn + 64 n\

12. - 25 z10 - 60 afyV - 36 ^V.

13. 64 oV - 240 afccy + 225 &y.

14. 49 m6 + 168 mW + 144aft

15. 100 a252 + 180 a&c2 + 81c4.

16. 144 a;V- 312 afyz3 + 169 z6.

1 7. - 121 a4m2 + 220 a2b2mn - 100 bAn\

18. 169 a862 + 364 aWd3 + 196 c4d6.

19. (x + y)2 + 22(x + y) + 121.

20. a2 - 8 a(m - *) + 16(m - n)2.

21. 9x2-6x(y-\-z) + (y + z)2.

22. (m - n)2 - 2(m - n)n + n\

23. 25(a + 6)2 + 40(a + b)c + 16 c2.

24. 36(a - x)2 - 84(a - »)y + 49 y\

25. 49 m2 + 42 m(m + x) + 9(ra + a;)2.

26. (a + 6)2 + 4(a + 6) (a - 6) + 4(a - 6)2.

27. 9(x + y)2-12(x + y)(x-y) + ±(x-y)2.

15. Case III. When the expression is in the form

a2 + b2 + c2 + 2 a& + 2 ac + 2 6c.

This expression is the square of a + 6 + c (i£ss. ^4Z^., § 187).

1. Factor 9^ + ^ + 4z2 — 6 a*/ + 12 zz - 4 ?/z.

We must show that this expression is in the form of the
square of a trinomial (Ess. Alg , § 187), and find its square
root.

8 Factoring.

Now 9 x2 is the square of 3 x-7 and hence the first term of
the trinomial is 3 x.

We next consider the terms involving the first power of x.

— 6 xy can be written in the form 2(3 x)(—y), and 12 xz
can be written in the form 2(3 x)(2 z).

This suggests that the second term of the trinomial is
— y, and the third term 2 z j and since the second, third,
and last terms of the given expression are in accordance
with this, the expression is a perfect square, and its square
root is 3 x — y -f- 2 z. Then,

9 x2 + y2 + 4 z2 - 6 xy + 12 xz - 4 $0 = (3 x - y + 2 z)2.

The square root of the expression is also — 3x + y — 2z; as may be
seen by writing it in the form
(- 3 x)2 + y2 + (_ 2 *)2 + 2(_ 3x)y + 2(- 3 «)(- 2 *) + 2 y(- 2 *).

EXERCISES IV.
Factor the following :

2. a2 + b2 + c? -2 ab - 2ac + 2 be.

3. a? + 4y2 + 9 + 4,xy + 6x + 12y.

4. l4-25m2 + 36n2-10m + 12n-60mn.

5. a2 + 81 b2 + 16 + 18 ab - 8 a - 72 6.

6. 9x2 + y2 + 25z2-6xy-30xz + 10yz.

7. 36m2+64n24-i»2 + 96mn — 12maj — 16na?.

8. 16 a4 + 9 64 + 81 c4 + 24 a262 + 72 a2c2 + 54 6V.

9. 25^+49^° + 36^-70a^ + 60a^-84^z4.

16. Case IV. TT^en £fte expression is tJie difference of two
perfect squares.
We have a2 - b2 = (a + 6) (a - 6). (#ss. .%., § 80.)

Then, to obtain the factors we have the following rule :

Extract the square roots of the first square and oftlie second
square ; add the results for one factor and subtract the second
result from the first for the other.

Case IV — Difference of Two Perfect Squares. 9

1. Factor 36 a2b4 - 49 c8.

By the rule, 36 a2b4 - 49 c6 = (6 ab2 + 7 c3) (6 ab2 -7 c3).

2. Factor (2 x - 3 y)2 - (x - #)2.
By the rule, (2 <c - 3 y)2 - (a; - y)2

= [(2x-3y) + (x-y)l[(2x-3y)-(x-y)l
= (2x-3y + x-y)(2x-3y-x + y)
= (3x-4y)(x-2y).

A polynomial of more than two terms may sometimes be
expressed as the difference of two squares, and factored by
the rule of Case IV.

3. Factor 2 mn + m2 — 1 + n2.

The first, second, and last terms may be grouped together
in the order m2 -|- 2 mn -f n2 ; which expression, by Sec. 13,
is the square of m + n.
Thus,

2 ran + m2 — 1 -f- n2 = (m2 + 2 ran + n2)— 1
= (m + w)3-l
= (m -f 71 + 1) (ra + 71 — 1).

4. Factor 12 2/ + x2 - 9 t/2 - 4.

122/ + ar2-92/2-4 = ar2--9 2/2 + 12 2/-4
= a2-(92/2-122/ + 4)
= ^-(3^-2)2,bySec. 13,
= [s+(3y-2)][>-(3y-2)]
==(a; + 32/_2)(iC-32/ + 2).

EXERCISES V.

Factor the following :

5. x2-^ 9. 169 aV- 400 b4y«.

6. 16 -a2. 10. 196 mV2 - 289 nV1.

7. 81ra4-25n«. 11. 361 a86» - 225.

8. 64 -- 121 aW. 12. 625 aKmw - 324 &un16.

io Factoring.

13. (a-fcy-c2. 16. 36m2-(2m-3)2.

14. (Sx + yy-x*. 17. (^-^-(m + ri)2.

15. a2-(m + n)2. 18. (4 a + x)2 - (6 + 3 y)\

19. 4(a-6)2-(c-<f)2.

20. (2m + n)2-(m + 2n)2.

21. (6 a + a)2 -(a -8 a)2.

22. 16(2 m + 7 a)2 - 49(3 n - 4 t/)2.

23. (9x-5yy-(6x + 7y)2.

24. 25(8 a - 3 b)2 - 9(4 a + 5 6)2.

25. a^--2^ + 2/2-9.

26. a2+b2-ci + 2ab.

27. 4m2 + n2— jp2 — 4mn.

28. x2-y2-2yz-z2.

29. 4-a2 + 2a6-62.

30. 16m2-n2-9p2-6np.

31. m2-2mn + n2-x2 + 2xy-y2.

32. a2 + 2a& + &a-c8-2cd-cP.

33. a2 + a2 - 62 - y2 + 2 aa - 2 &y.

34. aj2-2/2 + m2-l-2mx-2y.

35. 16a2-8a& + &2-c2--10ctf-25d2.

36. 28^-36z2 + 492/2 + 603-25 + 4z2.

17. Case V. When the expression is in the form

xA + axPy2 + y\

Certain trinomials of the above form may be factored by
expressing them as the difference of two perfect squares,
and then employing Sec. 16.

1. Factor a4 + a2b2 + b\

By Sec. 13, a trinomial is a perfect square if its first and
last terms are perfect squares, and positive, and its second
term plus or minus twice the product of their square roots.

Case V — Form xA + ax2y2 + y4. 1 1

The given expression may be made a perfect square by-
adding a?b2 to its second term ; and this can be done pro-
vided we subtract a2b2 from the result.

Thus,

a4 + aty + bA m (a4 + 2 aty + ft4) _ a262

= (a2 + b2)2 - a2b2, by Sec. 13,

= (a2 + &2 + a6)(a2 + b2 - aft), by Sec. 16,

= (a2 + ab + b2)(a2 - a& + J*2)-

2. Factor 9** -37 a? + 4.

The expression will be a perfect square if its second term
is -12 a;2.

Thus,

9 s4 - 37 x2 + 4 = (9 s4 - 12 s2 + 4) - 25 s2
= (3s2-2)2-(5s)2
= (3 a* _ 2 + 5 s) (3 a2 - 2 - 5 a?)
= (3 x2 + 5 s - 2) (3 s2 - 5 x - 2) .

The expression may also be factored as follows :
9a;4 - 37 x2 + 4 = (9s4 + 12a? + 4) - 49s2
= (3s2 + 2)2-(7s)2
= (3s2 + 7s + 2)(3s2-7s + 2).

EXERCISES VI.

Factor the following :

3. s4 + 5s2 + 9. 7. 9s4 -f 6s2?/2 + 49*/4.

4. a4 - 21 a2b2 + 36 64. 8. 16 a4 - 81 a2 + 16.

5. 4-33s2 + 4s4. 9. 64-64m2 + 25ra4.

6. 25m4-14m2n2 + H4. 10. 49 a4 - 127 aV + 81 s4.

Factor each of the following in two different ways (com-
pare Ex.2):

11. s4- 17 a? + 16. 13. 16 m4-104 m2x2+25x\

12. 9 -148 a2 + 64 a4. 14. 36 a4 - 97 a2m2 + 36 m4.

1 2 Factoring.

18. There are many expressions of the form xt+axPif+y*
which require radicals for their factoring.

Some of the most important applications of the method
in the Integral Calculus, are examples of this nature.

The present section should not be taken up until the
pupil is familiar with radicals. (See Ess. Alg., § 287.)

1. Factor z* -5a? + l-

x*-5x2 + l = (x*-2x2 + l)-3x2

m fy? L i)2 _ (aVS)2

= (x2 + xV3 - l)(x2 - xV3 - 1).

The expression may also be factored as follows ;
X4 _ 5^2 + i -.(gg* + 2 x2 + 1)- 7 x2
= (x2 + l)2-(fcV7)2
= (x2 4- xy/7 + l)(x2 - xV7 + 1).

Certain expressions of the form x4 + y* may be factored
by the above method.

2. Factor a** + l.

x* + l = (x4 + 2x2 + l)-2x2
= (^ + ±y __ (a. V2)2

= (x2 + zV2 + l)(rf - W2 + 1).

EXERCISES VII.
Factor the following :

3. tf + 7a? + 4:.

4. a4 + &4.

5. 9m4-llm2 + l.

6. 4a4 + 6a2 + 9.'

7. 36 a4 -92 a2 + 49.

8. 25m4 + 28raW-f 16n4.

Case VI — Form x2 + ax + l>. 13

19. Case VI. When the expression is in the form
x2 + ax -f b.

We have ar* -f- (m -f n)x -f mn = (a; + m)(a; + n), as may be
verified by multiplying x + m by a; -f- n.

If then a trinomial is in the form x2 + ax -f- b, and a and b
are, respectively, the sum and product of two numbers, the
factors are x plus one number, and x plus the other.

The numbers may be found by inspection.

1. Factor x2 + 14 x + 45.

We find two numbers whose sum is 14, and product 45.
By inspection, we determine that these numbers are 9 and 5.
Then, x2 + 14 x + 45 = (x + 9)(x + 5).

2. Factor x2 — 5 x + 4.

We find two numbers whose sum is — 5, and product 4.

Since the sum is negative, and the product positive, the
numbers must both be negative.

By inspection, we determine that the numbers are — 4
and — 1.

Then, a? _ 5z + 4 = (a> - 4)(a? - 1).

3. Factor x2 + 6 x - 16.

We find two numbers whose sum is 6, and product — 16.

Since the sum is positive, and the product negative, the
numbers must be of opposite sign, and the positive number
must have the greater absolute value.

By inspection, we determine that the numbers are -J- 8
and - 2. Then, x2 + 6x-16 = (x + 8)(x - 2).

4. Factor x4 - abx2 - 42 a2b2.

We find two numbers whose sum is — 1, and product — 42.

The numbers must be of opposite sign, and the negative
number must have the greater absolute value.

By inspection, we determine that the numbers are — 7
and + 6.

Then, xA - abx2 - 42 a2b2 = (ar> - 7 db)(a? + 6 ab).

14 Factoring.

5. Factor 1 + 2 a - 99 a2.

We find two numbers whose sum is 2, and product — 99.
By inspection, we determine that the numbers are + 11
and - 9.

Then, 1 + 2 a - 99 a2 = (1 + 11 a)(l - 9 a).

If the ic2 term is negative, the entire expression should be
enclosed in parentheses preceded by a — sign.

6. Factor 24 + 5 x — x2.

We have, 24 + 5«- ar* = -(ar*-5a;-24)

= - (x - 8)(« + 3)

= (8-&)(3 + a?).

In case the numbers are large, we may proceed as follows :
Required the numbers whose sum is — 26, and product — 192.
One number must be +, and the other — .

Taking in order, beginning with the factors -fix — 192, all possible
pairs of factors of — 192, one of which is + and the other — , we have :

-fix- 192,

+ 2 x - 96,

+ 3 x - 64,

+ 4x- 48,

-{-6x- 32.

Since the sum of -f 6 and — 32 is — 26, they are the numbers
required.

EXERCISES VIII.
Factor the following :

7.

a? + <Lx + 3.

14.

77 — 4n — ns.

8.

3^-7x + 10.

15.

a2-14a + 48.

9.

a2 + 7a-18.

16.

a? + 2Qx + 51.

10.

m2 — 14 m — 15.

17.

x*-12x-45.

11.

tf-16y + 55.

18.

«2 + 14n-32.

12.

rf + lSx + Sd.

19.

x*-llx + 52.

13.

28 + 3C-C2.

20.

84 + 5X-32.

Case VII — Form ax2 + bx + c. 15

21. a2 + 18 a + 56. 44. (z-t/)2-15(a;-2/)-16.

22. s/2 + 16?/ -57. 45. (m-n)2+21(m-n)-130.

23. x2- 10 x -75. 46. (a+z)2-28(a+a)+192.

24. m2 + 19m + 90. 47. a2 + 6 ax + 5 a2.

25. 95-14*i-n2. 48. tf-lxy-Zy2.

26. a>»-20a> + 96. 49. l + 5a-14a2.

27. a2 + 21 a + 98. 50. m2 - 17 mn + 66 n2.

28. a?-7»-78. 51. a2 + 12 a& + 27 62.

29. c2- 21c + 104. 52. or* - 14 mx + 40 m2.

30. 105-8m-m2. 53. l-9a-36a2.

31. «2-23x + 76. 54. m2 + 3 m?i - 54 n2.

32. a2 + a- 110. 55. a2 + 12 xy + 20 y2.

33. 7i2- 16 7i -80. 56. a2b2 - 17 abc + 60 c2.

34. a4 + 18a2 + 65. 57. l-13n-68?i2.

35. a;4 + llaj2-12. 58. a2 + 12 ax - 85 x2.

36. c6- 19 03 + 88. 59. 1 + 17 mn + 70 m2n2.

37. a2/ -13 ^-30. 60. a6-17afyz2 + 722/V.

38. a264-23a&2 + 112. 61. a2 + 6a&-9162.

39. n2x2 + 25nx + 154. 62. 1 - 3 a*/ - 108 afy2.

40. 126 + 15 2/4-2/8. 63. a2 - 32 a&c + 112 V<*.

41. a4a4 + 9aV-162. 64. x4y* + 7 ofy2* - 170 %\

42. m10 - 23 m5 + 120. 65. x2 - (2 m + 3 n)x + 6 mn.

43. (a+6)2+14(a+6)+24. 66. s2 - (a - 6)a? - ab.

20. Case VII. TT^en the expression is in the form

ax2 + foe + c.

If a is a perfect square and b divisible by Va, we may-
factor the expression directly by the method of Sec. 19.

1. Factor 9 x2 - 18 x + 5.

We have, 9a2- 18 a? + 5= (3<c)2- 6(3») + 5.

1 6 Factoring.

We find two numbers whose sum is — 6, and product 5.
The numbers are — 5 and — 1.
Then, 9a*-18a + 5=(3a>-5)(3a>-l).
If b is not divisible by Va, or if a is not a perfect square,
we multiply and divide the expression by a.

2. Factor 6 x2 + 5 a? — 4.

Multiplying and dividing the expression by 6, we have

6^{5x 1^36^ + 3Oa--24^(6a02 + 5(6a:)-24

6 6

The numbers are 8 and — 3.

Then, 6»' + 5»-4 = <6* + 8>(6*-3>.

2x3

Dividing the first factor by 2, and the second by 3, we have

6 x> + 5 x - 4 = (3 » + 4)(2 ft - 1).

In certain cases, the coefficient of cc2 may be made a per-
fect square by multiplying by a number less than itself.

3. Factor 8 x2 + 26 xy + 15 f.
Multiplying and dividing by 2, we have

8s? + 26sy + 15 y2 = 16 s2 + 52 ^ + 30 y2

_(43Q2 + 13y(4aQ + 30y2

2
_(4s + 10y)(4a? + 3y)

2
= (2a> + 52,)(4a + 32,).

4. Factor 2 + 5 a? - 3 a2.

2 + 5a>-3a2 = - (3 a2 - 5 a - 2)
_(3x)2-5(3x)-6

-3
= (3 g - 6) (3 x + 1)

-3
m (2 - x) (1 + 3 »).

Solutions Involving Radicals. 17

EXERCISES

IX.

Factor the following :

5.

2 x2 + 9 x + 9.

17.

12 x2 + 11 0 + 2.

6.

3 x2 - 11 x - 20.

18.

20 aV - 23 as + 6.

7.

4 or2 - 28 0 + 45.

19.

36 x2 + 12 a - 35.

8.

6 a2 + a; - 2.

20.

6 - » - 15 x2.

9.

5 a2 - 36 x + 36.

21.

5 4. 9 a _ 18 x>.

10.

16 x2 + 56 a; + 33.

22.

12 + 7 0 - 49 z2.

11.

8 n2 + 18 n - 5.

23.

24 jk2 - 17 no; + 3 n2.

12.

7 4. 3 x _ 4 a2.

24.

28x2-x-2.

13.

32 - 12 0 - 9 x2.

25.

21x2 + 23xy+6y2.

14.

6x2 + 7ax + 2a*.

26.

18 x2 - 27 abx- 35 a2b2.

15.

25x>-25mx-6

m2. 27.

5 _ 26 a2 - 24 a4.

16.

10 x2 - 39 0 + 14.

28.

7(a_6)2-30(a-6) + 8,

29. 12(a> + y)* + 6(* + y)+7.

30. 14 (ra- 1

n)2 + 39 a

(m _ n) + 10 a2.

31. oca2 — (ad + bc)x + bd.

21. It is not possible to factor every expression of the
form x2 + ax 4- b by the method of Sec. 19.

Thus, let it be required to factor x2 4- 18 x 4- 35.

We have to find two numbers whose sum is 18, and prod-
uct 35.

The only pairs of positive integral factors of 35 are 7 and
5, and 35 and 1 ; and in neither case is the sum 18.

22. We will now give a general method for factoring any
expression of the form ax2 4- bx 4- c.

This method requires a knowledge of quadratic equa-
tions ; and the present section, and the next, should not be
taken up until the pupil is familiar with quadratics. (See
Ess. Alg., § 284.)

We have, 0^4- bx + c = a(x2 + — 4--Y (1)

\ a a)

1 8 Factoring.

Now let rx and r2 denote the roots of the equation

a a
Then the equation can be written in the form

(x - rx) (x - r2) = 0. (Ess. Alg., § 283.)

Hence, the expression x2 -f — + - can be written

a a

(x - rx) (x - r2).

Substituting in (1), we have

ax2 + bx + c = a (a; — rx) (# — r2).

But rx and r2 are the roots of the equation x2 + — -f- - = 0,

a a

or asc2 -f- bx + c = 0 ; which, we observe, may be obtained by
placing the given expression equal to zero.
We then have the following rule :

Place the given expression equal to zero, and solve the equa-
tion thus formed.

Then, the required factors are the coefficient of x2 in the
given expression, x minus the first root, and x minus the second
root.

1. Factor 6a2 + 7^-3.

Solving the equation btf + lx — 3 = 0 (Ess. Alg., §265),
we have

„_-7±V49 + 72_-7±ll_l M 3
* 12 12 3 °r ~T

Then by the rule, 6^ + Ix- 3 = 6 (a> - £) (a? + f)

= 3(*-i) x (2)(» + f)
= (3a>-l)(2a> + 3).

2. Factor 4 + 13 x - 12 a?2.

Solving the equation 4 + 13 x — • 12 x2 = 0, we have

- 13 ± V169 + 192 = - 13 ± 19 = 1 op 4#
-24 -24 4 3*

Case VII — Form ax2 + bx 4- c. 19

Then, 4 + 13 a;- 12 ar5 = - 12 (a; 4- i)(s-|)

-*(*+i)x(-S)(#-D

= (l + 4z)(4-3a;).

3. Factor 2arJ-3a;2/ — 2y2 — 7a; + 4i/ + 6.
Placing the expression equal to zero, we have
2^-.3a;?/-22/2-7a; + 42/ + 6 = 0;
or, 2a? - (3 2/ + 7) a = 2y2 - 4# - 6.

Solving this,

3y + 7±V(3y-f7)2 + 162/2- S2y-4$

x ^

= 3y + 7±V25y2 + 10y + lr:=3,y + 7±(5y + l)
4 4

= J -r- or *—■ i — = 2y4-2 or — ^/ •

4 4 ' 2

Then, 2a2- 3a*/ - 2^/2 - 7x + 4y 4- 6

= 2[oJ-(22/4-2)]^-I^L?]
= (a?-2y-2)(2» + y-3).

EXERCISES X.

Factor the following :

4. a2 4- 14^4- 33. 6. ar>-a;-42.

5. ar2- 13a; 4- 40. 7. 9-8x-a?.

The student should now work the examples of Sec. 20 by
the method of Sec. 22.

Factor the following :

8. x2 -xy- 6tf -6x + 132/4- 5.

9. ar>-3a;2/-42/24-6a;-42/ + 8.
10. 3^-60^2/4- 5;?/2-2a;-22/-3.

20 Factoring.

11. 2a2 + 5a& + 262 + 7a + 56 + 3.

12. 3x2 + 7xy-6y2-10xz-8yz + 8z2.

13. 2-7y-7x + 3y2 + xy-4:X2.

23. In factoring an expression of the form ax2 + bx + c,
where a is a perfect square, if the method of § 20 can be
used, it is usually the shortest.

Another way of factoring an expression of this form is to
express it as the difference of two squares, and then apply
Sec. 16.

This method is shorter than that of Sec. 22.

As all the examples in the present section involve radicals, the pupil
should reserve it until he has studied that subject.

1. Factor a? -{- 2 a - 11.

The expression x2 + 2 x will become a perfect square by
adding to it the square of 1. (Ess. Alg., § 260.)
Then, ^ + 23-11 = ^ + 2a + 1-1-11
= 0 + 1)2-- 12
= (a+l + Vl2)(a;+l-Vl2),

by Sec. 16,
« (s 4- 1 + 2 V3) (x + 1 - 2 V3).

If the x2 term is negative, the entire expression should be
enclosed in parentheses preceded by a — sign.

2. Factor 4 + 12 a; -9 a2.

Wehave, 4 + 12a- 9a2 = - (9ar>- 12z - 4).

The expression 9 x2 — 12 x will become a perfect square

12

by adding to it the square of -9 or 2.

Then, 4 + 12a-9a2=- (9a2- 12a + 4-4-4)
= _[(3a__2)2~-8]
= (3a-2+V8)x(-l)(3a-2-V8)
m (2 V2 -2 + 3 x)(2V2 + 2 - 3 a).

Case VIII — Cube of a Binomial. 21

EXERCISES XL
Factor the following :

3. z2-2x-l. 6. 2-24:X-9x2.

4. 4*2 + 20^ + 19. 7. 36ar> + 72a + 29.

5. 16a,-2 - 16a + 1. 8. ll + 10x-25cc2.

24. Case VIII. TP7ien the expression is the cube of a
binomial.

1. Factor 8 a3 - 36 a262 + 54 a&4 - 27 66.

We must show that the expression is in the form of the
cube of a binomial (see Ess. Alg., § 188), and find its cube root.
We can write the expression as follows :

(2 a)3 - 3 (2 a)2(3 b2) + 3 (2 a) (3 ft2)2 - (3 by.

This shows that it is a perfect cube, and that its cube
root is 2 a — 3 b2.

Then, 8 a? - 36 a262 + 54 ab4 - 27 66 = (2 a - 3 62)3.

EXERCISES XII.

Factor the following :

2. a? + 3aj* + 3a; + l.

3. 8 -12 a-}- 6 a2 -a3.

4. l + 9m + 27m2 + 27m3.

5. 64?i3 -48n2 + 12n-l.

6. 8a3 + 36a2& + 54a&2 + 2763.

7. 125a*-75afy + 15a*/2--;?/3.

8. a6 + 18a463 + 108a2&6 + 216&9.

9. 125 m3 - 150 m2n + 60 mn2 - 8 n3.

10. 27 a363 - 108 a262c + 144 abc2 - 64 c3.

11. m6 + 21mV + 147m2a8 + 343a12.

22 Factoring.

25. Case IX. When the expression is the difference of two
perfect cubes.

The sum or difference of two perfect cubes is divisible by
the sum or difference, respectively, of their cube roots.

(Ess. Alg., § 85.)

In either case, the quotient may be obtained by aid of the
rules given in Ess. Alg., § 85.

1. Factor a6-27#9z3.

By Sec. 12, the cube root of a?6 is x2, and of 27 2/V is 3 ifz.

Then one factor is x2 — 3 y*z.

Dividing #6 — 27 t/V by x2 — 3 tfz, the quotient is

x4 + 3 x*tfz + 9 fz\ (Ess. Alg., § 85.)
Then, x« - 27 y9z* m (x2 - 3 2/V)(a4 + 3 rftfz + 9 tfz2).

2. Factor a6 + b\
One factor is a2 -f- b2.

Dividing a6 + b6 by a2 + &2, the quotient is a4 — a2b2 + 64.
Then, a6 + b* = (a2 + b2)(aA - a262 + &4).

3. Factor (x + a)3 - (x - a)3.
(as + a)3 - (» - a)3

= [(a?+a) — (»—«)] [(a+a)2+(a+a)(a-a) + (a;-a)2]
= (x+a— x-\-a)(x2 ^ax+af+x2— a2-^— 2ax-\-a2)
= 2a(3»2+a2).

EXERCISES XIII.
Factor the following :

4. m3 + ns. 7. 1-27ti3. 10. 64m3-n3.

5. 1-afy*. 8. a6 + l. 11. a363 + 216c3.

6. 8a3 + 1. 9. x* + yezQ. 12. 8m3 + 27n3.

13. 27^-125^. 15. 343 a3 -64 m6.

14. 64 + 125a363. 16. 125 x* + 512 yV.

Case X — Equal Odd Powers. 23

17. 216 aW- 343 ?i9. 20. m8-(m + 7i)8.

18. 729 asbs - 8 c3d3. 21. 27 (a - 6)3 + 8 68.

19. (x + yy + (x-y)3- 22. (2a + xy-(a + 2xy.

23. (5aj~22/)8-(3a;-42/)8.

26. Case X. When the expression is the sum or difference
of two equal odd powers of two numbers.

The sum or difference of two equal odd powers of two
numbers is divisible by the sum or difference, respectively,
of the numbers. (Ess. Alg., § 87.) The quotient may be
obtained by aid of the laws given in Ess. Alg., § 86.

1. Factor a5 + 32 b5.

We have, 32 b5 = (2 b)5.

Then, one factor is a + 2 b. (Ess. Alg., § 87.)
Dividing a5 + 32 b5 by a + 2 b, the quotient is
a4 - a8(2 b) + a2(2 b)2 - a(2 b)s + (2 b)\ (Ess. Alg., § 86.)
Then, a5+ 32 b5=(a + 2 6)(a4- 2 a36 + 4 a2&2- 8 a&3+ 16 64).

EXERCISES XIV.
Eactor the following :

2. rf + y5. 6. 1 + *7. 10. 32 a5 - 65.

3. a5-l. 7. m9 + 7i9. 11. 243^ + 2/*.

4. 1-mV. 8. a9-l. 12. m14 + 128n7.

5. a7-67. 9. ?i5 + 32. 13. 32 a5**5- 243 c10.

27. By application of the rules already given, an expres-
sion may often be resolved into more than two factors.

If the terms of the expression have a common factor, the
method of Sec. 4 should be applied first.

1. Factor 2 ax?y2 — 8 axy4.
By Sec. 4, 2 ax*y2 — 8 axy*

= 2axtf(x2-4:y2)

m 2 axy\x + 2 y)(x - 2 ?/), by Sec. 16.

24 Factoring.

2. Factor a6 - 66.

By Sec. 16, a6 - 66 = (a3 + &3)(a3 - &3).
Whence, by Sec. 25,

a6 - b* = (a + 6)(a2 - a& + 62)(a - 6)(a2 + a& + ft2).

3. Factor a8 — y8.

By Sec. 16, v*-tf = (a4 + i/4)(a;4 - y4)

= (** + ^)(«" + &)(*.+ y)(» - y).

4. Factor S(m + n)2 - 2(m2 - n2).

3(m + nf - 2(m2 - n2) = 3(m + ^)2 - 2(m + n)(m - n)
= (m + n) [3(m + n) — 2(m — n)]
=x(m4-w)(3w+3n-2m«f2n)
= (m + ?i)(ra + 5 w).

5. Factor a(a - 1) - 6(6 - 1).

a(a-l)-6(&-l) = a2-a-62 + 6
= a2-62-a + 6
= (a + 6)(a-6)-(a~6)
= (a-6)(a + &-l).

EXERCISES XV.
Factor the following :
i 6. a4 -625. 8. m16-l.

7. au-l. 9. a!8- 26 a8 -27.

10. (a2 + 4 a6 + 62)2 - (a2 + ft2)2.

11. J5^>-18#--6a3l+9#,

12. 81 m4- 256 n8. 13. a14-a>14.

14. x6-16^ + 64i/6. 15. (16m2 + n2)2-64mV.

16. 2a7x-8aW + 2a3x5--$ax7.

17. 9a2c2-16a2d2-3662c2 + 6462d2.

18. rf*-~2«F-+t 19. 729 -n6.

20. aV + ay-6W-ay- 81. 48 afy - 52 afy2 - 140 ay*.

Exercises in Case X.

25

22. 16 a7 -72 a6 + 108 a5 -54a4.

23. (m + nf - 2(m + nf + (m + n)2.

24. Eesolve a9 -f 512 into three factors by the method of
Sec. 25.

25. a2 — ra2 + a + m.

26. (^ + 4a)2-37(z2 + 4a)+160.

27. n10-1024. 28 m8 + m + ^ + a;.

29. a2c2-4 62c2-9a¥ + 36 62d2.

30. (m-w)(a?-ya) + (a + y)(m8-w2).

31. (x-l)3 + 6(a;-l)2 + 9(a;-l).

32. a2-462-a-26.

33. (m + %)(m2-^-(m-fa;)(m2-4

34. (a2 + 4y2-z2)2-16zy.

35. (a^-9x)2 + 4(^-9a;)-140.

36. aW + 27ay-8 68aj8-216ajy.

37. (2 a? -3)* -a?.

38. (m2 + m)2 + 2(m2 + m)(m + l) + (m + l)2.

39. 64 arte8 + 8 a8 -8^-1.

40. (4a2-62-9)2-36 62.

41. (x + 2y)*-x(i?-4ttf).

42. (l + aj8) + (l + a>)8.

43. (a2 + 6a + 8)2-14(a2+6a + 8)-15.

44. a4_9 + 2a(a2 + 3).

45. (a^ + y8) — ay(» + y).

46. (a8-8m8)-a(a-2m)2.

47. 9a2(3a + 4)2 + 6a(3a + 4) + l.

48. m8-m5 + 32m8-32.

49. a(a-c) — 6(6 — c).

50. m2(ra+p)+?i2(>i-p). 51. ^4-8^ + ^ + 8.

52. (27m8-a^) + (3m + a;)(9m2-12m^ + aj2).

53. (4a2 + 9)2-24a(4a2 + 9) + 144a2.

26 Factoring.

54. m9 + m6-64m3-64.

55. (tf + yy-xty^tf + y2).

56. a5 + a4b + a?b2 + a2b* + ab* + b5.

57. (8n3-27) + (2n-3)(4tt2 + 4n-6).

FACTORING BY SUBSTITUTION.

28. The Factor Theorem.

If any rational integral polynomial, involving x, becomes
zero when x is put equal to a, the polynomial has x — a as a
factor.
Let the polynomial be

Axn + Bx"-1 H h Mx + N.

Then, by hypothesis,

Aan + Ba"-1 H h Ma + jy= 0.

Since ^4an + J5an_1 + h Jfa + 2ST is equal to zero, if

we subtract it from the given polynomial, the latter will
not be changed; then, the given polynomial is equal to

A (xn — an) + B(xn~1 — an~l) -\ \- M(x — a). (1)

Each expression in parentheses is divisible by x — a {Ess.
Alg., § 87) ; and hence expression (1) has x — a as a factor.

29. 1. Factor z3 - 7 ar5 + 10 x + 6.

By Sec. 28, if the expression becomes 0 when x is put
equal to a, then x — a is a factor.

The positive and negative integral factors of 6 are 1, 2,
3, 6, - 1, - 2, - 3, and - 6.

It is best to try the numbers in their order of absolute
magnitude.

If x= 1, the expression becomes 1— 7+10+6.

If x=— 1, the expression becomes —1— 7—10+6.

If gas 2, the expression becomes 8—28—20+6.

If x=—2, the expression becomes —8—28—20+6.

If »= 3, the expression becomes 27—63+30+6, or 0.

Factoring by Substitution. 27

This shows that x — 3 is a factor.

Dividing the expression by x— 3, the quotient is 3^—4 x— 2.

Then, x3- 7 x2 + 10x + 6 = (*_ 3)(ar> - 4x- 2).

2. Prove that a is a factor of

(a + 6 + o)(a6 + 6c + ca) - (a + 6)(6 + o)(o + a).
Putting a = 0, the expression becomes

(b + c)bc — b(b + c)c, or 0.
Then, by Sec. 28, a is a factor of the expression.

3. Prove that m-\~n is a factor of

m4 — 4 msn -f- 2 mV + 5 mn3 — 2 n4.
Putting m = — n, we have

n4 + 47i4 + 27i4 — 5?i4-2ft4, or 0.
Then, m + n is a factor.

EXERCISES XVI.

Factor the following :

4. a? + 4aj2 + 7a>-12. 7. a8 - 9 a2 + 15 a; + 9.

5. a**--a?+6a?+14a+6. 8. ^-18 a + 8.

6. a? -a2 -11 a; -10. 9. aj3-5«2-8a; + 48.

10. a;4 + 8x3 + 13aj2-13a;-4.

11. 3aj*-8aj8 + 8aj8-14a? + 12.

Find, without actual division,

12. Whether x — 3 is a factor of x* - 6 x2 + 13 a; - 12.

13. Whether a; + 2 is a factor of Xs + 7 z2 — 6.

14. Whether a is a factor of x(y +z)2 +y(z+x)2 +z(x+y)2.

15. Whether a is a factor of a3(6 — c)3-f-63(c— a)3-fc3(a— 6)3.

16. Whether a;— y is a factor of (#— y)3-\-(y— z)3+(z— a)3.

17. Whether m+n is a factor of m(m+2 n)8—n(2 m+n)s.

18. Whether .a 4- b + c is a factor of

a(b + c) + b(c + a) + c(a + b) -f a2 + b2 + c2.

28 Factoring.

FACTORING OF SYMMETRICAL EXPRESSIONS.

30. Symmetry.

An expression containing two or more letters is said to be
symmetrical with respect to any two of them when they can
be interchanged without altering the value of the expression.

Thus, a + b + c is symmetrical with respect to a and b ; for, on
interchanging these letters, the expression becomes b + a + c, which
is equal to a + b + c.

An expression containing three or more letters is said to
be symmetrical with respect to them when it is symmetrical
with respect to any two of them.

Thus, ab + be -f ca is symmetrical with respect to the letters «,
6, and c ; for if a and b be interchanged, the expression becomes
ba + ac + c&, which is equal to ab + be + ca.

And in like manner, ab + be 4- ca is symmetrical with respect to
b and c, and with respect to c and a.

31. Cyclo-Symmetry.

An expression containing n letters, a, b, c, •••, m, n, is said
to be cyclo-symmetrical with respect to them when, if a is
substituted for b, b for c, •••, m for w, and n for a, the value
of the expression is not changed.

Thus, the expression (a — b)(b — c) (c — a) is cyclo-symmetrical
with respect to a, 6, and c; for if a is substituted for 5, 6 for c,
and c for a, the expression becomes (c — a) (a — 6) (6 — c), which
is equal to (a — &) (& — c)(c — a).

32. Factoring of Symmetrical Expressions.

The method of Sec. 28 is advantageous in the factoring of
symmetrical expressions.

1. Factor

The expression is symmetrical with respect to a, b, and c.
Being of the third degree, the only literal factors which

Factoring of Symmetrical Expressions. 29

it can have are three of the type a; three of the type a+b;
or a+b + c, and a factor of the second degree.
Putting a = 0, the expression becomes
&c2 + c&2 _ b2c _ C2bf or 0

Then, by Sec. 28, a is a factor ; and, by symmetry, b and
c are factors.

The expression, being of the third degree, can have no
other literal factor ; but it may have a numerical factor.

Let the given expression = mabc.

To determine m, let a = b = c = 1.

Then, 4 + 4 + 4 -2 - 2 -2 = m, or m = 6.

Whence, the given expression = 6 a&c.

2. Factor a2 (2/ ■+- z) + 2/2(z + #) + z2(x + y) + 3 #2/2.

The expression is symmetrical with respect to x, y} and 2.

The only literal factors which it can have are three of the
type x ; three of the type x + y \ or x + y -f- z, and a factor
of the second degree.

It is evident that neither x, y, nor z is a factor.

Putting x equal to — y, the expression becomes

y2(y + *) + y\z - y) - 3 A

which is not 0.

Then, x + y is not a factor; and, by symmetry, neither
y + z nor z -j- # is a factor.

Putting aj equal to — y — z, the expression becomes

= ^ + 3^ + Syz2 + z8 - f - 2s ~ 3^z - Syz2 = 0.
Then, x -f 2/ + 2 is a factor.

The other factor must be of the second degree; and, as it
is symmetrical with respect to x, y, and z, it must be of the

form

m (a? + #* + «■), or n(xy + yz + zx).

It is evident that the first of these cannot be a factor;
for, if it were, there would be terms involving ar*, y3, and z8
in the given expression.

30 Factoring.

Then, the given expression = n(x -f y + z) (xy + yz + zx).

To determine n, let x = 1, y = 1, and 2 = 0.

Then, 1+1 = 2n, andn = l.

Then, the given expression = (x + y + z) (xy + yz-\- zx).

3. Factor ab (a — b) + be (b — c) + ca (c — a).

The expression is cyelo-symmetrical with respect to a, b,
and c.

It is evident that neither a, b, nor c is a factor.

The expression becomes 0 when a is put equal to b.

Then, a — b is a factor ; and, by symmetry, b — c and
c — a are factors.

The expression can have no other literal factor, but may
have a numerical one.

Let the given expression = m(a — b) (b — c) (c — a).

To determine m, let a = 2, 6 = 1, and c = 0.

Then, 2 = — 2 m, and ra = — 1.

Then, the given expression = — (a — 6) (6 — c) (c — a).

EXERCISES XVII.

Factor the following :

4. a* + a?b + ab2 + b3.

5. ra3 + 2 ra2n -+- 2 ran2 -f- n8.

6. (a&+6c+ca)(a+6+c)-a2(6+c)-^2(c-f-a)-c2(a+6).

7. rf(y +*) +/ (r+«)+a?(a +ff)+2a>y*.

8. a(6 + c)2 + &(c+a)2 + c(a + 6)2-4a&c.

9. a2(&-c) + &2(c-a)+c2(a--&).

10. (a; + 2/ + ») (a?2/ + 2/« + ^) — (^ + 2/) (2/ + «)(^ + ^)-

11. ab(a + b) + bc(b + c) + ca(c + a) + 2 a&c.

12. (z + ^ + ^-^ + ^ + s3).

13. (tf + # + z)(#?/-f-2/z + 2a:) — ay*.

14. (x-yy + (y-zy + (z-xy.

15. a^ + a^ + SaY + a^ + S/4-

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