Charles Davies.

First lessons in algebra, embracing the elements of the science online

. (page 2 of 12)
Online LibraryCharles DaviesFirst lessons in algebra, embracing the elements of the science → online text (page 2 of 12)
Font size
QR-code for this ebook


8a 2 , and the terms crossed like the first term. Passing
then to the terms involving b 2 , we find their sum to be
— 5b 2 , after which we write — 3c 2 .

The marks are drawn across the terms, that none of them
may be overlooked and omitted.

(4) (5) (6)

a 6a 5a

a 5a 5b

2a Tla 5a-\-5b

(9) (10)

7abc -f- 9ax Sax +35

— 3abc—3ax 5ax — 9b
4abc-\-6ax I3ax — 6b 9a — 15c

Note. — If az=5, b=4, c=2, a?=l, what are the values
of the several sums above found.

(14)

3a/+ g +m
ag — 3af — m
ab— ag+3g
ab-\-4g

(16)

Sx 2 + 9acx+\3a 2 b 2 c 2

7x 2 —\3acx+\4a 2 b 2 c 2

— 4x 2 -{- 4acx — 20a 2 b 2 c 2

9x—9ab — Uc — 3x 2 + + 7a 2 6 2 c 2

(17) (18)

%2k-3c-7f+3g 19ah 2 +3a 3 b*-Sax 3

- 3h+Sc—2f—9g+5x — 17aA 2 — 9a 3 b*+9ax 3



(12)


(13)


9a+/


6ax — 8ac


6a+g


— 7 ax— 9ac


2a-f


ax-\-\7ac


a +g





(15)




7x+3ab-\-


3c


3x—3ab —


5c


5x — 9ab —


9c



\9h + 5c-9f—6g+5x 2ah 2 —6a 3 b*+ ax 3



22 FIRST LESSONS IN ALGEBRA.

( 19 ) ( 20 )

7a?— 9y+5*+3 — g Sa+ b

— a? — 3y —8—- g 2a— 5+ c

— a?+ y— 3*+l + 7g- — 3a+ b +2d
— 2x+6y+3z— 1 — # —65— 3c + 3d

ar+8y— 5^+9+ g — 5<? + 7c— Sd

4a?+3y+0 + 4 + 5§- 2a— 55+ 5c— 3d

21. Add together —b+3c—d—ll5e+6f—5g, 3b— 2c
—3d—e + 27f, 5c— 8J+3/— 7#, —75— 6c+17d+9c-5/
+ 11#, — 35-5d— 26+6/— 9g+h.

Ans. — 85— 109e+37/— lOg+k.

22. Add together the polynomials 7a 2 5— 3a5c— 85 2 c— 9c 3
+ ctf 2 , 8a5c— 5a 2 5-f 3c 3 — 4b 2 c + cd 2 and 4a 2 5 — 8c 3 +95 2 c
— 3d 3 . An*. 6a 2 5 + 5a5c— 35 2 c— 14c 3 +2cd 2 — 3d 3 .

23. What is the sum of 5a 2 bc+6bx— 4af, — 3a 2 5c— 65a?
+ 14a/, — a/+95a?+2a 2 5c, + 60/— 85a:+6a 2 5c.

Ans. l0a 2 bc+bx-\-\5af.

24. What is the sum of a 2 ra 2 +3a 3 m + 5, — 6a 2 n 2 — 6a 3 m— 5,
+ 95— 9a 3 m— 5a 2 rc 2 . A** — I0a 2 n 2 — I2a 3 m+9b

25. What is the sum of 4a 3 5 2 c— I6a*x— 9ax 3 d, +6a 3 b 2 c
— 6aa? 3 d+17a 4 a?, +\6ax 3 d— a±x— 9a 3 b 2 c.

Ans. a 3 b 2 c-\-ax 3 d.

26. What is the sum of —7^+35+4^—25, +3#
—35+25. Ans. 0.

27. What is the sum of a5 + 3#y— m— n, — 6#y— 3m
+ ll;i+ct?, + 3a?y + 4m — l0n-{-fg. Ans. ab-\-cd-\-fg.

28. What is the sum of 4xy+n-\-6ax-}-9am, —6xy-\-6n
— 6ax— 8am, 2xy — ln-\-ax — am. Ans. + aa\



SUBTRACTION. 23

29. Add the polynomials I9a 2 x 3 b— I2a 3 cb, 5a 2 x 3 b+l4a 3 cb

— Wax, —2a 2 x 3 b—12a 3 cb, and — I8a 2 x 3 b— I2a 3 cb+9ax.

Ans. 4a 2 x 3 b—22a 3 cb—ax.

30. Add together 3a+b+c, 5a+2b-{-3ac, a+c+ac,
and — 3a — 9ac — 8b. Ans. 6a—5b-\-2c—bac.

31. Add together . 5a 2 b-\-6cx-{-9bc 2 , 7cx—8a 2 b, and

— Ibex— 9bc 2 -{-2a 2 b. Ans. — a 2 b — 2cx.

32. Add together 8ax+5ab+3a 2 b 2 c 2 , — \8ax+6a 2 +\0ab
and lOax — 1 5a b— 6a 2 b 2 c 2 . Ans. — 3a 2 b 2 c 2 -\-6a 2 .

33. Add together 3a 2 +5a 2 b 2 c 2 — 9a 3 x, 7a 2 —8a 2 b 2 c 2 —l0a 3 x
and I0ab+16a 2 b 2 c 2 +19a 3 x. Ans. I0a 2 + I3a 2 b 2 c 2 +I0ab.



L



SUBTRACTION.



28. Subtraction, in Algebra, consists in finding the sim-
plest expression for the difference between two algebraic
quantities.

Thus, the difference between 6a and 3a is expressed bv

6a — 3a=z3a ;
and the difference between 7a 3 b and 3a 3 b by
7a 3 b-3a 3 b = 4a 3 L

In like manner the difference between 4a and 3b is
expressed by 4a— 3 b.

Hence, If the quantities are similar, subtract the coefficients ;
and if they are not similar, place the minus sign before the
quantity to be subtracted.

Quest. — 28. In what does subtraction in Algebra consist? How do
you find this difference when the quantities are positive and similar ?
When they are not similar, how do you express the difference ?



24



I


FIRST LESSONS IN ALGEBRA.






(1)


(2)


(3)


From


3ab


6 ax


9abc


take


2ab


3ax


7abc


Rem.


ab


3ax


2abc.




(4)


(5)


(6)


From


I6a 2 b 2 c


17 a 3 b 3 c l


24a 2 b 2 x


take


9a 2 b 2 c


3a 3 b 3 c


7a 2 b 2 x


Rem.


7a 2 b 2 c


I4a 3 b 3 c


17 a 2 b 2 x




(?)


(8)


(9)


From


3ax


4abx


2 am


take


8c
3ax — 8c


9ac
4abx — 9ac


ax


Rem.


2am — ax.



29. Let it be required to subtract from 4a

the binomial 2b — 3c

The difference may be put under the form 4a— (2b — 3c).
We must now remark that it is the difference between 2b"
and 3 c which is to be taken from 4a.

If then, we write 4a— 2b,

we shall have taken away too much by the units in 3c ;
hence, 3c must be added to give the true remainder, which
is 4a—2b+3c.

To illustrate this example by figures, suppose a— 5,
5 — 5, and c=3.

We shall then have 4a— 20

and 2b — 3c =10 — 9 — 1

which may be written . 4a— (2b — 3c) =20 — I =.19.



Quest. — 29. If 2b — 3c is to be taken from 4a, what is proposed to
be done ? If you subtract 26 from 4a, have you taken too much ? How
then must you supply the deficiency ?



SUBTRACTION.



25



Here it is required to subtract 1 from 20. If, then, we
subtract 25 = 10, from 4a = 20, it is plain that we shall
have taken too much by 3c =9, which must therefore be
added to give the true remainder.

30. Hence, for the subtraction of algebraic quantities,
we have the following general

I. Write the quantity to be subtracted under that from which
it is to be taken, placing the similar terms, if there are any,
under each other.

II. Change the signs of all the terms of the polynomial to
be subtracted, or conceive them to be changed, and then reduce
the polynomial result to its simplest form.

EXAMPLES.



(1)

From 6ac— bab-\- c 2
Take 3ac+3ab+7c
Rem. 3ac — 8ab+ c 2 -



« o 3



(i)

6ac—5ab-{- c 2
— 3ac — 3ab—7c



-7 c.






3ac— 8a& + c 2 — 7c.



(2)
From 6ax— a -{-3b 2
Take 9 ax— x-\- b 2
Rem. — 3ax— a-\-x-\-2b 2 .



(3)

6yx — 3x 2 -j-5b
yx — 3 + a



5yx—3x 2 +3 + 5b — a.



(4)

From 5a 3 — 4a 2 6-f 3b 2 c
Take — 2a 3 +3a 2 b— 8b 2 c
Rem. 7a 3 — 7a?b+\lb 2 c.



(5)

Aab— cd+3a 2
5ab— 4cd+3 a 2 +5b 2
- ab+37d~^bb 2 ~



Quest. — 30. Give the rule for the subtraction of algebraic quantities

3



26 FIRST LESSONS IN ALGEBRA.

6. From 6am-}- y take 3am— x. Ans. 3am -j- x-\-y.

7. From 3ax take 3ax—y. Ans. ~\-y.

8. From 7a 2 b 2 — x 2 take I8a 2 b 2 +x 2 .

Ans. —lla 2 b 2 —2x 2 .

9. From —7f+3m—8x take —6f—5m—2x+3d+8.

Ans. —f+8m—6x—3d—8.

10. From — a — 5b+7c — d take 4b—c + 2d+-2L

Ans. —a — 9b + 8c — 3d—2L

11. From . . — 3a+b— 8c+7e— 5/+3A— 7x— 13y take
£+2a - 9c+8e— 7x+7f— y— 31— k.

Ans. — 5a+&4-c — c— 12/+3A— 12y-f-3Z.

12. From a-j-5 take a— Z>. .4?is. 25.

13. From 2o?— 4a— 25 + 5 take 8 — 55 + a-f 6o\

Ans. — 4#— 5a+35— 3.

14. From 3a+5+c— d— 10 take c+2a— o\

.A/is. a-\-b — 10.

15. From 3a+b + c— c? — 10 take b— 19-f 3a.

JL/16-. c— a*-f 9.

16. From 2a£+& 3 — 4c + £c— 5 take 3a 2 — c+b 2 .

Ans. 2ab—3a 2 —3c+bc—b.

17. From a 3 +3b 2 c+ab 2 — abc take & 3 +a5 2 — abc.

Ans. a 3 +3b 2 c—P.

18. From 12a;+6a— 45+40 take 4b—3a+4x+6d—10.

Ans. 8x+9a—8b—6d+50.

19. From 2a;— 3a+45+6c— 50 take 9a + o?+65— 6c— 40.

.An*, a:— 12a— 25+12c— 10.

20. From 6a— 4b— \2c+\2x take 2a:— 8a+45— 6c.

.Aw*. 14a— 85— 6c-J- 10a?.

21. From 8a£c— 125 3 a+6c#— 7xy take 7cx—xy—l3b 3 a.

Ans. 8abc-\-b 3 a—cx—6xy.



SUBTRACTION,



27



3.1. By the rule for subtraction, polynomials may be
subjected to certain transformations.



For example,
becomes . . .
In like manner
becomes
or, again,
Also, . .
becomes
Also, . .
becomes



6a 2 — 3ab + 2b 2 —2bc,
6a 2 —(3ab — 2b 2 +2bc).
7 a 3 — 8a 2 b— 4b 2 c+6b 2 ,
7a 3 —(8a 2 b + 4b 2 c—6b 2 ) y
7a 3 — 8a 2 b-(4b 2 c-6b 2 ).
8a 3 — 7b 2 + c —d,
8a 3 -(7 b 2 — c +d).
9 b 3 — a + 3a 2 — d,
9b 3 — (a — 3a 2 +d).



32. Remark. — From what has been shown in addition
and subtraction, we deduce the following principles.

1st. In algebra, the words add and sum do not always, as
in arithmetic, convey the idea of augmentation; for a—b,
which results from the addition of —b to a, is properly
speaking, a difference between the number of units ex-
pressed by a, and the number of units expressed by b.
Consequently, this result is numerically less than a. To
distinguish this sum from an arithmetical sum, it is called
the algebraic sum.

Thus, the polynomial 2a 2 — 3a 2 b-\-3b 2 c is an algebraic
sum, so long as it is considered as the result of the union



Quest. — 31. How may you change the form of a polynomial ? 32. In
algebra do the words add and sum convey the same idea as in arithme-
tic ? What is the algebraic sum of 9 and — 4? Of 8 and — 2?
May an algebraic sum ever be negative ? What is the sum of 4 and
— 81 Do the words subtraction and difference in algebra always con-
vey the idea of diminution ? What is the algebraic difference between
8 and — 4? Between a and — hi



28 FIRST LESSONS IN ALGEBRA.

of the monomials 2a 2 , —3a 2 b, + 3b 2 c, with their respec-
tive signs ; and, in its proper acceptation, it is the arithmeti-
cal difference between the sum of the units contained in the
additive terms, and the sum of the units contained in the
sub tractive terms.

It follows from this, that an algebraic sum may, in the
numerical applications, be reduced to a negative number, or
a number affected with the sign — .

2nd. The words subtraction and difference do not always
convey the idea of diminution ; for, the numerical difference
between -\-a and — b being a-\-b, exceeds a. This result
is an algebraic difference, and can be put under the form of

a— ( — b) = a-{-b.



MULTIPLICATION.

33. If a man earns a dollars in one day, how much
will he earn in 6 days ? Here it is simply required to re-
peat the number a, 6 times, which gives 6a for the amount
earned.

1. What will ten yards of cloth cost at c dollars per yard ?

Ans. 10c dollars.

2. What will d hats cost at 9 dollars per hat ?

Ans. 9d dollars.

3. What will b cravats cost at 40 cents each ?

Ans. 40b cents.

4. What will b pair of gloves cost at a cents a pair ?



Quest. — 33. What is the object of multiplication in algebra 1 If a
man earns a dollars in one day, how much will he earn in 4 days 1 In
5 days 1 In 6 days 1



MULTIPLICATION. 29

Here it is plain that the cost will be found by repeating b
as many times as there are units in a : Hence, the cost is
ab cents. Ans. ab cents.

Note. — If we suppose a = 6, c=4, and d=z3, what would
be the numerical values of the above answers ?

5. If a man's income is 3a dollars a week, how much
will it be in 4b weeks. Here we must repeat 3a dollars as
many times as there are units in 4b weeks ; hence, the pro-
duct is equal to

3ax4b = 12ab.

If we suppose a=4 and b = 3 the product will be equal
to 144.

34. Remark.— It is plain that the product 12a b will not
be altered by changing the arrangement of the factors ; that
is, \2ab is the same as abx^-2, or as baxl2, or as
axl2xb (See Arithmetic, § 22).

35. Let us now multiply 3a 2 b 2 by 2a 2 b, which may be
placed under the form

3a 2 b 2 X 2a 2 b — 3 X 2aaaabbb ;

in which a is a factor four times, and b a factor three times:
hence (Art. 13).

3a 2 b 2 x2a 2 b = 3x2aaaabbb=6a*b 3 ,

in which, we multiply the co-efficients together and add the
exponents of the like letters.



Quest. — 34. Will a product be altered by changing the arrangement
of the factors 1 Is Sab the same as 3ba 1 Is it the same as a X 35 1
As b X 3a T 35. In multiplying monomials what do you do with the co-
efficients 1 What do you do with the exponents of the common letters ?
If a letter is found in one factor and not in the other, what do yon do ?

3*



30



FIRST LESSONS IN ALGEBRA.



Hence, for the multiplication of monomials, we have the
following

RULE.

I Multiply the co-efficients together.

II. Write after this product all the letters which are com-
mon to the multiplicand and multiplier, affecting each letter
with an exponent equal to the sum of the two exponents with
which this letter is affected in the two factors.

III. If a letter enters into but one of the factors, write it in
the product with the exponent with which it is affected in the

factor.

EXAMPLES.

1 . 8a 2 bc 2 X 7abd 2 cs 56a 3 b 2 c 2 d 2 .

2. 21 a 3 b 2 cd X 8abc 3 =z 1 68a 4 b 3 c*d.

3. 4abcx7df -=, 28abcdf.

(4) (5)

Multiply 3a 2 b 12a 2 x

by 2a 2 b \2x 2 y



6a*b 2



144a 2 x 3 y



(6;

6xy z
ay 2 z
6axy 3 z 2 .



(7)

a 2 xy

2xy 2

2a 2 x 2 y 3



(8)
3ab 2 c 3

9a 2 b 3 c

27a 3 b 5 c±



(9)
S7ax 2 y
3b 3 x*y 3



10. Multiply 5a 3 b 2 x 2 by 6cV.

11. Multiply 10aW by 7acd.

12. Multiply 9a 3 bxy by 9a 3 bxy.

13. Multiply 36a*b 1 c e d 5 by 20ab 2 c*d i .

14. Multiply 27 axyz by 9a 2 b 2 c 2 d 2 xyz.

Ans.

15. Multiply I3a?b 2 c by Babxy.



261 ab 3 x b y\

Ans. 30a 3 b 2 c 5 x 8 .

Ans. 70a 5 b 5 c*d.

Ans. 8la 6 b 2 x 2 y 2 .

Ans. 720a 9 bWd*.

243a 3 b 2 cH 2 x 2 y 2 z 2
Ans 104a 4 £ 3 c#y



MULTIPLICATION. 31

16. Multiply 20a 5 b*cd by I2a 2 x 2 y. Arts. 240a 1 b 5 cdx 2 y.

17. Multiply \4a±bWy by 20a 3 c 2 x 2 y.

Arts. 280a' ! b 6 c 2 d i x 2 y 2 .

18. Multiply 8a 3 b 3 y* by 7a*bxy 5 . Ans. 56a 7 6%y 9 .

19. Multiply Ibaxyz by 5a 5 bcdx 2 y 2 .

Ans. 37 5a e bcdx 3 y 3 z.

20. Multiply 5la 2 y 2 x 2 by 9a 2 bc 2 x 5 y.

Ans. 459a 4 ta 2 # 7 y s .

36. We will now proceed to the multiplication of poly-
nomials. Take the two polynomials a-\-b-\-c, and d-\-f
composed entirely of additive terms ; the product may be
presented under the form (a+b+c) (d+j). It is now re-
quired to take the multiplicand as many times as there are
units in d and/*.

Multiplicand a + b-\-c

Multiplier d+f

taken d times ad-\-bd-\-cd

taken/ times + af-\-bf+cf

entire product .... ad-{-bd-\-cd-\-af-\-bf-\-cf

Therefore, in order to multiply together two polynomials
composed entirely of additive terms :

Multiply successively each term of the multiplicand by each
term of the multiplier, and add together all the products,

EXAMPLES.

1. Multiply 3a 2 -f- 4ab + b 2

by 2a + 56

6a 3 + 8a 2 b+2ab 2
The product, after reducing, + I5a 2 b+20ab 2 +5b 3

becomes .... 6a 3 +23a 2 b + 22ab 2 + 5b 3 .



Quest. — 36. How do you multiply two polynomials composed of
additive terms 1



32 FIRST LESSONS IN ALGEBRA.

(2) (3)

x 2 +y 2 x 5 + xy* +7 ax

x -f-y ax -\-5ax



x 3 +xy 2 ax*+ ax 2 y®-\-7a 2 x 2

-\-x 2 y-\-y 3 -\-5ax 6 -)-5ax 2 y 6 -\-35a 2 x*

x 3 -\-xy 2 -\-x 2 y-\-y 3 6ax 6 +6ax 2 y 6 -\-42a 2 x 2 .

4. Multiply x 2 -\-2ax-\-a 2 by x+a.

Ans. x 3 -\-3ax 2 +3a 2 x+a*

5. Multiply x 3 -\-y 3 by a?+y.

Ans. x*-\-xy 3 -\-x 3 y-\-y*

6. Multiply 3ab 2 +6a 2 c 2 by 3ab 2 +3a 2 c 2 .

Ans. 9a 2 b±+27a 3 b 2 c 2 +18a /L c*

7. Multiply a 2 b 2 +c 2 d by a+b.

Ans. a 3 b 2 + ac 2 d-\- a 2 b 3 -f- &c 2 df.

8. Multiply 3aa; 2 +9a6 3 +cd 5 by 6a 2 c 2 .

Ans. I8a 3 c 2 x 2 + 54a 3 c 2 b 3 +6a 2 c 3 d\

9. Multiply 64a 3 x 3 + 27a 2 x+9ab by 8<* 3 ctf.

<*«*. 512a G cdx 3 +216a 5 cdx + 72a*bcd.

10. Multiply a 2 -f-2aa:-}-;r 2 by a+x.

Ans. a 3 +3a 2 x-\-3ax 2 -{-x 3 .

11. Multiply a 3 +3a 2 #+3a# 2 -f ff 3 by a+«.

.A/is. a 4 +4a 3 x+6a 2 x 2 +4ax 3 +x 4 .

37. To explain the most general case, multiply a—b
by c—d.

The required product is equal to a—b

a—b taken as many times as there c — d

are units in c—d. If then we mul- ac—bc

tiply by c, which gives ac — bc, we —ad-\-bd

have got too much by a—b taken ac—bc — ad -j-bd.
d times ; that is, we have ad—db



MULTIPLICATION. S3

too much. Changing the signs, and subtracting this from
the first product (Art. 30), we have

(a — b) (c — d)=:ac — be — ad-\-bd.

Let us suppose a = 10, b = 6, c = 5, and d=zl; in which
case we find the product

(a — b) (c — d)=zac — be — ad-\-bd=16.

Hence, we have the following rule for the signs.

When two terms of the multiplicand and multiplier are
affected with the same sign, the corresponding product is affect-
ed with the sign -f- ; and when they are affected with contrary
signs, the product is affected with the sign — .

Therefore we say in algebraic language, that + multi-
plied by + ) or - multiplied by — , gives +; —multi-
plied by +? or + multiplied by — , gives — .

Hence, for the multiplication of polynomials we have the
following

RULE.

Multiply all the terms of the multiplicand by each term of
the multiplier, observing that like signs give plus in the pro-
duct, and unlike signs minus. Then reduce the polynomial
result to its simplest form.

EXAMPLES-

1. Multiply .... 2ax — 3ab

by 3x — b.

The product ..... 6ax 2 — 9abx
becomes after .... — 2abx-\-3ab 2

reducing 6ax 2 — llabx+3ab 2 .

Qoest. — 37. What does -f multiplied by + give? -f- multiplied
by — 1 — multiplied by + ? — multiplied by — ? Give the rule for
the multiplication of polynomials ?



34 FIRST LESSONS IN ALGEBRA.

2. Multiply a 4 — 2b* by a—b.

Arts. a 5 —2ab 3 —a 4 b+2b 4 ,

3. Multiply a: 2 — 3a?— 7 by a?— 2.

Ans. a? 3 — 5a? 2 — a? -f- 14.

4. Multiply 3a 2 — 5a5-f2& 2 by a 2 — lab.

Ans. 3a*—26a 3 b+37a 2 b 2 —Uab*.
6. Multiply b 2 +b*+b Q by b 2 — 1. jt»*. £ 8 — ft*.

6. Multiply a? 4 — 2a? 3 y-f4a? 2 y 2 — 8a?y 3 -f-16y 4 by x+2y.

Ans. a? 5 + 32y 5 .

7. Multiply 4x 2 —2y by 2y. ^4»s. Sx 2 y — Ay*.

8. Multiply 2a?+4y by 2a?— Ay. Ans. 4a? 2 — 16y 2 .

9. Multiply x 3 +x 2 y+xy 2 -\-y 3 by a?— y.

Ans. x 4 — y 4 .

10. Multiply a? 2 +a?y+y 2 by a? 2 — xy+y 2 .

Ans. a? 4 + x 2 y 2 -\- y* \

11. Multiply 2a 2 — 3aa;-f 4a? 2 by 5a 2 — 6ax — 2a? 2 .

Ans. 10a 4 — 27a 3 a?+34a 2 a? 2 — 18aa? 3 — 8a? 4 .

12. Multiply 3a? 2 — 2a?y+5 by x 2 +2xy—3.

Ans. 3a? 4 +4a? 3 y — 4a? 2 — 4a? 2 y 2 +16a?y — 15.

13. Multiply 3a? 3 +2a? 2 y 2 + 3y 2 by 2a? 3 — 3a? 2 y 2 + 5y 3 .



A i 6a? 6 — 5a? 5 y 2 — 6ary+6a? 3 y 2 -f-
r\ t I5x 3 f— 9a? 2 y 4 +10a?V-M5?/



15x 3 f— 9a? 2 y 4 + lOa^y 5 -}-^.

14. Multiply 8aa: — 6ab—c by 2ax-\-ab+c.

Ans. 1 6a 2 x 2 —4a 2 bx — 6a 2 b 2 + 6acx — 7a5c — c 2 .

15. Multiply 3a 2 — 5£ 2 +3c 2 by a 2 — b 2 .

Ans. 3a 4 -8a 2 6 2 +3a 2 c 2 -f 56 4 -35 2 c 2 .

16. 3a 2 — 5bd+ cf

— _ _5a 2 +4^— 8c/:

Pro. red. — 1 5a 4 + 37a 2 bd—29a 2 cf—20b 2 d 2 +44bcdf— 8c 2 / 2 .



MULTIPLICATION. 35

8

38. To finish with what has reference to algebraic mul-
tiplication, we will make known a few results of frequent
use in Algebra.

Let it be required to form the square or second power of
the binomial (a-{-b). We have, from known principles,

(a+b) 2 =(a+b) (a+b)=za 2 +2ab+b 2 .

That is, the square of the sum of two quantities is equal to
the square of the first, plus twice the product of the first by the
second, plus the square of the second.

1. Form the square of 2a-\-3b. We have from the rule

(2a + 3b) 2 = 4a 2 -f VZab -f 9b 2 .

2. (5ab + 3ac) 2 =z25a 2 b 2 + 30a 2 bc+ 9a 2 c 2 .

3. (5a 2 +8a 2 b) 2 =25a 4 + 80a*b + 64aW.

4. (6ax+9a 2 x 2 ) 2 = 36a 2 x 2 +108a 3 x 3 +QlaW.

39. To form the square of a difference a—b, we have

(a-b) 2 =z(a-b) (a-b)=:a 2 -2ab+b 2 .

That is, the square of the difference between two quantities is
equal to the square of the first, minus twice the product of the
first by the second, plus the square of the second.
1 Form the square of 2a— b. We have

(2a-b) 2 =z4a 2 -4ab+b 2 .

2. Form the square of 4ac— be. We have

(4ac - be) 2 ss 1 6a 2 c 2 - Babe 2 + b 2 c 2 .

3. Form the square of 7a 2 b 2 — I2ab 3 . We have

(7a 2 b 2 —l2ab 3 ) 2 =z49a*b i -l68a 3 b 5 +U4a 2 b 6 .

Quest. — 38. What is the square of the sura of twd quantities equal to ?
39. What is the square of the difference of two quantities equal to 1



36 FIRST LESSONS IN ALGEBRA,

40. Let it be required to multiply a+h by a— b. We
have

(a+b)x(a-b)=:a 2 -b 2 .

Hence, the sum of two quantities, multiplied by their differ-
ence, is equal to the difference of their squares.

1. Multiply 2c+b by 2c— b. We have

(2c+&)X(2c — 5) = 4c 2 — b 2 .

2. Multiply 9ac-f 3bc by 9ac— 3bc. We have

(9ac+ 3bc)(9ac — 3bc) = 8 1 a 2 c 2 — 9b 2 c 2 .

3. Multiply 8a 3 +7ab 2 by 8a ? — 7«5 2 . We have

(8a 3 +7ab 2 )(8a 3 — 7ab 2 ) = 64a 6 — 49a 2 bK

41. It is sometimes convenient to find the factors of a
polynomial, or to resolve a polynomial into its factors.
Thus, if we have the polynomial

ac-\-ab-\-ad,

we see that a is a common factor to each of the terms :
hence, it may be placed under the form

a(c+b+d).

1. Find the factors of the polynomial a 2 b 2 -\-a 2 d-\-a 2 f

Ans. a 2 (b 2 +d+f).

2. Find the factors of 3a 2 b+6a 2 b 2 +b 2 d.

Ans. b(3a 2 +6a 2 b + bd).

3. Find the factors of 3a 2 b + 9a 2 c-{-l8a 2 xi/.

Ans. 3a 2 (b+3c + 6xy),

Quest. — 40. What is the sum of two quantities multiplied by their
difference equal to 1



DIVISION. 37

4. Find the factors of 8a 2 cx— I8acx 2 +2ac 5 y— 30a 6 c 9 a?.

Arts. 2ac(4ax—9x 2 +c*y— 15a 5 c 8 #).

5. Find the factors of a 2 +2ab+b 2 .

Ans. (a+b)x{a+b).

6. Find the factors of a 2 — b 2 . Ans. (a+b)x(a—b).

7. Find the factors of a 2 —2ab+b 2 .

Ans. (a— 6)x(a— £).



DIVISION.

42. Algebraic division has the same object as arithmeti-
cal, viz : having given a product, and one of its factors, to
find the other factor.

We will first consider the case of two monomials.

The division of 72a 5 by 8a 3 is indicated thus :

72a 5

8a 3 '

It is required to find a third monomial, which, multiplied

by the second, will produce the first. It is plain that the

third monomial is 9a 2 ; for by the rules of multiplication

8a 3 x9a 2 =72a 5 .

7 2a 5 „ , .
Hence, we have =9a 2 ,

a result which is obtained by dividing the coefficient of the
dividend by the coefficient of the divisor, and subtracting the
exponents of the like letter.

Quest. — 42. What is the object of division ill Algebra! Give the
rule for dividing monomials 1

4



38 FIRST LESSONS IN ALGEBRA.

A1SO, i^!L == 5fl3-lft2-l c = 5fl 2 iCf

for, 7 a b X 5a 2 bc = 35a 3 b 2 c.

i . 56a*b*c* m z

Hence, for the division of monomials we have the following

RULE.

I. Divide the coefficient of the dividend by the coefficient of
the divisor.

II. Write in the quotient, after the coefficient, all the letters
common to the dividend and divisor, and affect each with an
exponent equal to the excess of its exponent in the dividend
over that in the divisor.

III. Annex to these, those letters of the dividend, with their
respective exponents, which are not found in the divisor.

From these rules we find

48a 3 b 3 c 2 d A ' _ 150a 5 b 8 cd 3 m M ,
note =***** 30a 3 b*d 2 = 5abcd '

1. Divide I6x 2 by 8x. Ans. 2x.

2. Divide I5axy 3 by 3ay. Ans. 5xy 2 .

3. Divide 84ab 3 x by 12b 2 . Ans. 7abx.

4. Divide 36a*b 5 c 2 by 9a 3 b 2 c. Ans. 4ab 3 c.

5. Divide SSa 3 b 2 c by Sa 2 b. Ans. 11 abc.

6. Divide 99a±b*x 5 by lla 3 b 2 x*. Ans. 9ab 2 x.

7. Divide I08x e y 5 z 3 by 54x 5 z. Ans. 2xy 5 z 2 .

8. Divide 64x' r y 5 z G by 16# 6 y% 5 . Ans. 4xyz.

9. Divide 96a 7 5 6 c 5 by I2a 2 bc. Ans. 8a 5 b 5 c*.

10. Divide 54a 1 c 5 d G by 27 acd. Ans. 2a 6 c 4 d 5 .

11. Divide 38a 4 ZW by 2a 3 b 5 d. Ans. I9abd 3 .



DIVISION.



39



12. Divide 42a 2 b 2 c 2 by 7abc. Ans. 6abc.

13. Divide 64a 5 ¥c 8 by 32a*bc. Ans. 2ab 3 c\

14. Divide 128a 5 x 6 i/ by I6axy*. Ans. 8a^x 5 y 3 .

15. Divide 132bd 5 f Q by 2d±f. Ans. 66bdf 5 .

16. Divide 256a 4 6 9 c 8 d 7 by 16<z 3 6c 6 . Ans. \Gab*c 2 d?.

17. Divide 200a 8 m 2 7i 2 by 50a 7 mn. .Ans. 4amn,

18. Divide 300o: 3 y^ 2 by 60a?y 2 ^. Aws. 5x 2 y 2 z.

19. Divide 27 a 5 b 2 c 2 by 9dfe. ^.^. 3a 4 5c.

20. Divide 64a 3 y 6 z s by 32<zy 5 ;z 7 . Ans. 2a 2 yz.

21. Divide 88a 5 6 6 c 8 by lla 3 ^ 6 . Ans. 8a 2 b 2 c 2 .

43. It follows from the preceding rule, that the division
of monomials will be impossible,

1st. When the coefficients are not divisible by each other.

2nd. When the exponent of the same letter is greater
in the divisor than in the dividend.

3rd. When the divisor contains one or more letters which
are not found in the dividend.

When either of these three cases occurs, the quotient re-
mains under the form of a monomial fraction ; that is, a
monomial expression, necessarily containing the algebraic


2 4 5 6 7 8 9 10 11 12

Online LibraryCharles DaviesFirst lessons in algebra, embracing the elements of the science → online text (page 2 of 12)