Number systems - II
Prime Factors
A composite
number can be uniquely expressed as a product of prime factors. For example,
12=2x6=2x2x3=22x31. 20=4x5=2x2x5=22x51.
124=2x62=2x2x31=22x31 etc.
Every
composite number can be expressed in a similar manner in terms of its prime
factors.
Number of Factors
The number of factors of a
given composite number N (including 1 and the number itself) which can be resolved into its prime
factors as
N = am x
bn x CP...
where a, b, c
are prime numbers, are
(1+m)(1+n)(1+p)...
El. Find the total number of
factors of 240.
Sol. 240=2 x 2 x 2 x 2 x 3 x 5=
24 x 31 x 51
Comparing with the standard
format for the number N, we obtain
a=2,b=3,c=5,m=4,n=1,p=1.
.•. the total number of factors
of this number including 1 and itself are =(1+m)(1+n)(1+p).... =(1+4)x(1+1)x(1+1)=5
x 2 x 2 = 20
Factors
of 240 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120,
240 (Total 20 in number)
Sum of Factors
The sum of factors of the number N (as
defined above) is given by the formula
(a - 1)(b - 1)(c
—1)...
where a, b, c....., m, n, p... retain the same
meaning.
Sol. The sum
by the above formula
(25-1)(32 —1)52
—1) = 31x8x24 =744
(2-1) (3-1) (5-1) 1x2x4
We can see that: 1+2+3+4+5+6+8+10+12+15 +16+20+24+30+40+48+60+80+120+240=744.
Ø
The number of ways in
which a composite number N may be resolved into two factors
=1/2 (p + 1) (q + 1) (r + 1).. if N = ap
bq Cr is not a perfect square and
=1/ 2 [(p + 1) (q + 1) (r + 1) +
1] if N is a perfect square.
Ø The number of
ways in which a composite number can be resolved
into two factors which are prime to each other. If N = ap bq Cr ... then the
number of ways of resolving N into two factors prime to each other is
= 1/2 (1 + 1) (1 + 1) (1 + 1)... = 2n-1 where n is
the number of different prime factors of N.
Ø If P is a prime number,
the coefficient of every term in the expansion of (a + b)P except the first and the last is
divisible by P.
E3. In
how many ways can the number 7056 be resolved into
two
factors?
Sol.
N=7056=32x24x72=3Px2gx7r
Note: N is
perfect square. Number of ways in which it can be resolved into two factors =
1/2 {(p + 1) (q + 1) (r + 1) + 1}
=1/2 {(2+1)(4+1)(2+1)+1}
= 1/2x46=23
E4. Find
the number of ways in which N = 2778300 can be resolved into the factors prime to each other.
Sol. N=22x34x52x73
The
required number is the same as the number of ways of resolving 2 x 3 x 5 x
7 into two factors which is equal to 1/2(1+1)(1+1)(1+1)(1+1)=23=8.
HCF of Numbers
It is the highest common factor of two or more
given numbers. It is also called GCF (greatest common factor).
For
example HCF of 10 and 15 = 5, HCF of 55 and 200 = 5, HCF of 64 and 36 =
4 etc.
Factorization Method to find HCF
To find
the HCF of given numbers, first resolve the numbers into their prime factors. After expressing the numbers
in terms of the prime factors, the HCF is the product of common factors.
E5. Find
the HCF of 88, 24 and 124.
Sol. 88=2x44=2x2x22=2x 2x 2x11=23x111
24=2x12=2x2x6=2x2x2x3=23x31.
124=2x62=2x 2x
31=22x311.
=
HCF = 22 = 4.
E6. Find the HCF of 72,
60 and 96.
Sol.72=2x2x2x3x3=23x32.
60 = 2x2x3x5 = 22x31x51.
96=2x2x2x2x2x3=25x31.
HCF =
product of the highest number of common factors = 2 x 2 x 3= 12.
Division method to find H.C.F
By division method, we
start with the two numbers and proceed as shown below, till the remainder
becomes zero.
E7. Find the HCF of 12 and
48.
Sol. 12) 48 (4
48
00
HCF = 12.
Here,
when the remainder is not zero, divide the previous divisor with
that remainder and proceed in the same way until you get the remainder as zero. The
last divisor is the required HCF.
E8.
Find the HCF of 10 and 25. `
Sol.
10) 25 (2
20
5 ) 10 (2
10
00
HCF = 5.
If
there are more than two numbers, we will repeat the whole process with the HCF
obtained from two numbers as the divisor and so on. The last divisor will then be the
required HCF of the numbers
E9 . Find the HCF of 10,
25 and 30.
Sol. We can find the HCF of 10 and 25
i.e. 5. Now we have to find the HCF of 5 and
30 which is 5. So, the HCF of 10, 25 and 30 is 5.
Imp.
Ø If we have to find the
greatest number that will exactly divide p, q and r, then required number = HCF of p, q and r.
E10. Find the greatest number that will exactly divide
65, 52 and 78.
Sol.
Required number = HCF of 65, 52 and 78 = 13.
Ø If we
have to find the greatest number that will divide p, q and r
leaving remainders a, b and c respectively, then the required number = HCF of (p
— a), (q — b) and (r — c).
Ex11. Find the greatest number that will divide 65, 52
and 78 leaving
remainders 5,.2 and 8 respectively.
'Sol. Required
number = HCF of (65 — 5), (52 — 2) and (78 — 8) = HCF of 60, 30 and 70 = 10.
Ø
If we have tc find the greatest number that will divide p,
q and r
leaving the same remainder in each case, then required number = HCF of the
absolute values of (p — q), (q — r) and (r — p).
E12.
Find the greatest number that will divide 65, 81 and 145 leaving the same remainder in each
case.
Sol. Required number
= HCF of (81 —
65), (145 — 81) and (145-65) = HCF of 16, 64 and 80 = 16.
E13.
How many numbers below 90 and other than unity exist, such that the HCF
of that number and 90 is unity?
Sol. 90= 32 x 2 x 5
Number of multiples of 2 = 45
Number of multiples of 3 = 30
Number of multiples of 5 = 18
Number of multiples of 2 & 5 = 9
Number of multiples of 2 & 3 =15
Number
of multiples of 3 & 5 = 6
Number of multiples of 2,
3 & 5 = 3
Total=(24+6+
12)+(12+3+6+3)=42+24=66 24
numbers (including unity) compared with 90 have only'1'
as common factor. Hence
required result is 24 — 1 = 23.
E14.A riot
hit state was deployed with 3 different regiments of black commandos, each having 115, 161 and 253
commandos respectively. Each regiment has commandos domicile of one particular state only and each regiment came from
different states. The Commander-in-Charge further plans to split the regiments into smaller groups but in such a way
that all groups have same number of
commandos and each group has
commandos belonging to a particular state only. What is the minimum
number of groups that can be formed?
Sol.
115
= 23 x 5; 161 = 23 x 7; 233 = 23 x 11..•. HCF = 23 Hence minimum number of groups that
can be formed = 23
LCM of Numbers
Lowest common multiple of two
or more numbers is the smallest number which is exactly divisible by
all of them. e.g. LCM of 5,
7, 10 = 70,
LCM of 2, 4, 5 = 20,
LCM of 11, 10, 3 = 330.
Factorization Method to find LCM
To find the LCM of the given numbers, first resolve all
the numbers into their
prime factors and then the LCM is the product of highest powers of all the prime factors.
E15. Find the LCM of 40, 120, 380.
Sol.
40=4x
10=2x2x2x5=23x51;
120=4x
30=2 x 2 x 2x 5 x 3=23x 51 x 31; 380=2x190=2x2x95=2x2x5x19=22x51x191;
. Required LCM =23x51x31x 191 = 2280.
Division method to find LCM
Write
the given numbers separately. Then divide by 2 and write the result below
the numbers divisible by 2. If it is not divisible by 2 then try with
3, 5, 7.... etc. Leave the others (those not divisible) untouched. Do the same
for all steps till you get 1 as the remainder in each column.
E16. Find the LCM of 6, 10, 15, 24, 39.
Sol.
2 16 10 15 24 39
|
2
|
35
|
15
|
12
|
39
|
|
2
|
35
|
15
|
6
|
39
|
|
3
|
35
|
15
|
3
|
39
|
|
5
|
15
|
5
|
1
|
13
|
|
13
|
11
|
1
|
1
|
13
|
|
|
11
|
1
|
1
|
1
|
LCM=2x2x2x3x5x13=1560.
Imp.
Ø If we
have to find the least number which is exactly divisible by p, q and r, then the
required number = LCM of p ,q and r.
E17. Find the least number that is
exactly divisible by 6, 5 and 7.
Sol. Required number = LCM of 6, 5 and 7 =
210.
Ø If we
have to find the least number which when divided by p, q
and r leaves the remainders a, b and c respectively, then if it is observed that(p — a) = (q — b) = (r — c) = K (say), then
the required number
= (LCM of p, q and r) — (K).
E18. Find the least number which when divided by 6, 7 and
9 leaves the remainders
1, 2 and 4 respectively.
Sol. Here, (6 - 1) =
(7— 2) = (9-4) = 5. Required
number = (LCM of 6, 7 and 9) — 5 =126-5=121.
Ø If we
have to find the least number which when divided by p, q
and r leaves the same remainder 'a' each case, then required number = (LCM of
p, q and r) + a.
E19. Find the least number which when divided by 15, 20
and 30 leaves the
remainder 5 in each case.
Sol. Required number
= (LCM of 15, 20 and 30) + 5 =60+5= 65.
Ø LCM x HCF = Product of two
numbers. (Valid only for "two")
E 20.
Find the LCM of 25 and 35 if their HCF is 5.
Sol. LCM = Product of the numbers = 25 x
35 = 175
HCF 5
E21. By using the rule
that LCM = Product of two numbers + HCF, find LCM of 26 and 442.
Sol. HCF
of the two can be found as 26 = 13 x 2,
442 = 2 x 17 x 13
HCF =
26 LCM = 26 x 442/26 = 442.
HCF & LCM
of Decimals
E22.
Calculate the HCF and LCM of 0.6, 0.9, 1.5,
1.2 and 3.
Sol. The numbers can be written as
0.6,
0.9, 1.5, 1.2, 3.0
Consider them as 6, 9, 15, 12, 30→ HCF = 3 Required
HCF = 0.3 and LCM = 180. Required LCM = 18.0.
Imp.
If the first number in the
above example had been 0.61, then the equivalent integers would have been 61,
90, 150, 120 and 300 etc.
HCF & LCM of fractions
HCF of fractions = HCF of
numerators÷LCM of denominators
LCM of fractions = LCM of numerators÷HCF of denominators
E23.
Find the HCF and LCM of: 5/16 , 3 /4and 7/15
,
Sol.
HCF = (HCF of 5, 3, 7) = 1
(LCM of 16, 4,15) 240
LCM =(LCM of 5, 3, 7) =
105
(HCF of 16, 4,15) 1
Surds
Surds are irrational roots
of a rational number.
For example, √6 is a surd, since its
exact value can't be found .Similarly π, e, √7, √8, 3√9
, 4 √27 etc. are all surds.
Types of Surds
Pure Surds: The surds which are made up of only
an irrational number are known as pure surds. For example √6, √7, √8 etc.
E24. Convert
427 to a mixed surd.
Sol. 27 = 9x3 =3√3 .
E25.
Convert 2√8 to a pure surd.
Sol. 2√8 = √22 x 8 = √8 x
4 = √32
Rationalization
of Surds
In order to
rationalize a given surd, we multiply and divide it by the conjugate of denominator and then simplify
[conjugate of (a + √b) is (a - √b) and vice versa].
E26.
Rationalize 6 + √2
1 - √3
Sol. 6 + √2 = (6
+ √2) (1 +√ 3) = (6 + 6√3 +√2 +√ 6) = (
6 + 6√3 +√ 2 +√6)
1 –
√3 (1 – √3) (1 +√ 3) 1 – 3 -2
In order to rationalize
= X
√ a
+ √b +√ c
ü
Multiply and divide by √a +
√b - √c
ü Multiply and divide by (a + b - c) - 2√ab.
Variables
and Constants
A variable is a term which can assume any
value under given conditions. A constant is a term whose value does not change under
any conditions. The variables are generally represented by x, y, z etc. and the
constants are generally represented by a, b, c etc. `
Algebraic
Expressions
An algebraic expression is a combination of
variable and constant terms.
Thus 3x2 – 5xy + 2y4 , 2a3b5
, 5xy + 3z are example of algebraic
expression.
2a3 – c2
Terms
A term consists of products and quotients of
variables and constants. Thus 6x2y3, 5x/3y4,
—3x7 are terms. However,
6x2 + 7xy is an algebraic expression consisting of two terms.
Types of Terms
On the basis of the number of terms,
algebraic expressions are classified as follows.
Monomial: A monomial is an
algebraic expression consisting of only one term. Thus 7x3y4,
3xyz2, 4x2y are all monomials. Because of this
definition, monomials are sometimes simply called terms.
Binomial: A binomial is an
algebraic expression consisting of two terms. Thus 2x + 4y, 3x4 —
4xyz3 are examples of binomials.
Trinomial: A trinomial is an
algebraic expression consisting of three terms. For example, (5x2 —
3x +y), (3x3 — 5x2 + 2) etc.
Multinomial: A multinomial is an
algebraic expression consisting of more than one term. Thus 7x + 6y, 3x3
+ 6x2y — 7xy + 6, 7x + 5x2/y — 3x3/16 are all
multinomials.
Polynomial: A polynomial is a
multinomial that can only have rational coefficients and cannot have any
variable in the denominator.
Degree
The
degree of a monomial is the sum of all the exponents in the variables in the
term. Thus the degree of 4x3y2z is 3 + 2 + 1 = 6. The
degree of a constant, such as 6, 0, —'3, or π, is zero. The degree of a polynomial is the
same as that of the term having highest
degree and non zero coefficient. Thus 7x3y2 — 4xz5
+ 2x3y has terms of degree 5, 6 and 4 respectively; hence its degree
is 6.
Computation with Algebraic
Expressions
Addition of algebraic expressions
Addition of algebraic expressions is
achieved by combining like terms. In order
to accomplish this, the expressions may be arranged in rows with like
terms in the same column; which are then added.
E28. Add 7x + 3y3
— 4xy, 3x — 2y3 + 7xy and 2xy — 5x — 6y3.
Sol. 7x 3y3 —4xy
3x —2y3
7xy
—5x —6y3 2xy
Sum
= 5x —5y3 5xy
Hence the sum is 5x —
5y3 + 5xy.
Subtraction of algebraic
expressions
Subtraction of two
algebraic expressions is achieved by changing the sign of every term in the
expression which is being subtracted and adding this result to the other
expression.
E29 Subtract
2x2 — 3xy + 5y2 from 10x2 — 2xy — 3y2.
Sol. 10x2 —2xy —3y2
2x2 —3xy 5y2
Difference = 8x2
+ xy — 8y2
We may also write the
same as
(10x2
— 2xy — 3y2) — (2x2 — 3xy + 5y2)
= 10x2 —
2xy — 3y2 — 2x2 + 3xy — 5y2 = 8x2 +
xy — 8y2.
Multiplication of algebraic
expressions
Multiplication of
algebraic expressions is achieved by multiplying all the terms in the factors of the
expressions.
Ø To multiply two or more
monomials: Use the laws of exponents, the rules of signs and the commutative and associative properties
of multiplication.
E30. Multiply-3xzy3z,
2x4y and —4xy4z2.
Sol. (-3x2y3z)
(2x4y) (-4xy4z2).
Arranging according to commutative and associative laws.
{(-3) (2) (-4)) {(x2)(x4)(x)}
{(y3)(y)(y4)) {(z)(z2)}.
Combine
using rules of signs and laws of exponents to obtain
24x7 y8 z3.
Ø To multiply a
polynomial by a monomial: Multiply each term of the polynomial by the monomial
and combine results.
E31. Multiply
3xy — 4x3 + 2xy2 and 5x2 y4.
Sol. (5x2y4)(3xy —
4x3 + 2xy2)
=
(5x2 y4) (3xy) + (5x2 y4)
(-4x3) + (5x2 y4) (2xy2)
=
15x3 y5 — 20x5 y4 + 10x3 y6.
Ø
To multiply a polynomial by a polynomial, we multiply each term of
the two polynomials and combine the results. It is often useful to arrange the polynomials according to ascending (or
descending) powers of one of the variables involved.
E32 .Multiply —3x + 9 + x2 and 3 —
x.
Sol. Arranging in descending powers of x,
x2 —3x +9 ....(1)
—x +3 ....(2)
Multiplying (1) by —x, —x3 + 3x2
— 9x
Multiplying (1) by 3, 3x2 —9x + 27
Multiplying (1) by 3, 3x2 —9x + 27
Adding,
we get: —x3 + 6x2 — 18x + 27.
Division
of algebraic expressions is achieved by using the division laws of exponents.
Ø To
divide a monomial by a monomial, we find the quotient of the
numerical coefficients, the quotients of the variables and multiply these
quotients.
To
divide a polynomial by a polynomial
(a)
Arrange the
terms of both polynomials in descending (or ascending) powers of one of the variables common to both polynomials.
(b)
Divide the
first term in the dividend by the first term in the
divisor. This gives the first term of the quotient.
(c)
Multiply the
first term of the quotient by the divisor and subtract from the dividend, thus obtaining a new dividend.
(d) Use the dividend obtained in (c) to repeat
step (b) and (c) until a remainder is
obtained which is either of zero degree or lower than the degree of the
divisor.
(e)
The result can be
written as
Dividend = quotient + remainder
Divisor divisor
E34. Divide x2 + 2x4 - 3x3 + x - 2 by x2-3x+2.
Factorization of Algebraic Expressions
The factors of a given algebraic expression consist of two
or more
algebraic expressions which when multiplied together produce the given expression.
Different Types with Examples
The factorization of algebraic expressions can be done in
many ways.
It depends on the terms contained in the expression. Based on this,
following are the standard types of factorization. You are expected to understand each of
them thoroughly.
Ø Common Monomial Factor
ac + ad = a(c + d)
E35.
Factorize 3x2 + 6x3 + 12x4
Sol. 3x2 + 6x3 + 12x4 = 3x2(1 + 2x + 4x2)
E36.
Factorize 9s3 t + 15s2 t3 – 3s2 t2..
Sol. 9s3 t + 15s2 t3 - 3s2 t2 = 3s2 t(3s + 5t2 - t).
Ø Difference of two Squares
a2 - b2 = (a + b)(a - b)
E37. Factorize 1 - x8.
Sol.
Since (1 + x4)(1 - x4) = (1 + x4)(1 + x2)(1 - x2)
So 1 - x8 = (1 + x4)(1 + x2)(1 + x)(1 - x)
E38.
Factorize 3x2 - 12.
Sol. 3x2- 12
= 3(x2-4) = 3(x + 2)(x - 2).
Ø Perfect Square Trinomials
a2 + 2ab + b2 = (a + b)2
a2 - 2ab +
b2 = (a - b)2
E39.
Factorize
9x4 - 24x2y + 16y2.
Sol. 9x4 - 24x2y + 16y2 = (3x2 - 4y)2
E40. Factorize 4m6 n6 + 32m4 n4 + 64m2 n2.
Sol. 4m6 n6 + 32m4 n4 + 64m2 n2
= 4m2 n2
(m4 n4 + 8m2 n2
+ 16) = 4m2 n2
(m2 n2 + 4)2.
Ø
Other Trinomials
X2 + (a+b)x + ab = (x+a) (x+b)
acx2 + (ad + bc)x + bd = (ax
+ b)(cx + d)
E41. Factorize x2 - 7xy + 12y2.
Sol. x2 - 7xy + 12y2 = x2
- 3xy - 4xy + 12y2
= (x - 3y)(x - 4y).
E42. Factorize y4 + 7y2 +12
Sol. y4 + 7y2 + 12 = y4
+ 4y2 + 3y2 + 12 = (y2 +4)(y2+3).
E43.
Factorize s2t2 - 2st3 -
63t4.
Sol. s2
t2 - 2st3 - 63t4
=
t2(s2 - 2st - 63t2) = t2(s - 9t)(s
+ 7t).
E44.
Factorize each of the given algebraic
expressions.
(a) x2
- 7x + 6 (b) x2 + 8x
(c) 6x2 - 7x - 5 (d) x2
+ 2xy - 8y2
Sol. (a) x2-7x+ 6 = (x - 1)(x - 6)
(b)
x2 + 8x =
x(x + 8)
(c)
6x2 - 7x - 5 = (3x - 5)(2x
+ 1)
(d)
x2 + 2xy -
8y2 = (x + 4y)(x - 2y).
Sum or the Difference of
two Cubes
a3 + b3
= (a + b)(a2
- ab + b2) a3 - b3 = (a - b)(a2
+ ab + b2)
E45. Factorize a9 + b9.
Sol. a9
+ b9
= (a3)3 + (b3)3 =
(a3 + b3)[(a3)2 - a3
b3 + (b3)2]
=
(a + b)(a2 - ab + b2)(a6 - a3 b3
+ b6)
E46.
Factorize
x6 - 7x3 - 8.
Sol. X6-7X3-8 = X6-8x3+X3-8
= (x3 - 8)(x3 + 1)
= (x3 - 23)(X3 + 1)
=
(x - 2)(x2 + 2x + 4)(x + 1)(x2 - x + 1)
Ø Grouping of
Terms
ac+bc+ad+bd=c(a+b)+d(a+b)=(a+b)(c+d)
E47. Factorize x3 + x2y + xy2
+ y3
Sol. x3 + x2y + xy2 + y3
= x2 (x + y) + y2(x + y)
= (x + y)(x2
+ y2)
E48.
Factorize
x3 y3 - y3 + 8x3 - 8.
Sol. X3 y3-y3+8X3-8
= y3(x3 - 1) + 8(X3
- 1)
= (X3 - 1)(y3
+ 8)
= (x - 1)(x2 + x + 1)(y +
2)(y2 - 2y + 4)
Factors of an
± bn
an + bn has a + b as a factor if and only if n is a positive odd integer.
an + bn
= (a + b)(an-1 - a n-2
b + an-3 b2 - ... - abn-2 + bn-1).
E49. Factorize
x3 + 8y6.
Sol. X3 + 8y6
= x3 + (2y2)3 = (x + 2y2)
[x2 - x(2y2) + (2y2)2]
= (x + 2y2) (x2 - 2xy2
+ 4y4)
E50. Factorize
z5 + 32.
Sol. z5+32 = z5+25
= (z + 2)(z4 - 2z3
+ 22 z2 - 23
z + 24) =
(z + 2)(z4 - 2z3 + 4z2 - 8z + 16)
an - bn has a - b as a factor if and
only if n is a positive integer.
an - bn =(a -
b)(an-1+ an-2b + an-3b2 + ... + abn-2+
bn-1)
E51. Factorize
27x3 - y3.
Sol.
27X3
- y3
= (3x)3 - y3
= (3x - y)[(3x)2 + (3x)y + y2]
= (3x - y)(9x2 + 3xy + y2).
E52. Factorize
y7 - z7.
Sol. y7 - z7
= (y - z)(y6 + y5 z + y4 z2 + y3 z3 + y2 z4
+ yz5 + z6).
HCF of Polynomials
The HCF of two or
more given polynomials is the polynomial of highest
degree and largest numerical coefficients which is a factor of all the
given polynomials.
Method
to find HCF of polynomial
The
following method is suggested for finding the HCF of several polynomials
(a) Write each polynomial as a product of prime factors.
(b) The HCF
is the product obtained by taking each factor to the lowest power to which
it occurs in both the polynomials. For example,
The HCF of 23 32
(x - y)3(x + 2y)2;
2233
(x - y)2 (x + 2y)3
and
32(x
- y)2 (x + 2y) is 32(x - y)2 (x + 2y).
Imp.
Two or
more polynomials are relatively prime if their HCF is 1.
LCM of Polynomials
The LCM of two or
more given polynomials is the polynomial of lowest degree and smallest numerical coefficients for
which each of the given polynomials
will be its factor.
Method to find LCM of polynomial
The following
procedure is suggested for determining the LCM of several
polynomials.
(a)
Write
each polynomial as a product of prime factors.
(b)
The
LCM is the product obtained by taking each factor to the highest power to which it occurs. For example, The LCM of 2332 (x — y)3(x
+ 2y)2;
2233(x
— y)2 (x + 2y)3; 32(x — y)2(x + 2y)
is
2333 (X — y)3
(x + 2y)3.
E53. Find the HCF and LCM of (I) 9X4y2
and 12X3y3
(ii) 6x —
6y and 4x2 — 4y2.
Sol. (i) 9x4y2
= 32 x x4 x y2 and
12X3y3 = 22 x
3 x X3 x
y3.
HCF = 3x3y2 and
LCM = 32 x 22 X x4 X
y3 =36X4y3
So, HCF = 3x3y2, LCM = 36x4y3
(ii) 6x —
6y = 2 x 3 x (x — y) and
4x2 — 4y2 = 22(x2— y2)
= 22 x (x + y) (x — y) HCF = 2(x — y) and
LCM = 22 x 3
x (x — y)(x + y) = 12(x — y) (x + y). So,
HCF = 2(x — y), LCM = 12(x + y) (x — y).
Special Products
The
following are some of the products which occur frequently in mathematics and the student should become familiar
with them as soon as possible.
Product of a monomial and a binomial a (c + d) = ac + ad.
Product of the sum and the
difference of two terms (a + b)(a — b) = a2
— b2.
Square of a
binomial
(a + b)2 = a2 + 2ab + b2.
(a — b)2 = a2
— 2ab + b2
Product of two binomials
(x + a)(x + b) = x2 + (a +
b)x + ab.
(ax + b)(cx + d) = acx2
+ (ad + bc)x + bd. ,(a+b)(c+d)=ac+bc+ad+bd.
Cube of a binomial
(a + b)3 = a3
+ 3a2b + 3ab2 + b3.
(a — b)3 = a3
— 3a2b + 3ab2 — b3.
Square of a trinomial
(a + b + c)2 = a2
+ b2 + c2 + 2(ab + ac + bc).
Products
that give answers of the form a n ± bn.
It may be verified by
multiplication that (a — b)(a2
+ ab + b2) = a3 — b3
(a — b)(a3 + a2b + ab2
+ b3) = a4 — b4
(a — b)(a4 + a3b + a2b2
+ ab3 + b4) = a5 — b5
(a — b) (a 5 + a 4 b + a 3 b 2 + a 2 b 3
+ a b 4 + b 5 ) = a 6 — b 6 and so on These may
be summarized as
(a — b)(an-1 + an-2b
+ an-3 b2 + ... + ab2 + bn-1)
= an —
bn
where
n is any positive integer (1, 2, 3, 4, ...).
Similarly, it may be
verified that (a + b)(a2 - ab
+ b2) = a3 + b3
(a + b)(a4 — a3b + a2b2
— ab3 + b4) = a5 + bs
(a + b)(a6 — a5b
+ a4b2 — a3 b3 + a2 b4
— abs + b6) =
a7 + b7 and so on.
These may
be summarized as
(a + b)(an-1-
an-2b + an-3 b2 — ... — abn-2 + bn-1)
= an + bn Where n is any positive odd integer
(1, 3, 5, 7, ...).
E54. Simplify
(3xy + 1)(2x2 — 3y).
Sol. (3xy)(2x2) +
(3xy)(-3y) + (1)(2x2) + (1)(-3y) =
6x3y — 9xy2 + 2x2 — 3y.
E55.Simplify (x + y + z + 1)2.
Sol. [(x + y) + (z + 1)]2
=
(x + y)2 + 2(x + y)(z + 1) + (z + 1)2
= x2 + 2xy + y2 + 2x2 + 2x +
2yz + 2y + z2 + 2z + 1.
Binary Operations
Operations other
than the four fundamental operations are caller binary operations. i.e. such operations don't precisely
exist it mathematics but
following examples will clarify this.
E56.If A * B = A + B + A x B — A2
x B2, find the value of 2*3.
Sol. By
definition 2*3 = 2 + 3 + 2 x 3 — 22 x 32 = —25
E57.If A ¥ B = HCF of A
and B,
A # B =
LCM of A and B
A $ B =
Quotient when A is divided by B, then find the value of[10#{(4Y12)¥(7#11)}]$6.
Sol. HCF of 4 and 12 = 4 LCM of 7 and 11 = 77
HCF of 4 and 77 = 1 LCM of 10 and 1 = 10
The quotient when 10 is divided by 6 = 1 and the remainde
is 4.
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