2.4 Multiplication of algebraic expressions:
Like the way we follow the
rules of addition/subtraction of numbers and for addition/subtraction of
algebraic expressions, we follow the same rules for multiplication of algebraic
expressions: (The associative and commutative properties of multiplications):
1. The ‘numerical co efficient’ of the product of two
monomials is the product of their numerical co-efficients (6a*5 = (6*5)*a = 30a
2. The ‘variable part’ of two monomials is equal to the
product of the variables in the given monomials (6a*5b = (6*5)*a*b = 30a*b =
30ab)
2.4 Problem 1:
Multiply (1/10)*(x^{5}y^{2}) by 10x^{3}y
Solution:
(1/10)*(x^{5}y^{2})
* 10x^{3}y
=(1/10)*10
*(x^{5}y^{2}) *x^{3}y ( By grouping co efficient and
same variables together)
= 1*(x^{5} *x^{3})
* y^{2}y
= (x^{5+3 )} * y^{2+1}
= x^{8} y^{3}
2.4 Problem 2:
Find the product of -3x^{2}y,
4xy^{2}z and (5/4)z
Solution:
Let us first, multiply the
first two terms:
-3x^{2}y* 4xy^{2}z
= (-3*4) (x^{2}*x)
(y*y^{2})z
(By grouping co efficient and same
variables together)
= -12 x^{3} y^{3}z (By first law of exponent)
Let us take all the three
terms:
(-3x^{2}y* 4xy^{2}z) * (5/4)z
= -12 x^{3} y^{3}z*
(5/4)z
= (-12)*(5/4) x^{3}
y^{3}z*z
= -15 x^{3} y^{3}
z^{2}
2.4.1 Multiplication of monomial by a binomial
We
know 24 = 2*12
= 2*(8+4) = 2*8+2*4 = 16+8:
Similarly
a*(b +c) = a*b + a*c = ab+ac (
Distributive property)
1. To multiply monomial and
a binomial, multiply each term of the binomial by the monomial and then
simplify.
2.4.1 Problem 1:
Find the product of -2pq and (-11p^{2}q-q^{2})
Solution:
-2pq *(-11p^{2}q-q^{2})
= (-11p^{2}q)*
(-2pq) -(q^{2})* (-2pq) ( Multiply first term
with each of the term in (-11p^{2}q-q^{2})
= (-11)*(-2)p^{2}*p*q*q
-(1*-2)*p*q^{2}* q
= 22p^{3}q^{2}+2pq^{3}
2.4.2 Multiplication of two binomials
Let us see how we can
multiply 12 by 8 by a different method though we know 12*8 = 96
We know 12 = 8+4 and 8= 6+2
_{} 12*8
= (8+4)*(6+2)
= 8*(6+2) + 4*(6+2)
= (8*6+8*2)+ (4*6 +4*2)
= 48+16+24+8 = 96
Similarly
(a+b)*(c+d)
= a*(c+d)+b*(c+d)
= ac+ad+bc+bd
In general to multiply two
binomials, multiply each term of one binomial with the every term of the other
binomial and then simplify. The same method holds good for multiplication of
trinomials.
2.4.2 Problem 1:
Find the product of 2x^{2}-3x
+1 and (x-3)
Solution:
(x-3)* (2x^{2}-3x
+1)
= x*(2x^{2}-3x +1) -3*(2x^{2}-3x +1) (Multiply each term of
first binomial with every other term of second trinomial)
= (2x^{2}*x-3x*x +1*x)+( -3*2x^{2}-3*-3x -3*1) (Simplify terms)
= (2x^{3}-3x^{2}+x)
+ (- 6 x^{2}+9x-3)
=2x^{3} -3x^{2}- 6 x^{2}+x +9x -3 (By grouping like terms to be together)
=2x^{3} -9x^{2}+10x
-3
2.4.3 Identities and formulae
Definition :
An ‘identity’
is a statement true for all values of
variables in a statement. We list below algebraic formulae which are
true for all values of the variables (a,b,c
or x)
We
have seen earlier that
(a+b)*(c+d) = a*(c+d)+b*(c+d)
= ac+ad+bc+bd ------------------->(1)
Let us
replace a by x, c by x, b by a and d by b in the above statement (1)
We get
(x+a)*(x+b)
= x*x+ xb+ax+ab
= x^{2}+xa+xb+ab
= x^{2}+x(a+b)+ab
2.4.3 Problem 1 : Don’t you feel that
finding an answer to 102*106 is
little difficult?
Solution:
Note 102
can be written as 100+2 and 106 can be written as 100+6
So (100+2)*(100+6) is of
the form (x+a)*(x+b) with
x=100, a= 2 and b=6. By using the formula we get
102*106
= (100+2)*(100+6)
= 100^{2}+ 100*(2+6)+ 2*6
= 10000+800+12 = 10812
Was this not easy?
2.4.3 Problem 2:
Find 97*95
Solution:
97 can be written as 100-3
and 95 can be written as 100-5
So (100-3)*(100-5) is of
the form (x+a)*(x+b) with
x=100, a= -3 and b=-5. Hence we get
97*95
= (100-3)*(100-5)
= 100^{2}+
100*(-3+-5)+ (-3*-5)
= 10000-800+15 = 9215
2.4.3 Problem 3: Find 103*96
Solution:
103 can be written as 100+3
and 96 can be written as 100-4
So (100+3)*(100-4) is of
the form (x+a)*(x+b) with
x=100, a= 3 and b=-4. Hence we get
103*96
= (100+3)*(100-4)
= 100^{2}+
100*(3+-4)+ (3*-4)
= 10000-100-12 = 9888
Let us replace c by a and d
by b in the statement (1) arrived
earlier( i.e. (a+b)*(c+d) = ac+ad+bc+bd )
We get
(a+b)*(a+b) = aa+ab+ba+bb
= a^{2}+ 2ab+b^{2}
_{} (a+b)^{2}= a^{2}+ 2ab+b^{2}
2.4.3 Problem 4: Find (10.1)^{2}
Solution:
10.1 can be
written as 10+0.1
So (10.1)^{2} is of
the form (a+b)^{2}^{
}with a=10 and b=0.1 . Hence we get
(10.1)^{2}
= 10^{2}+ 2*10*0.1+
(0.1)^{2}
= 100+2+0.01 = 102.01
2.4.3 Problem 5: What
should be added to 4x^{2}+12xy+8y^{2
}to make it a perfect square
Solution:
4x^{2} is of the
form a^{2} with a = 2x.
But 8y^{2} is not a
square, however 9y^{2} is of the form b^{2} with b = 3y
Therefore 2ab = 2*2x*3y =
12xy
_{}4x^{2}+12xy+ 9 y^{2} is a perfect square
Hence we need to add y^{2}
to the given problem to make it a perfect square
However there are many
solutions for this problem….
Exercise:Verify that
(2x+3y)^{2} = 4x^{2}+12xy+ 9 y^{2}
Let us
replace b by -b and c by a and d by -b in the statement (1)
We get
(a-b)*(a-b)
= a*a+ a(*-b)
+ (–b)*a +b*(-b)
= a^{2}-ab-ab+ b^{2}
= a^{2}-2ab+ b^{2}
2.4.3 Problem 6 :Find (4.9)^{2}
Solution:
4.9 can be written as 5-0.1
So (4.9)^{2} is of
the form (a-b)^{2}^{ }with a=5 and
b=0.1 .Hence we get
4.9^{2}
= 5^{2}+-2*5*0.1+
(0.1)^{2}
= 25 -1 +.01 = 24.01
2.4.3 Problem 7 :Find (x-1/x)^{2}
Solution:
This
is of the form (a-b)^{2} with a=x and b=1/x
.Hence we get
_{}(x-1/x)^{2}= x^{2}-2x(1/x)+ (1/x)^{2}
= x^{2}-2+ 1/x^{2}
Let us replace c by a and d by -b in the statement (1)
We get
(a+b)*(a-b) = aa+a*(-b)+ba+b*(-b)
= a^{2}-ab+ab-b^{2
}(_{} -ab+ab=0)
= a^{2}-b^{2}
2.4.3 Problem 8: Find 9.5*10.5^{}
Solution:
9.5 can be written as
10-0.5 and 10.5 can be written as 10+0.5
So 9.5* 10.5 is of the form
(a+b)(a-b) ^{ }with a=10 and b=0.5. Hence we get
9.5*10.5^{}
= 10^{2}- (0.5)^{2}
= 100-0.25 = 99.75
2.4.3 Problem 9: Evaluate (x+2)(x-2)( x^{2}+4)^{}
Solution:
Let us evaluate first two
terms which are of the form (a+b)*(a-b):
(x+2)(x-2) = ( x^{2}-4)
_{} (x+2)(x-2)(
x^{2}+4)= ( x^{2}-4) * ( x^{2}+4)
= ( x^{4}-16)
(_{}Square of x^{2 }is
x^{4})
Let us
replace b by b+c, c by
a+b and d
by c in the statement (1)
(a+(b+c))*((a+b)+c))
= a(a+b) + ac+ (b+c)(a+b)+
(b+c)c
= (a.a+ab)+ac+(ba+ b.b+ca+cb)+(
= a^{2}+ab+ac+ba+ b^{2}+ca+cb+
= a^{2} + b^{2}+
c^{2}+ab+ba+ac+ca+ cb+bc
(Rearrange the terms)
= a^{2} + b^{2}+
c^{2}+2ab+2bc+2ca
_{}(a+b+c)^{2}^{
} = a^{2} + b^{2}+ c^{2}+2ab+2bc+2ca
2.4.3 Problem 10: Evaluate 173^{2}
Solution:
173 can be written as
100+70+3. Hence 173^{2} is of the form (a+b+c)^{2}^{ } with a=100,b=70 and c=3
_{}173^{2}= 100^{2}+70^{2}+3^{2}+
2*100*70 +2*70*3+2*3*100
=
10000+4900+9+14000+420+600 =
29929
2.4.3 Problem 11: Simplify (x^{2} + y^{2}- z^{2})^{2} -(x^{2} - y^{2}-+z^{2})^{2}
Solution:
(x^{2} + y^{2}- z^{2})^{2} is of the form (a+b+c)^{2} with a = x^{2} , b = y^{2} and c = - z^{2}
Therefore by formula
(identity)
(x^{2}
+ y^{2}- z^{2})^{2} = (x^{4} + y^{4}+z^{4}
+ 2 x^{2} y^{2} - 2 y^{2} z^{2}-2 z^{2}
x^{2})
Similarly
(x^{2}
- y^{2}+z^{2})^{2} = (x^{4} + y^{4}+z^{4}
- 2 x^{2} y^{2} - 2 y^{2} z^{2}+2 z^{2}
x^{2})
_{}(x^{2} + y^{2}- z^{2})^{2} -(x^{2} - y^{2}-+z^{2})^{2}
= (x^{4}
+ y^{4}+z^{4} + 2 x^{2} y^{2} - 2 y^{2}
z^{2}-2 z^{2} x^{2}) -(x^{4} + y^{4}+z^{4}
- 2 x^{2} y^{2} - 2 y^{2} z^{2}+2 z^{2}
x^{2})
= (x^{4} + y^{4}+z^{4}
+ 2 x^{2} y^{2} - 2 y^{2} z^{2}-2 z^{2}
x^{2}) -x^{4} - y^{4}-z^{4} +2 x^{2} y^{2}
+ 2 y^{2} z^{2}-2 z^{2} x^{2}
= 4x^{2} y^{2}
-4 z^{2} x^{2}
=4x^{2}
(y^{2} -z^{2})
Let
us expand (a+b)^{3}
(a+b)^{3}
= (a+b)*(a+b)*(a+b)
= (a+b)*( a^{2}+ 2ab+b^{2}) (_{} (a+b)^{2}= a^{2}+
2ab+b^{2})
= a*( a^{2}+
2ab+b^{2})+b*( a^{2}+ 2ab+b^{2}) (Each term of the
algebraic term is multiplied by every
term of another algebraic term)
= (a^{3}+ 2a^{2}b+ab^{2})+(ba^{2}+ 2ab^{2}+b^{3}) ( By simplification)
= a^{3}+ 3a^{2}b+
3ab^{2}+b^{3}( By adding like terms)
= a^{3}+ 3ab(a+b)+b^{3}(By taking
common factor out)
It is interesting to note
that Bhaskaracharya has given this formula.( Lilavati
Shloka 27)
Exercise
: Arrive at (a-b)^{3}= a^{3}-3ab(a-b)-b^{3}
2.4.3 Problem 12: Evaluate 51^{3}
Solution:
51 can be
written as 50+1.
So 51^{3
} is of the form (a+b)^{3} with a =50 and b =1
_{}51^{3} = 50^{3}+ 3*50*1(50+1)+1^{3}
= 125000+7650+1 =132651
2.4.3 Problem 13: Evaluate (x+1/x)^{3}^{}
Solution:
(x+1/x)^{3} is of the form (a+b)^{3}
_{}(x+1/x)^{3}^{}
= x^{3}+ 3x*1/x(x+1/x)+(1/x)^{3}
= x^{3}+ 3(x+1/x)+(1/x^{3})
= x^{3}+ 3x+3/x+1/x^{3}
2.4.3 Problem 14 : Evaluate 9.9^{3}
Solution:
9.9 can be written as
10-0.1
So 9.9^{3 }is of the form (a-b)^{3} ^{ }with a=10 and b=0.1. Hence we get^{}
(9.9)^{3}= 10^{3}-3*10*0.1*(10-0.1)-(0.1)^{3}
= 1000-3*9.9- 0.001
= 1000-29.7-.001 =970.299
2.4.3 Problem 15: Evaluate (x/2-y/3)^{3}
Solution:
(x/2-y/3)^{3} is of the form (a-b)^{3}^{}
_{}(x/2-y/3)^{3}
= (x/2)^{3}- 3(x/2)*(y/3)(x/2-y/3)-(y/3)^{3}
= x^{3}/8 –
(xy/2)*(x/2-y/3)-y^{3}/27 ( By simplification)
= x^{3}/8 – x^{2}y/4
+xy^{2}/6-y^{3}/27
2.4.3 Problem 16: Show that (a+b) (a^{2}+b^{2}-ab) = a^{3}+b^{3}
Solution:
(a+b) (a^{2}+b^{2}-ab)
= a(a^{2}+b^{2}-ab)
+b(a^{2}+b^{2}-ab)
= a*a^{2}+a*b^{2}-a*ab + b*a^{2}+b*b^{2}-b*ab
=a^{3}+ab^{2}-a^{2}b
+ ba^{2}+b^{3}-ab^{2} (Eliminate positive and negative
equal terms)
=a^{3}+b^{3}
2.4.3 Problem 17: If
a+b+c = 12 and a^{2}+b^{2}+ c^{2}
=50 find ab+bc+ca
Solution:
We know
(a+b+c)^{2} = (a^{2}+b^{2}+ c^{2})+2ab+2bc+2ca
By substituting values we have
12^{2}= 50+2(ab+bc+ca)
_{} 144-50 = 2(ab+bc+ca)
I.e. ab+bc+ca
= 47
2.4.3 Problem 18: If
x^{3}+y^{3}+ z^{3} = 3xyz and x+y+z=0
find the value of
(x+y)^{2}/xy +(y+z)^{2}/yz+(z+x)^{2}/zx
Solution:
Since x+y+z=0
We have x+y
= -z, y+z = -x and z+x = -y
_{} (x+y)^{2}/xy +(y+z)^{2}/yz+(z+x)^{2}/zx
= z^{2}/xy+x^{2}/yz+y^{2}/zx
= z^{3}/xyz+x^{3}/xyz+y^{3}/zxy
( Make xyz as common denominator)
= (x^{3}+y^{3}+
z^{3})/xyz
= 3xyz/xyz
=3
2.4 Summary of
learning
No |
Formula/Identity |
Expansion |
1 |
(a+b)(c+d) |
ac+ad+bc+bd |
2 |
(x+a)*(x+b) |
x^{2}+x(a+b)+ab |
3 |
(a+b)^{2} |
a^{2}+b^{2}+2ab |
4 |
(a-b)^{2} |
a^{2}+b^{2}-2ab |
5 |
(a+b)(a-b) |
a^{2}-b^{2} |
6 |
(a+b+c)^{2} |
a^{2}+b^{2}+
c^{2}+2ab+2bc+2ca |
7 |
(a+b)^{3} |
a^{3}+b^{3}+3ab(a+b) |
8 |
(a-b)^{3} |
a^{3}-b^{3}-3ab(a-b) |