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) |