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)*(x5y2) by 10x3y

Solution:

(1/10)*(x5y2) * 10x3y

=(1/10)*10 *(x5y2) *x3y ( By grouping co efficient and same  variables together)

= 1*(x5 *x3) * y2y

= (x5+3 ) * y2+1

= x8 y3

2.4 Problem 2: Find the product of   -3x2y, 4xy2z and (5/4)z

Solution:

Let us first, multiply the first two terms:

-3x2y* 4xy2z

= (-3*4) (x2*x) (y*y2)z     (By grouping co efficient and same  variables together)

= -12 x3 y3z            (By first law of exponent)

Let us take all the three terms:

(-3x2y* 4xy2z) * (5/4)z

= -12 x3 y3z* (5/4)z

= (-12)*(5/4) x3 y3z*z

= -15 x3 y3 z2

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 (-11p2q-q2)

Solution:

-2pq *(-11p2q-q2)

= (-11p2q)* (-2pq) -(q2)* (-2pq) ( Multiply first term with each of the term in  (-11p2q-q2)

= (-11)*(-2)p2*p*q*q  -(1*-2)*p*q2* q

= 22p3q2+2pq3

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)

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   2x2-3x +1 and (x-3)

Solution:

(x-3)* (2x2-3x +1)

= x*(2x2-3x +1) -3*(2x2-3x +1) (Multiply each term of first binomial with every other term of second trinomial)

= (2x2*x-3x*x +1*x)+( -3*2x2-3*-3x -3*1) (Simplify terms)

= (2x3-3x2+x) + (- 6 x2+9x-3)

=2x3 -3x2- 6 x2+x +9x -3 (By grouping like terms to be together)

=2x3 -9x2+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)

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

= x2+xa+xb+ab

= x2+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)

= 1002+ 100*(2+6)+ 2*6

= 10000+800+12 = 10812

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)

= 1002+ 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)

= 1002+ 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

= a2+ 2ab+b2 (a+b)2= a2+ 2ab+b2

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

= 102+ 2*10*0.1+ (0.1)2

= 100+2+0.01 = 102.01

2.4.3 Problem  5: What should be added to 4x2+12xy+8y2  to make it a perfect square

Solution:

4x2 is of the form a2 with a = 2x.

But 8y2 is not a square, however 9y2 is of the form b2 with b = 3y

Therefore 2ab = 2*2x*3y = 12xy 4x2+12xy+ 9 y2 is a perfect square

Hence we need to add y2 to the given problem to make it a perfect square

However there are many solutions for this problem….

Exercise:Verify that (2x+3y)2 = 4x2+12xy+ 9 y2

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)

= a2-ab-ab+ b2

= a2-2ab+ b2

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

4.92

= 52+-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= x2-2x(1/x)+ (1/x)2

= x2-2+ 1/x2

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)

= a2-ab+ab-b2 ( -ab+ab=0)

= a2-b2

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

= 102- (0.5)2

= 100-0.25  = 99.75

2.4.3 Problem 9: Evaluate (x+2)(x-2)( x2+4)

Solution:

Let us evaluate first two terms which are of the form (a+b)*(a-b):

(x+2)(x-2) = ( x2-4) (x+2)(x-2)( x2+4)= ( x2-4) * ( x2+4)

= ( x4-16) ( Square of x2  is x4)

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)+(bc+ c.c)

= a2+ab+ac+ba+ b2+ca+cb+bc+ c2

= a2 + b2+ c2+ab+ba+ac+ca+ cb+bc (Rearrange the terms)

= a2 + b2+ c2+2ab+2bc+2ca (a+b+c)2  = a2 + b2+ c2+2ab+2bc+2ca

2.4.3 Problem 10: Evaluate 1732

Solution:

173 can be written as 100+70+3. Hence 1732 is of the form (a+b+c)2  with a=100,b=70 and c=3 1732= 1002+702+32+ 2*100*70 +2*70*3+2*3*100

= 10000+4900+9+14000+420+600 =  29929

2.4.3 Problem 11: Simplify (x2 + y2- z2)2 -(x2 - y2-+z2)2

Solution:

(x2 + y2- z2)2  is of the form (a+b+c)2 with a = x2 , b = y2 and c = - z2

Therefore by formula (identity)

(x2 + y2- z2)2 = (x4 + y4+z4 + 2 x2 y2 - 2 y2 z2-2 z2 x2)

Similarly

(x2 - y2+z2)2 = (x4 + y4+z4 - 2 x2 y2 - 2 y2 z2+2 z2 x2) (x2 + y2- z2)2 -(x2 - y2-+z2)2

= (x4 + y4+z4 + 2 x2 y2 - 2 y2 z2-2 z2 x2) -(x4 + y4+z4 - 2 x2 y2 - 2 y2 z2+2 z2 x2)

= (x4 + y4+z4 + 2 x2 y2 - 2 y2 z2-2 z2 x2) -x4 - y4-z4 +2 x2 y2 + 2 y2 z2-2 z2 x2

= 4x2 y2 -4 z2 x2

=4x2 (y2 -z2)

Let us expand (a+b)3

(a+b)3

= (a+b)*(a+b)*(a+b)

= (a+b)*( a2+ 2ab+b2) ( (a+b)2= a2+ 2ab+b2)

= a*( a2+ 2ab+b2)+b*( a2+ 2ab+b2) (Each term of the algebraic term is multiplied by  every term of another algebraic term)

= (a3+ 2a2b+ab2)+(ba2+ 2ab2+b3) ( By simplification)

= a3+ 3a2b+ 3ab2+b3( By adding like terms)

= a3+ 3ab(a+b)+b3(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= a3-3ab(a-b)-b3

2.4.3 Problem 12: Evaluate 513

Solution:

51  can be written as 50+1.

So 513  is of the form (a+b)3 with a =50 and b =1 513 = 503+ 3*50*1(50+1)+13

= 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

= x3+ 3x*1/x(x+1/x)+(1/x)3

= x3+ 3(x+1/x)+(1/x3)

= x3+ 3x+3/x+1/x3

2.4.3 Problem 14 : Evaluate  9.93

Solution:

9.9 can be written as 10-0.1

So 9.93 is of the form (a-b)3  with a=10 and b=0.1. Hence we get

(9.9)3= 103-3*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

= x3/8 – (xy/2)*(x/2-y/3)-y3/27 ( By simplification)

= x3/8 – x2y/4 +xy2/6-y3/27

2.4.3 Problem 16: Show that (a+b) (a2+b2-ab) = a3+b3

Solution:

(a+b) (a2+b2-ab)

= a(a2+b2-ab) +b(a2+b2-ab)

= a*a2+a*b2-a*ab + b*a2+b*b2-b*ab

=a3+ab2-a2b + ba2+b3-ab2 (Eliminate positive and negative equal terms)

=a3+b3

2.4.3 Problem 17: If a+b+c = 12 and a2+b2+ c2 =50 find ab+bc+ca

Solution:

We know

(a+b+c)2 = (a2+b2+ c2)+2ab+2bc+2ca By substituting values we have

122= 50+2(ab+bc+ca) 144-50 = 2(ab+bc+ca)

I.e.  ab+bc+ca = 47

2.4.3 Problem 18: If x3+y3+ z3 = 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

= z2/xy+x2/yz+y2/zx

= z3/xyz+x3/xyz+y3/zxy ( Make xyz as common denominator)

= (x3+y3+ z3)/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) x2+x(a+b)+ab 3 (a+b)2 a2+b2+2ab 4 (a-b)2 a2+b2-2ab 5 (a+b)(a-b) a2-b2 6 (a+b+c)2 a2+b2+ c2+2ab+2bc+2ca 7 (a+b)3 a3+b3+3ab(a+b) 8 (a-b)3 a3-b3-3ab(a-b)