## RD Sharma Solutions for Class 8 Chapter 6 Algebraic Expressions and Identities Free Online

EXERCISE 6.1 PAGE NO: 6.2
1. Identify the terms, their coefficients for each of the following expressions:
(i) 7x2yz – 5xy
(ii) x2 + x + 1
(iii) 3x2y2 – 5x2y2z2 + z2
(iv) 9 – ab + bc – ca
(v) a/2 + b/2 – ab
(vi) 0.2x – 0.3xy + 0.5y
Solution:
(i)
7x2yz – 5xy
The given equation has two terms that are:
7x2yz and – 5xy
The coefficient of 7x2yz is 7
The coefficient of – 5xy is – 5
(ii) x2 + x + 1
The given equation has three terms that are:
x2, x, 1
The coefficient of x2 is 1
The coefficient of x is 1
The coefficient of 1 is 1
(iii) 3x2y2 – 5x2y2z2 + z2
The given equation has three terms that are:
3x2y, -5x2y2z2 and z2
The coefficient of 3x2y is 3
The coefficient of -5x2y2z2 is -5
The coefficient of z2 is 1
(iv) 9 – ab + bc – ca
The given equation has four terms that are:
9, -ab, bc, -ca
The coefficient of 9 is 9
The coefficient of -ab is -1
The coefficient of bc is 1
The coefficient of -ca is -1
(v) a/2 + b/2 – ab
The given equation has three terms that are:
a/2, b/2, -ab
The coefficient of a/2 is 1/2
The coefficient of b/2 is 1/2
The coefficient of -ab is -1
(vi) 0.2x – 0.3xy + 0.5y
The given equation has three terms that are:
0.2x, -0.3xy, 0.5y
The coefficient of 0.2x is 0.2
The coefficient of -0.3xy is -0.3
The coefficient of 0.5y is 0.5
2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any category?
(i) x+y
(ii) 1000
(iii) x+x2+x3+x4
(iv) 7+a+5b
(v) 2b-3b2
(vi) 2y-3y2+4y3
(vii) 5x-4y+3x
(viii) 4a-15a2
(ix) xy+yz+zt+tx
(x) pqr
(xi) p2q+pq2
(xii) 2p+2q
Solution:
(i) x+y
The given expression contains two terms x and y
∴ It is Binomial
(ii) 1000
The given expression contains one term 1000
∴ It is Monomial
(iii) x+x2+x3+x4
The given expression contains four terms
∴ It belongs to none of the categories
(iv) 7+a+5b
The given expression contains three terms
∴ It is Trinomial
(v) 2b-3b2
The given expression contains two terms
∴ It is Binomial
(vi) 2y-3y2+4y3
The given expression contains three terms
∴ It is Trinomial
(vii) 5x-4y+3x
The given expression contains three terms
∴ It is Trinomial
(viii) 4a-15a2
The given expression contains two terms
∴ It is Binomial
(ix) xy + yz + zt + tx
The given expression contains four terms
∴ It belongs to none of the categories
(x) pqr
The given expression contains one term
∴ It is Monomial
(xi) p2q+pq2
The given expression contains two terms
∴ It is Binomial
(xii) 2p+2q
The given expression contains two terms
∴ It is Monomial

EXERCISE 6.2 PAGE NO: 6.5
1. Add the following algebraic expressions:
(i) 3a2b, -4a2b, 9a2b
(ii) 2/3a, 3/5a, -6/5a
(iii) 4xy2 – 7x2y, 12x2y -6xy2, -3x2y + 5xy2
(iv) 3/2a – 5/4b + 2/5c, 2/3a – 7/2b + 7/2c, 5/3a + 5/2b – 5/4c
(v) 11/2xy + 12/5y + 13/7x, -11/2y – 12/5x – 137xy
(vi) 7/2x3 – 1/2x2 + 5/3, 3/2x3 + 7/4x2 – x + 1/3, 3/2x2 -5/2x -2
Solution:
(i) 3a2b, -4a2b, 9a2b
Let us add the given expression
3a2b + (-4a2b) + 9a2b
3a2b – 4a2b + 9a2b
3a2b
(ii) 2/3a, 3/5a, -6/5a
Let us add the given expression
2/3a + 3/5a + (-6/5a)
2/3a + 3/5a – 6/5a
Let us take LCM for 3 and 5 which is 15
(2×5)/(3×5)a + (3×3)/(5×3)a – (6×3)/(5×3)a
10/15a + 9/15a – 18/15a
(10a+9a-18a)/15
a/15
(iii) 4xy2 – 7x2y, 12x2y -6xy2, -3x2y + 5xy2
Let us add the given expression
4xy2 – 7x2y + 12x2y – 6xy2 – 3x2y + 5xy2
Upon rearranging
4x2 + 12x2y – 3x2y – 7x2y – 6xy2 + 5xy2
3xy2 + 2x2y
(iv) 3/2a – 5/4b + 2/5c, 2/3a – 7/2b + 7/2c, 5/3a + 5/2b – 5/4c
Let us add the given expression
3/2a – 5/4b + 2/5c + 2/3a – 7/2b + 7/2c + 5/3a + 5/2b – 5/4c
Upon rearranging
3/2a + 2/3a + 5/3a – 5/4b – 7/2b + 5/2b + 2/5c + 7/2c – 5/4c
By taking LCM for (2 and 3 is 6), (4 and 2 is 4), (5,2 and 4 is 20)
(9a+4a+10a)/6 + (-5b-14b+10b)/4 + (8c+70c-25c)/20
23a/6 – 9b/4 + 53c/20
(v) 11/2xy + 12/5y + 13/7x, -11/2y – 12/5x – 13/7xy
Let us add the given expression
11/2xy + 12/5y + 13/7x + -11/2y – 12/5x – 13/7xy
Upon rearranging
11/2xy – 13/7xy + 13/7x – 12/5x + 12/5y -11/2y
By taking LCM for (2 and 7 is 14), (7 and 5 is 35), (5 and 2 is 10)
(11xy-12xy)/14 + (65x-84x)/35 + (24y-55y)/10
51xy/14 – 19x/35 – 31y/10
(vi) 7/2x3 – 1/2x2 + 5/3, 3/2x3 + 7/4x2 – x + 1/3, 3/2x2 -5/2x – 2
Let us add the given expression
7/2x3 – 1/2x2 + 5/3 + 3/2x3 + 7/4x2 – x + 1/3 + 3/2x2 -5/2x – 2
Upon rearranging
7/2x3 + 3/2x3 – 1/2x2 + 7/4x2 + 3/2x2 – x – 5/2x + 5/3 + 1/3 – 2
10/2x3 + 11/4x2 – 7/2x + 0/6
5x3 + 11/4x2 -7/2x
2. Subtract:
(i) -5xy from 12xy
(ii) 2a2 from -7a2
(iii) 2a-b from 3a-5b
(iv) 2x3 – 4x2 + 3x + 5 from 4x3 + x2 + x + 6
(v) 3/2y3 – 2/7y2 – 5 from 1/3y3 + 5/7y2 + y – 2
(vi) 3/2x – 5/4y – 7/2z from 2/3x + 3/2y – 4/3z
(vii) x2y – 4/5xy2 + 4/3xy from 2/3x2y + 3/2xy2 – 1/3xy
(viii) ab/7 -35/3bc + 6/5ac from 3/5bc – 4/5ac
Solution:
(i) -5xy from 12xy
Let us subtract the given expression
12xy – (- 5xy)
5xy + 12xy
17xy
(ii) 2a2 from -7a2
Let us subtract the given expression
2a2 + (-7a2)
-2a2 + 7a2
-9a2
(iii) 2a-b from 3a-5b
Let us subtract the given expression
-(2a – b)+ (3a – 5b)
-2a + b+ 3a – 5b
a – 4b
(iv) 2x3 – 4x2 + 3x + 5 from 4x3 + x2 + x + 6
Let us subtract the given expression
– (2x3 – 4x2 + 3x + 5) + (4x3 + x2 + x + 6)
– 2x3 + 4x2 – 3x – 5 + 4x3 + x2 + x + 6
2x3 + 5x2 – 2x + 1
(v) 3/2y3 – 2/7y2 – 5 from 1/3y3 + 5/7y2 + y – 2
Let us subtract the given expression
1/3y3 + 5/7y2 + y – 2 – 2/3y3 + 2/7y2 + 5
Upon rearranging
1/3y3 – 2/3y3 + 5/7y2 + 2/7y2 + y – 2 + 5
By grouping similar expressions we get,
-1/3y3 + 7/7y2 + y + 3
-1/3y3 + y2 + y + 3
(vi) 3/2x – 5/4y – 7/2z from 2/3x + 3/2y – 4/3z
Let us subtract the given expression
2/3x + 3/2y – 4/3z – (3/2x – 5/4y – 7/2z)
Upon rearranging
2/3x – 3/2x + 3/2y + 5/4y – 4/3z + 7/2z
By grouping similar expressions we get,
LCM for (3 and 2 is 6), (2 and 4 is 4), (3 and 2 is 6)
(4x-9x)/6 + (6y+5y)/4 + (-8z+21z)/6
-5x/6 + 11y/4 + 13z/6
(vii) x2y – 4/5xy2 + 4/3xy from 2/3x2y + 3/2xy2 – 1/3xy
Let us subtract the given expression
2/3x2y + 3/2xy2 – 1/3xy – (x2y – 4/5xy2 + 4/3xy)
Upon rearranging
2/3x2y – x2y + 3/2xy2 + 4/5xy2 – 1/3xy – 4/3xy
By grouping similar expressions we get,
LCM for (3 and 1 is 3), (2 and 5 is 10), (3 and 3 is 3)
-1/3x2y + 23/10xy2 – 5/3xy
(viii) ab/7 -35/3bc + 6/5ac from 3/5bc – 4/5ac
Let us subtract the given expression
3/5bc – 4/5ac – (ab/7 – 35/3bc + 6/5ac)
Upon rearranging
3/5bc + 35/3bc – 4/5ac – 6/5ac – ab/7
By grouping similar expressions we get,
LCM for (5 and 3 is 15), (5 and 5 is 5)
(9bc+175bc)/15 + (-4ac-6ac)/5 – ab/7
184bc/15 + -10ac/5 – ab/7
– ab/7 + 184bc/15 – 2ac
3. Take away:
(i) 6/5x2 – 4/5x3 + 5/6 + 3/2x from x3/3 – 5/2x2 + 3/5x + 1/4
(ii) 5a2/2 + 3a3/2 + a/3 – 6/5 from 1/3a3 – 3/4a2 – 5/2
(iii) 7/4x3 + 3/5x2 + 1/2x + 9/2 from 7/2 – x/3 – x2/5
(iv) y3/3 + 7/3y2 + 1/2y + 1/2 from 1/3 – 5/3y2
(v) 2/3ac – 5/7ab + 2/3bc from 3/2ab -7/4ac – 5/6bc
Solution:
(i) 6/5x2 – 4/5x3 + 5/6 + 3/2x from x3/3 – 5/2x2 + 3/5x + 1/4
Let us subtract the given expression
1/3x3 – 5/2x2 + 3/5x + 1/4 – (6/5x2 – 4/5x3 + 5/6 + 3/2x)
Upon rearranging
1/3x3 + 4/5x3 – 5/2x2 – 6/5x2 + 3/5x – 3/2x + 1/4 – 5/6
By grouping similar expressions we get,
LCM for (3 and 5 is 15), (2 and 5 is 10), (5 and 2 is 10), (4 and 6 is 24)
17/15x3 – 37/10x2 – 9/10x – 14/24
17/15x3 – 37/10x2 – 9/10x – 7/12
(ii) 5a2/2 + 3a3/2 + a/3 – 6/5 from 1/3a3 – 3/4a2 – 5/2
Let us subtract the given expression
1/3a3 – 3/4a2 – 5/2 – (5/2a2 + 3/2a3 + a/3 – 6/5)
Upon rearranging
1/3a5 – 3/2a3 – 3/4a2 – 5/2a2 – a/3 – 5/2 + 6/5
By grouping similar expressions we get,
LCM for (3 and 2 is 6), (4 and 2 is 4), (2 and 5 is 10)
(2a3 – 9a3)/6 – (3a2 – 10a2)/4 – a/3 + (-25+12)/10
-7/6a3 – 13/4a2 – a/3 – 13/10
(iii) 7/4x3 + 3/5x2 + 1/2x + 9/2 from 7/2 – x/3 – x2/5
Let us subtract the given expression
7/2 – x/3 – 1/5x2 – (7/4x3 + 3/5x2 + 1/2x + 9/2)
Upon rearranging
-7/4x3 – 1/5x2 – 3/5x2 – x/3 – x/2 + 7/2 – 9/2
By grouping similar expressions we get,
LCM for (3 and 2 is 6)
-7/4x3 – 4/5x2– (2x-3x)/6 + (7-9)/2
-7/4x3 – 4/5x2 – 5/6x – 1
(iv) y3/3 + 7/3y2 + 1/2y + 1/2 from 1/3 – 5/3y2
Let us subtract the given expression
1/3 – 5/3y2 – (1/3y3 + 7/3y2 + 1/2y + 1/2)
Upon rearranging
-1/3y3 – 5/3y2 – 7/3y2 – 1/2y + 1/3 – 1/2
By grouping similar expressions we get,
LCM for (3 and 3 is 3), (3 and 2 is 6)
-1/3y3 + (-5y2 – 7y2)/3 – 1/2y + (2-3)/6
-1/3y3 – 12/3y2 – 1/2y – 1/6
(v) 2/3ac – 5/7ab + 2/3bc from 3/2ab -7/4ac – 5/6bc
Let us subtract the given expression
3/2ab – 7/4ac – 5/6bc – (2/3ac – 5/7ab + 2/3bc)
Upon rearranging
3/2ab + 5/7ab – 7/4ac – 2/3ac – 5/6bc – 2/3bc
By grouping similar expressions we get,
LCM for (2 and 7 is 14), (4 and 3 is 12), (6 and 3 is 6)
(21ab+10ab)/14 – (21ac-8ac)/12 – (5bc-4bc)/6
31/14ab – 29/12ac – 3/2bc
4. Subtract 3x – 4y – 7z from the sum of x – 3y + 2z and -4x + 9y – 11z.
Solution:
The sum of x – 3y + 2z and -4x + 9y – 11z is
(x – 3y + 2z) + (-4x + 9y – 11z)
Upon rearranging
x – 4x – 3y + 9y + 2z – 11z
-3x + 6y – 9z
Now, Let us subtract the given expression from -3x + 6y – 9z
(-3x + 6y – 9z) – (3x – 4y – 7z)
Upon rearranging
-3x – 3x + 6y + 4y – 9z + 7z
-6x + 10y – 2z
5. Subtract the sum of 3l – 4m – 7n2 and 2l + 3m – 4n2 from the sum of 9l + 2m – 3n2 and -3l + m + 4n2….
Solution:
Sum of 3l – 4m – 7n2 and 2l + 5m – 4n2
3l – 4m – 7n2 + 2l + 3m – 4n2
Upon rearranging
3l + 2l – 4m + 3m – 7n2 – 4n2
5l – m – 11n2 ……………………..equation (1)
Sum of 9l + 2m – 3n2 and -3l + m + 4n2
9l + 2m – 3n2 + (-3l + m + 4n2)
Upon rearranging
9l – 3l + 2m + m – 3n2 + 4n2
6l + 3m + n2 ……………………….equation (2)
Let us subtract equation (i) from (ii), we get
6l + 3m + n2 – (5l – m – 11n2)
Upon rearranging
6l – 5l + 3m + m + n2 + 11n2
l + 4m + 12n2
6. Subtract the sum of 2x – x+ 5 and -4x – 3 + 7x2 from 5.
Solution:
Sum of 2x – x+ 5 and -4x – 3 + 7x2 is
2x – x+ 5 + (-4x – 3 + 7x2)
2x – x+ 5 – 4x – 3 + 7x2
Upon rearranging
– x2 + 7x2 + 2x – 4x + 5 – 3
6x2 -2x + 2 ………….equation (i)
Let us subtract equation (i) from 5 we get,
5 – (6x2 -2x + 2)
5 – 6x2 + 2x – 2
3 + 2x – 6x2
7. Simplify each of the following:
(i) x2 – 3x + 5 – 1/2(3x2 – 5x + 7)
(ii) [5 – 3x + 2y – (2x – y)] – (3x – 7y + 9)
(iii) 11/2x2y – 9/4xy2 + 1/4xy – 1/14y2x + 1/15yx2 + 1/2xy
(iv) (1/3y2 – 4/7y + 11) – (1/7y – 3 + 2y2) – (2/7y – 2/3y2 + 2)
(v) -1/2a2b2c + 1/3ab2c – 1/4abc2 – 1/5cb2a2 + 1/6cb2a – 1/7c2ab + 1/8ca2b
Solution:
(i) x2 – 3x + 5 – 1/2(3x2 – 5x + 7)
Upon rearranging
x2 – 3/2x2 – 3x + 5/2x + 5 – 7/2
By grouping similar expressions we get,
LCM for (1 and 2 is 2)
(2x2 – 3x2)/2 – (6x + 5x)/2 + (10-7)/2
-1/2x2 – 1/2x + 3/2
(ii) [5 – 3x + 2y – (2x – y)] – (3x – 7y + 9)
5 – 3x + 2y – 2x + y – 3x + 7y – 9
Upon rearranging
– 3x – 2x – 3x + 2y + y + 7y + 5 – 9
By grouping similar expressions we get,
-8x + 10y – 4
(iii) 11/2x2y – 9/4xy2 + 1/4xy – 1/14y2x + 1/15yx2 + 1/2xy
Upon rearranging
11/2x2y + 1/15x2y – 9/4xy2 – 1/14xy2 + 1/4xy + 1/2xy
By grouping similar expressions we get,
LCM for (2 and 15 is 30), (4 and 14 is 56), (4 and 2 is 4)
(165x2y + 2x2y)/30 + (-126xy2 – 4xy2)/56 + (xy + 2xy)/4
167/30x2y – 130/56xy2 + 3/4xy
167/30x2y – 65/28xy2 + 3/4xy
(iv) (1/3y2 – 4/7y + 11) – (1/7y – 3 + 2y2) – (2/7y – 2/3y2 + 2)
Upon rearranging
1/3y2 – 2y2 – 2/3y2 – 4/7y – 1/7y – 2/7y + 11 + 3 – 2
By grouping similar expressions we get,
LCM for (3, 1 and 3 is 3), (7, 7 and 7 is 7)
(y2 – 6y2 + 2y2)/3 – (4y – y – 2y)/7 + 12
-3/3y2 – 7/7y + 12
-y2 – y + 12
(v) -1/2a2b2c + 1/3ab2c – 1/4abc2 – 1/5cb2a2 + 1/6cb2a – 1/7c2ab + 1/8ca2b
Upon rearranging
-1/2a2b2c – 1/5a2b2c + 1/3ab2c + 1/6ab2c – 1/4abc2 – 1/7abc2 + 1/8a2bc
By grouping similar expressions we get,
LCM for (2 and 5 is 10), (3 and 6 is 6), (4 and 7 is 28)
-7/10a2b2c + 1/2ab2c – 11/28abc2 + 1/8a2bc

EXERCISE 6.3 PAGE NO: 6.13
Find each of the following products:
1. 5x2 × 4x3
Solution:
Let us simplify the given expression
5 × x × x × 4 × x × x × x
5 × 4 × x1+1+1+1+1
20 × x5
20x5
2. -3a2 × 4b4
Solution:
Let us simplify the given expression
– 3 × a2 × 4 × b4
-12 × a2 × b4
-12a2b4
3. (-5xy) × (-3x2yz)
Solution:
Let us simplify the given expression
(-5) × (-3) × x × x2 × y × y × z
15 × x1+2 × y1+1 × z
15x3y2z
4. 1/2xy × 2/3x2yz2
Solution:
Let us simplify the given expression
1/2 × 2/3 × x × x2 × y × y × z2
1/3 × x1+2 × y1+1 × z2
1/3x3y2z2
5. (-7/5xy2z) × (13/3x2yz2)
Solution:
Let us simplify the given expression
-7/5 × 13/3 × x × x2 × y2 × y × z × z2
-91/15 × x1+2 × y2+1 × z1+2
-91/15x3y3z3
6. (-24/25x3z) × (-15/16xz2y)
Solution:
Let us simplify the given expression
-24/25 × -15/16 × x3 × x × z × z2 × y
18/20 × x3+1 × z1+2 × y
9/10x4z3y
7. (-1/27a2b2) × (9/2a3b2c2)
Solution:
Let us simplify the given expression
-1/27 × 9/2 × a2 × a3 × b2 × b2 × c2
-1/6 x a2+3 × b2+2 × c2
-1/6a5b4c2
8. (-7xy) × (1/4x2yz)
Solution:
Let us simplify the given expression
-7 × 1/4 × x × y × x2 × y × z
-7/4 × x1+2 × y1+1 × z
-7/4x3y2z
9. (7ab) × (-5ab2c) × (6abc2)
Solution:
Let us simplify the given expression
7 × -5 × 6 × a × a × a × b × b2 × b × c × c2
210 × a1+1+1 × b1+2+1 × c1+2
210a3b4c3
10. (-5a) × (-10a2) × (-2a3)
Solution:
Let us simplify the given expression
(-5) × (-10) × (-2) × a × a2 × a3
-100 × a1+2+3
-100a6
11. (-4x2) × (-6xy2) × (-3yz2)
Solution:
Let us simplify the given expression
(-4) × (-6) – (-3) × x2 × x × y2 × y × z2
– 72 × x2+1 × y2+1 × z2
-72x3y3z2
12. (-2/7a4) × (-3/4a2b) × (-14/5b2)
Solution:
Let us simplify the given expression
-2/7 × -3/4 × -14/5 × a4 × a2 × b × b2
-6/10 × a4+2 × b1+2
-3/5a6b3
13. (7/9ab2) × (15/7ac2b) × (-3/5a2c)
Solution:
Let us simplify the given expression
7/9 × 15/7 × -3/5 × a × a × a2 × b2 × b × c2 × c
– a1+1+2 × b2+1 × c2+1
-a4b3c3
14. (4/3u2vw) × (-5uvw2) × (1/3v2wu)
Solution:
Let us simplify the given expression
4/3 × -5 × 1/3 × u2 × u × u × v × v × v2 × w × w2 × w
-20/9 × u2+1+1 × v1+1+2 × w1+2+1
-20/9u4v4w4
15. (0.5x) × (1/3xy2z4) × (24x2yz)
Solution:
Let us simplify the given expression
0.5 × 1/3 × 24 × x × x × y2 × y × x2 × z4 × z
12/3 × x1+1+2 × y2+1 × z4+1
4x4 × y3 × z5
4x4y3z5
16. (4/3pq2) × (-1/4p2r) × (16p2q2r2)
Solution:
Let us simplify the given expression
4/3 × 1/4 × 16 × p × p2 × p2 × q2 × q2 × r × r2
-16/3 × p1+2+2 × q2+2 × r1+2
-16/3p5q4r3
17. (2.3xy) × (0.1x) × (0.16)
Solution:
Let us simplify the given expression
2.3 × 0.1 × 0.16 × x × x × y
0.0368 × x1+1 × y
0.0368x2y
Express each of the following products as a monomials and verify the result in each case for x=1:
18. (3x) × (4x) × (-5x)
Solution:
Let us simplify the given expression
3 × 4 × -5 × x × x × x
-60 × x1+1+1
-60x3
19. (4x2) × (-3x) × (4/5x3)
Solution:
Let us simplify the given expression
4 × -3 × 4/5 × x2 × x × x3
-48/5 × x2+1+3
-485x6
20. (5x4) × (x2)3 × (2x) 2
Solution:
Let us simplify the given expression
5 × x4 × x6 × 4 × x2
5 × 4 × x4 × x6 × x2
20 × x4+6+2
20x12
21. (x2)3 × (2x) × (-4x) × (5)
Solution:
Let us simplify the given expression
x6 × 2 × x × -4 × x × 5
2 × -4 × 5 × x6 × x × x
-40 × x6+1+1
-40x8
22. Write down the product of -8x2y6 and -20xy verify the product for x = 2.5, y = 1
Solution:
Let us simplify the given expression
-8 × -20 × x2 × x × y6 × y
160 × x2+1 × y6+1
160x3y7
Now let us verify when, x = 2.5 and y = 1
For 160x3y7
160 (2.5)3 × (1)7
16 × 15.625
250
For -8x2y6 and -20xy
-8 × 2.52 × 16 × -20 × 1 × 2.5
250
Hence, the given expression is verified.
23. Evaluate (3.2x6y3× (2.1x2y2) when x = 1 and y = 0.5
Solution:
Let us simplify the given expression
3.2 × 2.1 × x6 × x2 × y3 × y2
6.72 × x6+2 × y3+2
6.72x8y5
Now let us substitute when, x = 1 and y = 0.5
For 6.72x8y5
6.72 × 18 × 0.55
0.21
24. Find the value of (5x6) × (-1.5x2y3) × (-12xy2) when x = 1, y = 0.5
Solution:
Let us simplify the given expression
5 × -1.5 × -12 × x6 × x2 × x × y3 × y2
90 × x6+2+1 × y3+2
90x9y5
Now let us substitute when, x = 1 and y = 0.5
For 90x9y5
90 × (1)9× (0.5)5
2.8125
45/16
25. Evaluate (2.3a5b2) × (1.2a2b2) when a = 1 and b = 0.5
Solution:
Let us simplify the given expression
2.3a5b2 × 1.2a2b2
2.3 × 1.2 × a5 × a2 × b2 × b2
2.76 × a5+2 × b2+2
2.76a7b4
Now let us substitute when, a = 1 and b = 0.5
For 2.76 a7 b4
2.76 (1)7 (0.5)4
2.76 × 1 × 0.0025
0.1725
6.9/40
26. Evaluate (-8x2y6) × (-20xy) for x = 2.5 and y = 1
Solution:
Let us simplify the given expression
-8 × – 20 × x2 × x × y6 × y
160x2+1y6+1
160x3y7
Now let us substitute when, x = 2.5 and y = 1
160x3y7
160 × (2.5)3 × (1)7
2500
Express each of the following products as a monomials and verify the result for x = 1, y = 2:
27. (-xy3) × (yx3) × (xy)
Solution:
Let us simplify the given expression
-x × y3 × y × x3 × x × y
-x1+3+1 × y3+1+1
-x5y5
Now let us substitute when, x = 1 and y = 2
-x5y5
-15 × 25
-32
28. (1/8x2y4) × (1/4x4y2) × (xy) × 5
Solution:
Let us simplify the given expression
1/8 × 1/4 × 5 × x2 × x4 × x × y4 × y2 × y
5/32 × x2+4+1 × y4+2+1
5/32x7y7
Now let us substitute when, x = 1 and y = 2
5/32 × 16 × 26
5/32 × 64
5 × 2
10
29. (2/5a2b) × (-15b2ac) × (-1/2c2)
Solution:
Let us simplify the given expression
2/5 × -15 × -1/2 × a2 × a × b × b2 × c × c2
3 × a2+1 × b1+2 × c1+2
3a3b3c3
30. (-4/7a2b) × (-2/3b2c) × (-7/6c2a)
Solution:
Let us simplify the given expression
-4/7 × -2/3 × -7/6 × a2 × a × b × b2 × c × c2
-4/9 × a2+1 × b2+1 × c1+2
-4/9a3b3c3
31. (4/9abc3) × (-27/5a3b2) × (-8b3c)
Solution:
Let us simplify the given expression
4/9 × -27/5 × -8 × a × a3 × b × b2 × b3 × c3 × c
96/5 × a1+3 × b1+2+3 × c3+1
96/5a4b6c4
Evaluate each of the following when x = 2, y = -1.
32. (2xy) × (x2y/4) × (x2) × (y2)
Solution:
Let us simplify the given expression
2 × 1/4 × x × x2 × x2 × y × y2 × y
1/2x1+2+2y1+2+1
1/2x5y4
Now let us substitute when, x = 2 and y = -1
For 1/2x5y4
1/2 × (2)5 × (-1)4
1/2 × 32 × 1
16
33. (3/5x2y) × (-15/4xy2) × (7/9x2y2)
Solution:
Let us simplify the given expression
3/5 × -15/4 × 7/9 × x2 × x × x2 × y × y2 × y2
-7/4 × x2+1+2 × y1+2+2
7/4x5y5
Now let us substitute when, x = 2 and y = -1
For -7/4x5y5
-7/4 × (2)5 (-1)5
-7/4 × 32 × -1
56

EXERCISE 6.4 PAGE NO: 6.21
Find the following products:
1. 2a(3a + 5b)
Solution:
Let us simplify the given expression
2a3 (3a + 5b)
(2a3 × 3a) + (2a3 × 5b)
6a3+1 + 10a3b
6a4 + 10a3b
2. -11a (3a + 2b)
Solution:
Let us simplify the given expression
-11a (3a + 2b)
(-11a × 3a) + (-11a × 2b)
-33a2 – 22ab
3. -5a (7a – 2b)
Solution:
Let us simplify the given expression
-5a (7a – 2b)
(-5a × 7a) – (-5a × 2b)
-35a2 + 10ab
4. -11y(3y + 7)
Solution:
Let us simplify the given expression
-11y2 (3y + 7)
(-11y2 × 3y) + (-11y2 × 7)
-33y3 – 77y2
5. 6x/5(x3 + y3)
Solution:
Let us simplify the given expression
6/5x (x3 + y3)
(6/5x × x3) + (6/5x × y3)
6/5x4 + 6/5xy3
6. xy (x3 – y3)
Solution:
Let us simplify the given expression
xy (x3 – y3)
(xy × x3) – (xy × y3)
x4y – xy4
7. 0.1y (0.1x5 + 0.1y)
Solution:
Let us simplify the given expression
0.1y (0.1x5 + 0.1y)
(0.1y × 0.1x5) + (0.1y × 0.1y)
0.01x5y + 0.01y2
8. (-7/4ab2c – 6/25a2c2) (-50a2b2c2)
Solution:
Let us simplify the given expression
(-7/4ab2c – 6/25a2c2) (-50a2b2c2)
(-7/4ab2c × -50a2b2c2) – (6/25a2c2 × -50a2b2 × c2)
350/4a3b4c3 + 12a4b2c4
175/2a3b4c3 + 12a4b2c4
9. -8/27xyz (3/2xyz2 – 9/4xy2z3)
Solution:
Let us simplify the given expression
-8/27xyz (3/2xyz2 – 9/4xy2z3)
(-8/27xyz × 3/2xyz2) – (-8/27xyz × 9/4xy2z3)
-4/9x2y2z3 + 2/3x2y3z4
10. -4/27xyz (9/2x2yz – 3/4xyz2)
Solution:
Let us simplify the given expression
-4/27xyz (9/2x2yz – 3/4xyz2)
(-4/27xyz × 9/2x2yz) – (-4/27xyz × 3/4xyz2)
-2/3x3y2z2 + 1/9x2y2z3
11. 1.5x (10x2y – 100xy2)
Solution:
Let us simplify the given expression
1.5x (10x2y – 100xy2)
(1.5x 10x2y) – (1.5x × 100xy2)
15x3y – 150x2y2
12. 4.1xy (1.1x – y)
Solution:
Let us simplify the given expression
4.1xy (1.1x – y)
(4.1xy × 1.1x) – (4.1xy × y)
4.51x2y – 4.1xy2
13. 250.5xy (xz + y/10)
Solution:
Let us simplify the given expression
250.5xy (xz + y/10)
(250.5xy × xz) + (250.5xy × y/10)
250.5x2yz + 25.05xy2
14. 7/5x2y (3/5xy2 + 2/5x)
Solution:
Let us simplify the given expression
7/5x2y (3/5xy2 + 2/5x)
(7/5x2y × 3/5xy2) + (7/5x2y × 2/5x)
21/25x3y3 + 14/25x3y
15. 4/3a (a2 + b2 – 3c2)
Solution:
Let us simplify the given expression
4/3a (a2 + b2 – 3c2)
(4/3a × a2) + (4/3a × b2) – (4/3a × 3c2)
4/3a3 + 4/3ab2 – 4ac2
16. Find the product 24x2 (1-2x) and evaluate its value for x = 3
Solution:
Let us simplify the given expression
24x2 (1 – 2x)
(24x2× 1) – (24x2× 2x)
24x2 – 48x3
Now let us evaluate the expression when x = 3
24x2 – 48x3
24 × (3)2 – 48 × (3)3
24 × (9) – 48 × (27)
216 – 1296
-1080
17. Find the product -3y (xy+y2) and evaluate its value for x = 4 and y = 5
Solution:
Let us simplify the given expression
-3y (xy+y2)
(-3y × xy) + (-3y × y2)
-3xy2 – 3y3
Now let us evaluate the expression when x = 4 and y = 5
-3xy2 – 3y3
-3 × (4) × (5)2 – 3 × (5)3
-300 – 375
-675
18. Multiply -3/2x2y3 by (2x-y) and verify the answer for x = 1 and y = 2
Solution:
Let us simplify the given expression
-3/2x2y3 by (2x-y)
(-3/2x2y3 × 2x) – (-3/2x2y3 × y)
-3x3y3 + 3/2x2y4
Now let us evaluate the expression when x = 1 and y = 2
-3x3y3 + 3/2x2y4
-3 × (1)4 × (2)3 + 3/2 × (1)2 × (2)4
– 3 × (8) + 3 (8)
-24+24
0
19. Multiply the monomial by the binomial and find the value of each for x = -1, y = 0.25 and z = 0.005:
(i) 15y2 (2 – 3x)
(ii) -3x (y+ z2)
(iii) z2 (x – y)
(iv) xz (x2 + y2)
Solution:
(i) 15y2 (2 – 3x)
Let us simplify the given expression
30y2 – 45xy2
By evaluating the values in the expression x = -1, y = 25/100 and z = 5/1000
30 × (25/100)2 – 45 × (-1) × (25/100)2
30 (1/16) + 45 (1/16)
15/8 + 45/16
(30+45)/16
75/16
(ii) -3x (y+ z2)
Let us simplify the given expression
-3xy2 + -3xz2
By evaluating the values in the expression x = -1, y = 25/100 and z = 5/1000
-3× (-1) × (25/100)2 – 3 × (-1) × (5/1000)2
(3×25×25/100×100) + (3×5×5/1000×1000)
3/16 + 3/40000
39/200
(iii) z2 (x – y)
Let us simplify the given expression
z2x – z2y
By evaluating the values in the expression x = -1, y = 25/100 and z = 5/1000
z2 (x – y)
(5/1000)2 (-1 – 25/100)
(1/40000) (-100-25/100)
(1/40000) (-125/100)
(1/40000) (-5/4)
-5/160000
-1/32000
(iv) xz (x2 + y2)
Let us simplify the given expression
x3z + xzy2
By evaluating the values in the expression x = -1, y = 25/100 and z = 5/1000
x3z + xzy2
(-1)3 × (5/1000) + (-1) × (5/1000) × (25/100)2
-1/200 – 1/16 × 1/200
-1/200 – 1/3200
By taking LCM as 3200
(-16 -1)/3200
-17/3200
20. Simplify:
(i) 2x2 (x3 – x) – 3x (x4 + 2x) – 2 (x4 – 3x2)
(ii) x3y (x2 – 2x) + 2xy (x3 – x4)
(iii) 3a2 + 2 (a+2) – 3a (2a+1)
(iv) x (x+4) + 3x (2x2 -1) + 4x2 + 4
(v) a (b-c) – b (c-a) – c (a-b)
(vi) a (b-c) +b (c-a) + c (a-b)
(vii) 4ab (a-b) – 6a(b-b2) -3b2 (2a2 -a) + 2ab (b-a)
(viii) x2 (x+ 1) – x3 (x + 1) – x (x3 – x)
(ix) 2a2 + 3a (1 – 2a3) + a (a + 1)
(x) a2 (2a – 1) + 3a + a3 – 8
(xi) 3/2x2 (x2 – 1) + 1/4x2 (x2 + x) – 3/4x (x3 – 1)
(xii) a2b (a-b2) + ab2(4ab – 2a2) – a3b(1-2b)
(xiii) a2b (a3– a + 1) – ab(a4 – 2a2 + 2a) – b(a3– a2 – 1)
Solution:
(i) 2x2 (x3 – x) – 3x (x4 + 2x) – 2 (x4 – 3x2)
Let us simplify the given expression
2x5 – 2x3 – 3x5 – 6x2 – 2x4 + 6x2
By grouping similar expressions we get,
2x5 – 3x5 – 2x3 – 2x4 – 6x2 + 6x2
-x5 – 2x4 – 2x3
(ii) x3y (x2 – 2x) + 2xy (x3 – x4)
Let us simplify the given expression
x5y – 2x4y + 2x4y – 2x5y
By grouping similar expressions we get,
-x5y – 2x5y
-x5y
(iii) 3a2 + 2 (a+2) – 3a (2a+1)
Let us simplify the given expression
3a2 + 2a + 4 – 6a2 – 3a
By grouping similar expressions we get,
3a2 – 6a2 + 2a – 3a + 4
-3a2 – a + 4
(iv) x (x+4) + 3x (2x2 -1) + 4x2 + 4
Let us simplify the given expression
x2 + 4x + 6x3 – 3x + 4x2 + 4
By grouping similar expressions we get,
6x3 + 5x2 + x + 4
(v) a (b-c) – b (c-a) – c (a-b)
Let us simplify the given expression
ab – ac – bc + ab – ca + bc
By grouping similar expressions we get,
2ab – 2ac
(vi) a (b-c) +b (c-a) + c (a-b)
Let us simplify the given expression
ab – ac + bc – ab + ac – bc
By grouping similar expressions we get,
0
(vii) 4ab (a-b) – 6a(b-b2) -3b2 (2a2 -a) + 2ab (b-a)
Let us simplify the given expression
4a2b – 4ab2 – 6a2b + 6a2b2 – 6a2b2 + 3ab2 + 2ab2 – 2a2b
By grouping similar expressions we get,
4a2b – 6a2b– 2a2b – 4ab2 + 3ab2 + 2ab2 + 6a2b2 – 6a2b2
-4a2b + ab2
(viii) x2 (x+ 1) – x3 (x + 1) – x (x3 – x)
Let us simplify the given expression
x4 + x2 – x4 – x3 – x4 + x2
By grouping similar expressions we get,
x4 – x4 – x4 – x3 + x2 + x2
– x4 – x3 + 2x2
(ix) 2a2 + 3a (1 – 2a3) + a (a + 1)
Let us simplify the given expression
2a2 + 3a – 6a4 + a2 + a
By grouping similar expressions we get,
-6a4 + 3a2 + 4a
(x) a2 (2a – 1) + 3a + a3 – 8
Let us simplify the given expression
2a3 – a2 + 3a + a3 – 8
By grouping similar expressions we get,
3a3 – a2 + 3a – 8
(xi) 3/2x2 (x2 – 1) + 1/4x2 (x2 + x) – 3/4x (x3 – 1)
Let us simplify the given expression
3/2x4 – 3/2x2 + 1/4x4 + 1/4x3 – 3/4x4 + 3/4x
By grouping similar expressions we get,
3/2x4 + 1/4x4 – 3/4x4 – 3/2x+ 1/4x3 + 3/4x
4/4x4 + 1/4x3 – 3/2x2 + 3/4x
x4 + 1/4x3 – 3/2x2 + 3/4x
(xii) a2b (a-b2) + ab2(4ab – 2a2) – a3b(1-2b)
Let us simplify the given expression
a3b – a2b3 + 4a2b3 – 2a3b2 – a3b + 2a3b2
By grouping similar expressions we get,
-a2b3 + 4a2b3
3a2b3
(xiii) a2b (a3– a + 1) – ab(a4 – 2a2 + 2a) – b(a3– a2 – 1)
Let us simplify the given expression
a5b – a3b + a2b – a5b + 2a3b – 2a2b – ba3 + a2b + b
By grouping similar expressions we get,
a5b – a5b – a3b + 2a3b – ba3 + a2b – 2a2b + a2b + b
b

EXERCISE 6.5 PAGE NO: 6.30
Multiply:
1. (5x + 3) by (7x + 2)
Solution:
Now let us simplify the given expression
(5x + 3) × (7x + 2)
5x (7x + 2) + 3 (7x + 2)
35x2 + 10x + 21x + 6
35x2 + 31x + 6
2. (2x + 8) by (x – 3)
Solution:
Now let us simplify the given expression
(2x + 8) × (x – 3)
2x (x – 3) + 8 (x – 3)
2x2 – 6x + 8x – 24
2x2 + 2x – 24
3. (7x + y) by (x + 5y)
Solution:
Now let us simplify the given expression
(7x + y) × (x + 5y)
7x (x + 5y) + y (x + 5y)
7x2 + 35xy + xy + 5y2
7x2 + 36xy + 5y2
4. (a – 1) by (0.1a2 + 3)
Solution:
Now let us simplify the given expression
(a – 1) × (0.1a2 + 3)
a (0.1a2 + 3) -1 (0.1a2 + 3)
0.1a3 + 3a – 0.1a2 – 3
0.1a3 – 0.1a2 + 3a – 3
5. (3x2 + y2) by (2x2 + 3y2)
Solution:
Now let us simplify the given expression
(3x2 + y2) × (2x2 + 3 y2)
3x2 × (2x2 + 3y2) + y2 × (2x2 + 3y2)
6x4 + 9x2y2 + 2x2y2 + 3y4
6x4 + 11x2y2 + 3y4
6. (3/5x + 1/2y) by (5/6x + 4y)
Solution:
Now let us simplify the given expression
(3/5x + 1/2y) × (5/6x + 4y)
3/5x × (5/6x + 4y) + 1/2y × (5/6x + 4y)
15/30x2 + 12/5xy + 5/12xy + 4/2y2
1/2x2 + 169/60xy + 2y2
7. (x6 – y6) by (x2 + y2)
Solution:
Now let us simplify the given expression
(x6 – y6) × (x2 + y2)
x6 × (x2 + y2) – y6 × (x2 + y2)
x8 + x6y2 – x2y6 – y8
8. (x2 + y2) by (3a + 2b)
Solution:
Now let us simplify the given expression
(x2 + y2) × (3a + 2b)
x2 × (3a + 2b) + y2 × (3a + 2b)
3ax2 + 3ay2 + 2bx2 + 2by2
9. (- 3d – 7f) by (5d + f)
Solution:
Now let us simplify the given expression
(- 3d – 7f) × (5d + f)
-3d (5d + f) – 7f (5d + f)
– 15d2 – 3df – 35df – 7f2
– 15d2 – 38df – 7f2
10. (0.8a – o.5b) by (1.5a – 3b)
Solution:
Now let us simplify the given expression
(0.8a – 0.5b) × (1.5a – 3b)
0.8a (1.5a – 3b) – 0.5b (1.5a – 3b)
1.2a2 – 2.4ab – 0.75ab + 1.5b2
1.2a2 – 3.15ab + 1.5b2
11. (2x2y2 – 5xy2) by (x2 – y2)
Solution:
Now let us simplify the given expression
(2x2y2 – 5xy2) × (x2 – y2)
2x2y2 (x2 – y2) – 5xy2 (x2 – y2)
2x4y2 – 5x3y2 – 2x2y4 + 5xy4
12. (x/7 + x2/2) by (2/5 + 9x/4)
Solution:
Now let us simplify the given expression
(x/7 + x2/2) × (2/5 + 9x/4)
x/7 (2/5 + 9x/4) + x2/2 (2/5 + 9x/4)
2x/35 + (9 x2)/28 + x2/5 + (9 x3)/8
9/8×3 + 73/140x2 + 2/35x
13. (-a/7 + a2/9) by (b/2 – b2/3)
Solution:
Now let us simplify the given expression
(-a/7 + a2/9) × (b/2 – b2/3)
-a/7 (b/2 – b2/3) + a2/9 (b/2 – b2/3)
-ab/14 + ab2/21 + a2b/18 – a2b2/27
14. (3x2y – 5xy2) by (1/5x2 + 1/3y2)
Solution:
Now let us simplify the given expression
(3x2y – 5xy2) × (1/5x2 + 1/3y2)
3x2y (1/5x2 + 1/3y2) – 5xy2 (1/5x2 + 1/3y2)
3/5x4y + 3/3x2y3 – x3y2 + 5/3xy4
3/5x4y + x2y3 – x3y2 + 5/3xy4
15. (2x2 – 1) by (4x3 + 5x2)
Solution:
Now let us simplify the given expression
(2x2 – 1) × (4x3 + 5x2)
2x2 (4x3 + 5x2) – 1 (4x3 + 5x2)
8x5 + 10x4 – 4x3 – 5x2
16. (2xy + 3y2) by (3y2 – 2)
Solution:
Now let us simplify the given expression
(2xy + 3y2) × (3y2 – 2)
2xy (3y2 – 2) + 3y2 (3y2 – 2)
6xy3 – 4xy + 9y4 – 6y2
Find the following products and verify the results for x = -1, y = -2:
17. (3x – 5y) (x + y)
Solution:
Now let us simplify the given expression
(3x – 5y) × (x + y)
(3x – 5y) × (x + y)
x (3x – 5y) + y (3x – 5y)
3x2 – 5xy + 3xy – 5y2
3x2 – 2xy – 5y2
Let us substitute the given values x = – 1 and y = – 2, then
(3x – 5y) × (x + y)
[3 (-1) – 5 (-2)] × [(-1) + (-2)]
(-3+10) × (-1-2)
7×-3
-21
3x2 – 2xy – 5y2
3 (-1)2 – 2 (-1) (-2) – 5 (-2)2
3 – 4 – 20
– 21
∴ the given expression is verified.
18. (x2y – 1) (3 – 2x2y)
Solution:
Now let us simplify the given expression
(x2y – 1) × (3 – 2x2y)
x2y (3 – 2x2y) – 1 (3 – 2x2y)
3x2y – 2x4y2 – 3 + 2x2y
5x2y – 2x4y2 – 3
Let us substitute the given values x = – 1 and y = – 2, then
(x2y – 1) × (3 – 2x2y)
[(-1)2 (-2) – 1] × [3 – 2 (-1)2 (-2)
(-2 – 1) × (3 + 4)
-3 × 7
-21
5x2y – 2x4y2 – 3
[-2 (-1)4 (-2)2 + 5 (-1)2 (2) – 3]
– 8 – 10 – 3
-21
∴ the given expression is verified.
19. (1/3x – y2/5) (1/3x + y2/5)
Solution:
Now let us simplify the given expression
(1/3x – y2/5) × (1/3x + y2/5)
(1/3x) 2 – (y2/5)2
(1/3x – y2/5) (1/3x + y2/5)
1/9x2 – 1/25y4
Let us substitute the given values x = – 1 and y = – 2, then
(1/3x – y2/5) × (1/3x + y2/5)
(1/3(-1) – (-2)2/5) × (1/3(-1) + (-2)2/5)
(-17/15) × (7/15)
-119/225
1/9x2 – 1/25y4
1/9 (-1)2 – 1/25 (-2)4
1/9 -16/25
-119/225
∴ the given expression is verified.
Simplify:
20. x2 (x + 2y) (x – 3y)
Solution:
Now let us simplify the given expression
x2 (x + 2y) (x – 3y)
x2 (x2 – 3xy + 2xy – 3y2)
x2 (x2 – xy – 6y2)
x4 – x3y – 6x2y2
21. (x2 – 2y2) (x + 4y)x2y2
Solution:
Now let us simplify the given expression
(x2 – 2y2) (x + 4y)x2y2
(x3 + 4x2y – 2xy2 – 8y3) × x2y2
x5y2 + 4x4y3 – 2x3y4 – 8x2y5
22. a2b2 (a + 2b) (3a + b)
Solution:
Now let us simplify the given expression
a2b2 (a + 2b) (3a + b)
a2b2 (3a2 + ab + 6ab + 2b2)
a2b2 (3a2 + 7ab + 2b2)
3a4b2 + 7a3b3 + 2a2b4
23. x2 (x – y) y2 (x + 2y)
Solution:
Now let us simplify the given expression
x2 (x – y) y2 (x + 2y)
x2y2 (x2 + 2xy – xy – 2y2)
x2y2 (x2 + xy – 2y2)
x4y2 + x3y3 – 2x2y4
24. (x3 – 2x2 + 5x – 7) (2x – 3)
Solution:
Now let us simplify the given expression
(x3 – 2x2 + 5x – 7) (2x – 3)
2x4 – 4x3 + 10x2 – 14x – 3x3 + 6x2 – 15x + 21
2x4 – 7x3 + 16x2 – 29x + 21
25. (5x + 3) (x – 1) (3x – 2)
Solution:
Now let us simplify the given expression
(5x + 3) (x – 1) (3x – 2)
(5x2 – 2x – 3) (3x – 2)
15x3 – 6x2 – 9x – 10x2 + 4x + 6
15x3 – 16x2 – 5x + 6
26. (5 – x) (6 – 5x) (2 – x)
Solution:
Now let us simplify the given expression
(5 – x) (6 – 5x) (2 – x)
(x2 – 7x + 10) (6 – 5x)
-5x3 + 35x2 – 50x + 6x2 – 42x + 60
60 – 92x + 41x2 – 5x3
27. (2x2 + 3x – 5) (3x2 – 5x + 4)
Solution:
Now let us simplify the given expression
(2x2 + 3x – 5) (3x2 – 5x + 4)
6x4 + 9x3 – 15x2 – 10x3 – 15x2 + 25x + 8x2 + 12x – 20
6x4 – x3 – 22x2 + 37x – 20
28. (3x – 2) (2x – 3) + (5x – 3) (x + 1)
Solution:
Now let us simplify the given expression
(3x – 2) (2x – 3) + (5x – 3) (x + 1)
6x2 – 9x – 4x + 6 + 5x2 + 5x – 3x – 3
11x2 – 11x + 3
29. (5x – 3) (x + 2) – (2x + 5) (4x – 3)
Solution:
Now let us simplify the given expression
(5x – 3) (x + 2) – (2x + 5) (4x – 3)
5x2 + 10x – 3x – 6 – 8x2 + 6x – 20x + 15
-3x2 – 7x + 9
30. (3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)
Solution:
Now let us simplify the given expression
(3x + 2y) (4x + 3y) – (2x – y) (7x – 3y)
12x2 + 9xy + 8xy
12x2 + 9xy + 8xy + 6y2 – 14x2 + 6xy + 7xy – 3y2
-2x2 + 3y2 + 30xy
31. (x2 – 3x + 2) (5x – 2) – (3x2 + 4x – 5) (2x – 1)
Solution:
Now let us simplify the given expression
(x2 – 3x + 2) (5x – 2) – (3x2 + 4x – 5) (2x – 1)
5x3 – 15x2 + 10x – 2x2 + 6x – 4 – (6x3 + 8x2 – 10x – 3x2 – 4x + 5)
5x3 – 6x3 – 15x2 – 2x2 – 5x2 + 16x + 14x – 4 – 5
– x3 – 22x2 + 30x – 9
32. (x3 – 2x2 + 3x – 4) (x – 1) – (2x – 3) (x2 – x + 1)
Solution:
Now let us simplify the given expression
(x3 – 2x2 + 3x – 4) (x – 1) – (2x – 3) (x2 – x + 1)
x4 – 2x3 + 3x2 – 4x – x3 + 2x2 – 3x + 4 – (2x3 – 2x2 + 2x – 3x2 + 3x – 3)
x4 – 3x3 + 5x2 – 7x + 4 – 2x3 + 5x2 – 5x + 3
x4 – 5x3 + 10x2 – 12x + 7

EXERCISE 6.6 PAGE NO: 6.43
1. Write the following squares of binomials as trinomials:
(i) (x + 2)2
(ii) (8a + 3b)2
(iii) (2m + 1)2
(iv) (9a + 1/6)2
(v) (x + x2/2)2
(vi) (x/4 – y/3)2
(vii) (3x – 1/3x)2
(viii) (x/y – y/x)2
(ix) (3a/2 – 5b/4)2
(x) (a2b – bc2)2
(xi) (2a/3b + 2b/3a)2
(xii) (x2 – ay)2
Solution:
(i) (x + 2)2
Let us express the given expression in trinomial
x2 + 2 (x) (2) + 22
x2 + 4x + 4
(ii) (8a + 3b)2
Let us express the given expression in trinomial
(8a)2 + 2 (8a) (3b) + (3b)2
64a2 + 48ab + 9b2
(iii) (2m + 1)2
Let us express the given expression in trinomial
(2m)2 + 2 (2m) (1) + 12
4m2 + 4m + 1
(iv) (9a + 1/6)2
Let us express the given expression in trinomial
(9a)2 + 2 (9a) (1/6) + (1/6)2
81a2 + 3a + 1/36
(v) (x + x2/2)2
Let us express the given expression in trinomial
(x)2 + 2 (x) (x2/2) + (x2/2)2
x2 + x3 + 1/4x4
(vi) (x/4 – y/3)2
Let us express the given expression in trinomial
(x/4)2 – 2 (x/4) (y/3) + (y/3)2
1/16x2 – xy/6 + 1/9y2
(vii) (3x – 1/3x)2
Let us express the given expression in trinomial
(3x)2 – 2 (3x) (1/3x) + (1/3x)2
9x2 – 2 + 1/9x2
(viii) (x/y – y/x)2
Let us express the given expression in trinomial
(x/y)2 – 2 (x/y) (y/x) + (y/x)2
x2/y2 – 2 + y2/x2
(ix) (3a/2 – 5b/4)2
Let us express the given expression in trinomial
(3a/2)2 – 2 (3a/2) (5b/4) + (5b/4)2
9/4a2 – 15/4ab + 25/16b2
(x) (a2b – bc2)2
Let us express the given expression in trinomial
(a2b)2 – 2 (a2b) (bc2) + (bc2)2
a4b2 – 2a2b2c2 + b2c4
(xi) (2a/3b + 2b/3a)2
Let us express the given expression in trinomial
(2a/3b)2 + 2 (2a/3b) (2b/3a) + (2b/3a)2
4a2/9b2 + 8/9 + 4b2/9a2
(xii) (x2 – ay)2
Let us express the given expression in trinomial
(x2)2 – 2 (x2) (ay) + (ay)2
x4 – 2x2ay + a2y2
2. Find the product of the following binomials:
(i) (2x + y) (2x + y)
(ii) (a + 2b) (a – 2b)
(iii) (a2 + bc) (a2 – bc)
(iv) (4x/5 – 3y/4) (4x/5 + 3y/4)
(v) (2x + 3/y) (2x – 3/y)
(vi) (2a3 + b3) (2a3 – b3)
(vii) (x4 + 2/x2) (x4 – 2/x2)
(viii) (x3 + 1/x3) (x3 – 1/x3)
Solution:
(i) (2x + y) (2x + y)
Let us find the product of the given expression
2x (2x + y) + y (2x + y)
4x2 + 2xy + 2xy + y2
4x2 + 4xy + y2
(ii) (a + 2b) (a – 2b)
Let us find the product of the given expression
a (a – 2b) + 2b (a – 2b)
a2 – 2ab + 2ab – 4b2
a2 – 4b2
(iii) (a2 + bc) (a2 – bc)
Let us find the product of the given expression
a2 (a2 – bc) + bc (a2 – bc)
a4 – a2bc + bca2 – b2c2
a4 – b2c2
(iv) (4x/5 – 3y/4) (4x/5 + 3y/4)
Let us find the product of the given expression
4x/5 (4x/5 + 3y/4) – 3y/4 (4x/5 + 3y/4)
16/25x2 + 12/20yx – 12/20xy – 9y2/16
16/25x2 – 9/16y2
(v) (2x + 3/y) (2x – 3/y)
Let us find the product of the given expression
2x (2x – 3/y) + 3/y (2x – 3/y)
4x2 – 6x/y + 6x/y – 9/y2
4x2 – 9/y2
(vi) (2a3 + b3) (2a3 – b3)
Let us find the product of the given expression
2a3 (2a3 – b3) + b3 (2a3 – b3)
4a6 – 2a3b3 + 2a3b3 – b6
4a6 – b6
(vii) (x4 + 2/x2) (x4 – 2/x2)
Let us find the product of the given expression
x4 (x4 – 2/x2) + 2/x2 (x4 – 2/x2)
x8 – 2x2 + 2x2 – 4/x4
(x8 – 4/x4)
(viii) (x3 + 1/x3) (x3 – 1/x3)
Let us find the product of the given expression
x3 (x3 – 1/x3) + 1/x3 (x3 – 1/x3)
x6 – 1 + 1 – 1/x6
x6 – 1/x6
3. Using the formula for squaring a binomial, evaluate the following:
(i) (102)2
(ii) (99)2
(iii) (1001)2
(iv) (999)2
(v) (703)2
Solution:
(i) (102)2
We can express 102 as 100 + 2
So, (102)2 = (100 + 2)2
Upon simplification we get,
(100 + 2)2 = (100)2 + 2 (100) (2) + 22
= 10000 + 400 + 4
= 10404
(ii) (99)2
We can express 99 as 100 – 1
So, (99)2 = (100 – 1)2
Upon simplification we get,
(100 – 1)2 = (100)2 – 2 (100) (1) + 12
= 10000 – 200 + 1
= 9801
(iii) (1001)2
We can express 1001 as 1000 + 1
So, (1001)2 = (1000 + 1)2
Upon simplification we get,
(1000 + 1)2 = (1000)2 + 2 (1000) (1) + 12
= 1000000 + 2000 + 1
= 1002001
(iv) (999)2
We can express 999 as 1000 – 1
So, (999)2 = (1000 – 1)2
Upon simplification we get,
(1000 – 1)2 = (1000)2 – 2 (1000) (1) + 12
= 1000000 – 2000 + 1
= 998001
(v) (703)2
We can express 700 as 700 + 3
So, (703)2 = (700 + 3)2
Upon simplification we get,
(700 + 3)2 = (700)2 + 2 (700) (3) + 32
= 490000 + 4200 + 9
= 494209
4. Simplify the following using the formula: (a – b) (a + b) = a2 – b2 :
(i) (82)2 – (18)2
(ii) (467)2 – (33)2
(iii) (79)2 – (69)2
(iv) 197 × 203
(v) 113 × 87
(vi) 95 × 105
(vii) 1.8 × 2.2
(viii) 9.8 × 10.2
Solution:
(i) (82)2 – (18)2
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
(82)2 – (18)2 = (82 – 18) (82 + 18)
= 64 × 100
= 6400
(ii) (467)2 – (33)2
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
(467)2 – (33)2 = (467 – 33) (467 + 33)
= (434) (500)
= 217000
(iii) (79)2 – (69)2
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
(79)2 – (69)= (79 + 69) (79 – 69)
= (148) (10)
= 1480
(iv) 197 × 203
We can express 203 as 200 + 3 and 197 as 200 – 3
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
197 × 203 = (200 – 3) (200 + 3)
= (200)2 – (3)2
= 40000 – 9
= 39991
(v) 113 × 87
We can express 113 as 100 + 13 and 87 as 100 – 13
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
113 × 87 = (100 – 13) (100 + 13)
= (100)2 – (13)2
= 10000 – 169
= 9831
(vi) 95 × 105
We can express 95 as 100 – 5 and 105 as 100 + 5
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
95 × 105 = (100 – 5) (100 + 5)
= (100)2 – (5)2
= 10000 – 25
= 9975
(vii) 1.8 × 2.2
We can express 1.8 as 2 – 0.2 and 2.2 as 2 + 0.2
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
1.8 × 2.2 = (2 – 0.2) ( 2 + 0.2)
= (2)2 – (0.2)2
= 4 – 0.04
= 3.96
(viii) 9.8 × 10.2
We can express 9.8 as 10 – 0.2 and 10.2 as 10 + 0.2
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
9.8 × 10.2 = (10 – 0.2) (10 + 0.2)
= (10)2 – (0.2)2
= 100 – 0.04
= 99.96
5. Simplify the following using the identities:
(i) ((58)2 – (42)2)/16
(ii) 178 × 178 – 22 × 22
(iii) (198 × 198 – 102 × 102)/96
(iv) 1.73 × 1.73 – 0.27 × 0.27
(v) (8.63 × 8.63 – 1.37 × 1.37)/0.726
Solution:
(i) ((58)2 – (42)2)/16
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
((58)2 – (42)2)/16 = ((58-42) (58+42)/16)
= ((16) (100)/16)
= 100
(ii) 178 × 178 – 22 × 22
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
178 × 178 – 22 × 22 = (178)2 – (22)2
= (178-22) (178+22)
= 200 × 156
= 31200
(iii) (198 × 198 – 102 × 102)/96
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
(198 × 198 – 102 × 102)/96 = ((198)2 – (102)2)/96
= ((198-102) (198+102))/96
= (96 × 300)/96
= 300
(iv) 1.73 × 1.73 – 0.27 × 0.27
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
1.73 × 1.73 – 0.27 × 0.27 = (1.73)2 – (0.27)2
= (1.73-0.27) (1.73+0.27)
= 1.46 × 2
= 2.92
(v) (8.63 × 8.63 – 1.37 × 1.37)/0.726
Let us simplify the given expression using the formula (a – b) (a + b) = a2 – b2
We get,
(8.63 × 8.63 – 1.37 × 1.37)/0.726 = ((8.63)2 – (1.37)2)/0.726
= ((8.63-1.37) (8.63+1.37))/0.726
= (7.26 × 10)/0.726
= 72.6/0.726
= 100
6. Find the value of x, if:
(i) 4x = (52)2 – (48)2
(ii) 14x = (47)2 – (33)2
(iii) 5x = (50)2 – (40)2
Solution:
(i) 4x = (52)2 – (48)2
Let us simplify to find the value of x by using the formula (a – b) (a + b) = a2 – b2
4x = (52)2 – (48)2
4x = (52 – 48) (52 + 48)
4x = 4 × 100
4x = 400
x = 100
(ii) 14x = (47)2 – (33)2
Let us simplify to find the value of x by using the formula (a – b) (a + b) = a2 – b2
14x = (47)2 – (33)2
14x = (47 – 33) (47 + 33)
14x = 14 × 80
x = 80
(iii) 5x = (50)2 – (40)2
Let us simplify to find the value of x by using the formula (a – b) (a + b) = a2 – b2
5x = (50)2 – (40)2
5x = (50 – 40) (50 + 40)
5x = 10 × 90
5x = 900
x = 180
7. If x + 1/x =20, find the value of x2 + 1/ x2.
Solution:
We know that x + 1/x = 20
So when squaring both sides, we get
(x + 1/x)2 = (20)2
x2 + 2 × x × 1/x + (1/x)2 = 400
x2 + 2 + 1/x2 = 400
x2 + 1/x2 = 398
8. If x – 1/x = 3, find the values of x2 + 1/ x2 and x4 + 1/ x4.
Solution:
We know that x – 1/x = 3
So when squaring both sides, we get
(x – 1/x)2 = (3)2
x2 – 2 × x × 1/x + (1/x)2 = 9
x2 – 2 + 1/x2 = 9
x2 – 1/x2 = 9+2
x2 – 1/x2 = 11
Now again when we square on both sides we get,
(x2 – 1/x2)2 = (11)2
x4 – 2 × x2 × 1/x2 + (1/x2)2 = 121
x4 – 2 + 1/x4 = 121
x4 – 1/x4 = 121+2
x4 – 1/x4 = 123
∴ x2 – 1/x2 = 11
x4 – 1/x4 = 123
9. If x2 + 1/x2 = 18, find the values of x + 1/ x and x – 1/ x.
Solution:
We know that x2 + 1/x2 = 18
When adding 2 on both sides, we get
x2 + 1/x2 + 2 = 18 + 2
x2 + 1/x2 + 2 × x × 1/x = 20
(x + 1/x)2 = 20
x + 1/x = √20
When subtracting 2 from both sides, we get
x2 + 1/x2 – 2 × x × 1/x = 18 – 2
(x – 1/x)2 = 16
x – 1/x = √16
x – 1/x = 4
10. If x + y = 4 and xy = 2, find the value of x+ y2
Solution:
We know that x + y = 4 and xy = 2
Upon squaring on both sides of the given expression, we get
(x + y)2 = 42
x2 + y2 + 2xy = 16
x2 + y2 + 2 (2) = 16   (since xy=2)
x2 + y2 + 4  = 16
x2 + y2 = 16 – 4
x2 + y2 =12
11. If x – y = 7 and xy = 9, find the value of x2+y2
Solution:
We know that x – y = 7 and xy = 9
Upon squaring on both sides of the given expression, we get
(x – y)2 = 72
x2 + y2 – 2xy = 49
x2 + y2 – 2 (9) = 49   (since xy=9)
x2 + y2 – 18  = 49
x2 + y2 = 49 + 18
x2 + y2 =67
12. If 3x + 5y = 11 and xy = 2, find the value of 9x+ 25y2
Solution:
We know that 3x + 5y = 11 and xy = 2
Upon squaring on both sides of the given expression, we get
(3x + 5y)2 = 112
(3x)2 + (5y)2 + 2(3x)(5y) = 121
9x2 + 5y2 + 2 (15xy) = 121   (since xy=2)
9x2 + 5y2 + 2(15(2)) = 121
9x2 + 5y2 + 60 = 121
9x2 + 5y2 = 121-60
9x2 + 5y2 = 61
13. Find the values of the following expressions:
(i) 16x2 + 24x + 9 when x = 7/4
(ii) 64x2 + 81y2 + 144xy when x = 11 and y = 4/3
(iii) 81x2 + 16y2 – 72xy when x = 2/3 and y = ¾
Solution:
(i) 16x2 + 24x + 9 when x = 7/4
Let us find the values using the formula (a + b)2 = a2 + b2 + 2ab
(4x)2 + 2 (4x) (3) + 32
(4x + 3)2
Evaluating when x = 7/4
[4 (7/4) + 3]2
(7 + 3)2
100
(ii) 64x2 + 81y2 + 144xy when x = 11 and y = 4/3
Let us find the values using the formula (a + b)2 = a2 + b2 + 2ab
(8x)2 + 2 (8x) (9y) + (9y)2 (8x + 9y)
Evaluating when x = 11 and y = 4/3
[8 (11) + 9 (4/3)]2
(88 + 12)2
(100)2
10000
(iii) 81x2 + 16y2 – 72xy when x = 2/3 and y = ¾
Let us find the values using the formula (a + b)2 = a2 + b2 + 2ab
(9x)2 + (4y)2 – 2 (9x) (4y)
(9x – 4y)2
Putting x = 2/3 and y = 3/4
[9 (2/3) – 4 (3/4)]2
(6 – 3)2
32
9
14. If x + 1/x = 9 find the value of x4 + 1/ x4.
Solution:
We know that x + 1/x = 9
So when squaring both sides, we get
(x + 1/x)2 = (9)2
x2 + 2 × x × 1/x + (1/x)2 = 81
x2 + 2 + 1/x2 = 81
x2 + 1/x2 = 81 – 2
x2 + 1/x2 = 79
Now again when we square on both sides we get,
(x2 + 1/x2)2 = (79)2
x4 + 2 × x2 × 1/x2 + (1/x2)2 = 6241
x4 + 2 + 1/x4 = 6241
x4 + 1/x4 = 6241- 2
x4 + 1/x4 = 6239
∴ x4 – 1/x4 = 6239
15. If x + 1/x = 12 find the value of x – 1/x.
Solution:
We know that x + 1/x = 12
So when squaring both sides, we get
(x + 1/x)2 = (12)2
x2 + 2 × x × 1/x + (1/x)2 = 144
x2 + 2 + 1/x2 = 144
x2 + 1/x2 = 144 – 2
x2 + 1/x2 = 142
When subtracting 2 from both sides, we get
x2 + 1/x2 – 2 × x × 1/x = 142 – 2
(x – 1/x)2 = 140
x – 1/x = √140
16. If 2x + 3y = 14 and 2x – 3y = 2, find value of xy. [Hint: Use (2x+3y) 2 – (2x-3y) 2 = 24xy]
Solution:
We know that the given equations are
2x + 3y = 14… equation (1)
2x – 3y = 2… equation (2)
Now, let us square both the equations and subtract equation (2) from equation (1), we get,
(2x + 3y) 2 – (2x – 3y) 2 = (14)2 – (2)2
4x2 + 9y2 + 12xy – 4x2 – 9y2 + 12xy = 196 – 4
24 xy = 192
xy = 8
∴ the value of xy is 8.
17. If x+ y2 = 29 and xy = 2, find the value of
(i) x + y
(ii) x – y
(iii) x+ y4
Solution:
(i) x + y
We know that

x2 y2 = 29
x2 + y2 + 2xy – 2xy = 29
(x + y) 2 – 2 (2) = 29
(x + y) 2 = 29 + 4
x + y = ± √33
(ii) x – y
We know that
x2 + y2 = 29
x2 + y2 + 2xy – 2xy = 29
(x – y)2 + 2 (2) = 29
(x – y)2 + 4 = 29
(x – y)2 = 25
(x – y) = ± 5

(iii)
x+ y4
We know that
x+ y2 = 29
Squaring both sides, we get
(x2 + y2)2 = (29)2
x4 + y4 + 2x2y2 = 841
x4 + y4 + 2 (2)2 = 841
x4 + y4 = 841 – 8
x4 + y4 = 833
18. What must be added each of the following expression to make it a whole square?
(i) 4x2 – 12x + 7
(ii) 4x2 – 20x + 20
Solution:
(i) 4x2 – 12x + 7
(2x)2 – 2 (2x) (3) + 32 – 32 + 7
(2x – 3)2 – 9 + 7
(2x – 3)2 – 2
∴ 2 must be added to the expression to make it a whole square.
(ii) 4x2 – 20x + 20
(2x)2 – 2 (2x) (5) + 52 – 52 + 20
(2x – 5)2 – 25 + 20
(2x – 5)2 – 5
∴ 5 must be added to the expression to make it a whole square.
19. Simplify:
(i) (x – y) (x + y) (x2 + y2) (x4 + y4)
(ii) (2x – 1) (2x + 1) (4x2 + 1) (16x4 + 1)
(iii) (7m – 8n) 2 + (7m + 8n) 2
(iv) (2.5p – 1.5q) 2 – (1.5p – 2.5q) 2
(v) (m2 – n2m) 2 + 2m3n2
Solution:
(i) (x – y) (x + y) (x2 + y2) (x4 + y4)
B7 grouping the values
(x2 – y2) (x2 + y2) (x4 + y4)
[(x2)2 – (y2)2] (x4 + y4)
(x4 – y4) (x4 – y4)
[(x4)2 – (y4)2]
x8 – y8
(ii) (2x – 1) (2x + 1) (4x2 + 1) (16x4 + 1)
Let us simplify the expression by grouping
[(2x)2 – (1)2] (4x2 + 1) (16x4 + 1)
(4x2 – 1) (4x2 + 1) (16x4 + 1) 1
[(4x2)2 – (1)2] (16x4 + 1) 1
(16x4 – 1) (16x4 + 1) 1
[(16x4)2 – (1)2] 1
256x8 – 1
(iii) (7m – 8n)2 + (7m + 8n)2
Upon expansion
(7m)2 + (8n)2 – 2(7m)(8n) + (7m)2 + (8n)2 + 2(7m)(8n)
(7m)2 + (8n)2 – 112mn + (7m)2 + (8n)2 + 112mn
49m2 + 64n2 + 49m2 + 64n2
By grouping the similar expression we get,
98m2 + 64n2 + 64n2
98m2 + 128n2
(iv) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
Upon expansion
(2.5p)2 + (1.5q)2 – 2 (2.5p) (1.5q) – (1.5p)2 – (2.5q)2 + 2 (1.5p) (2.5q)
6.25p2 + 2.25q2 – 2.25p2 – 6.25q2
By grouping the similar expression we get,
4p2 – 6.25q2 + 2.25q2
4p2 – 4q2
4 (p2 – q2)
(v) (m2 – n2m)2 + 2m3n2
Upon expansion using (a + b) 2 formula
(m2)2 – 2 (m2) (n2) (m) + (n2m) 2 + 2m3n2
m4 – 2m3n2 + (n2m)2 + 2m3n2
m4+ n4m2 – 2m3n2 + 2m3n2
m4+ m2n4
20. Show that:
(i) (3x + 7)2 – 84x = (3x – 7)2
(ii) (9a – 5b)2 + 180ab = (9a + 5b)2
(iii) (4m/3 – 3n/4)2 + 2mn = 16m2/9 + 9n2/16
(iv) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Solution:
(i) (3x + 7)2 – 84x = (3x – 7)2
Let us consider LHS (3x + 7)2 – 84x
By using the formula (a + b)2 = a2 + b2 + 2ab
(3x)2 + (7)2 + 2 (3x) (7) – 84x
(3x)2 + (7)2 + 42x – 84x
(3x)2 + (7)2 – 42x
(3x)2 + (7)2 – 2 (3x) (7)
(3x – 7)2 = R.H.S
Hence, proved
(ii) (9a – 5b)2 + 180ab = (9a + 5b)2
Let us consider LHS (9a – 5b)2 + 180ab
By using the formula (a + b)2 = a2 + b2 + 2ab
(9a)2 + (5b)2 – 2 (9a) (5b) + 180ab
(9a)2 6 (5b)2 – 90ab + 180ab
(9a)2 + (5b)2 + 9ab
(9a)2 + (5b)2 + 2 (9a) (5b)
(9a + 5b)2 = R.H.S
Hence, proved
(iii) (4m/3 – 3n/4)2 + 2mn = 16m2/9 + 9n2/16
Let us consider LHS (4m/3 – 3n/4)2 + 2mn
(4m/3)2 + (3n/4)2 – 2mn + 2mn
(4m/3)2 + (3n/4)2
16/9m2 + 9/16n2 = R.H.S
Hence, proved
(iv) (4pq + 3q)2 – (4pq – 3q)2 = 48pq2
Let us consider LHS (4pq + 3q)2 – (4pq – 3q)2
(4pq)2 + (3q)2 + 2 (4pq) (3q) – (4pq)2 – (3q)2 + 2(4pq)(3q)
24pq2 + 24pq2
48pq2 = RHS
Hence, proved
(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Let us consider LHS (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)
By using the identity (a – b) (a + b) = a2 – b2
We get,
(a2 – b2) + (b2 – c2) + (c2 – a2)
a2 – b2 + b2 – c2 + c2 – a2
0 = R.H.S
Hence, proved

EXERCISE 6.7 PAGE NO: 6.47
1. Find the following products:
(i) (x + 4) (x + 7)
(ii) (x – 11) (x + 4)
(iii) (x + 7) (x – 5)
(iv) (x – 3) (x – 2)
(v) (y2 – 4) (y2 – 3)
(vi) (x + 4/3) (x + 3/4)
(vii) (3x + 5) (3x + 11)
(viii) (2x2 – 3) (2x2 + 5)
(ix) (z2 + 2) (z2 – 3)
(x) (3x – 4y) (2x – 4y)
(xi) (3x2 – 4xy) (3x2 – 3xy)
(xii) (x + 1/5) (x + 5)
(xiii) (z + 3/4) (z + 4/3)
(xiv) (x2 + 4) (x2 + 9)
(xv) (y2 + 12) (y2 + 6)
(xvi) (y2 + 5/7) (y2 – 14/5)
(xvii) (p2 + 16) (p2 – 1/4)
Solution:
(i) (x + 4) (x + 7)
Let us simplify the given expression
x (x + 7) + 4 (x + 7)
x2 + 7x + 4x + 28
x2 + 11x + 28
(ii) (x – 11) (x + 4)
Let us simplify the given expression
x (x + 4) – 11 (x + 4)
x2 + 4x – 11x – 44
x2 – 7x – 44
(iii) (x + 7) (x – 5)
Let us simplify the given expression
x (x – 5) + 7 (x – 5)
x2 – 5x + 7x – 35
x2 + 2x – 35
(iv) (x – 3) (x – 2)
Let us simplify the given expression
x (x – 2) – 3 (x – 2)
x2 – 2x – 3x + 6
x2 – 5x + 6
(v) (y2 – 4) (y2 – 3)
Let us simplify the given expression
y2 (y2 – 3) – 4 (y2 – 3)
y4 – 3y2 – 4y2 + 12
y4 – 7y2 + 12
(vi) (x + 4/3) (x + 3/4)
Let us simplify the given expression
x (x + 3/4) + 4/3 (x + 3/4)
x2 + 3x/4 + 4x/3 + 12/12
x2 + 3x/4 + 4x/3 + 1
x2 + 25x/12 + 1
(vii) (3x + 5) (3x + 11)
Let us simplify the given expression
3x (3x + 11) + 5 (3x + 11)
9x2 + 33x + 15x + 55
9x2 + 48x + 55
(viii) (2x2 – 3) (2x2 + 5)
Let us simplify the given expression
2x2 (2x2 + 5) – 3 (2x2 + 5)
4x4 + 10x2 – 6x2 – 15
4x4 + 4x2 – 15
(ix) (z2 + 2) (z2 – 3)
Let us simplify the given expression
z2 (z2 – 3) + 2 (z2 – 3)
z4 – 3z2 + 2z2 – 6
z4 – z2 – 6
(x) (3x – 4y) (2x – 4y)
Let us simplify the given expression
3x (2x – 4y) – 4y (2x – 4y)
6x2 – 12xy – 8xy + 16y2
6x2 – 20xy + 16y2
(xi) (3x2 – 4xy) (3x2 – 3xy)
Let us simplify the given expression
3x2 (3x2 – 3xy) – 4xy (3x2 – 3xy)
9x4 – 9x3y – 12x3y + 12x2y2
9x4 – 21x3y + 12x2y2
(xii) (x + 1/5) (x + 5)
Let us simplify the given expression
x (x + 1/5) + 5 (x + 1/5)
x2 + x/5 + 5x + 1
x2 + 26/5x + 1
(xiii) (z + 3/4) (z + 4/3)
Let us simplify the given expression
z (z + 4/3) + 3/4 (z + 4/3)
z2 + 4/3z + 3/4z + 12/12
z2 + 4/3z + 3/4z + 1
z2 + 25/12z + 1
(xiv) (x2 + 4) (x2 + 9)
Let us simplify the given expression
x2 (x2 + 9) + 4 (x2 + 9)
x4 + 9x2 + 4x2 + 36
x4 + 13x2 + 36
(xv) (y2 + 12) (y2 + 6)
Let us simplify the given expression
y2 (y2 + 6) + 12 (y2 + 6)
y4 + 6y2 + 12y2 + 72
y4 + 18y2 + 72
(xvi) (y2 + 5/7) (y2 – 14/5)
Let us simplify the given expression
y2 (y2 – 14/5) + 5/7 (y2 – 14/5)
y4 – 14/5y2 + 5/7y2 – 2
y4 – 73/35y2 – 2
(xvii) (p2 + 16) (p2 – 1/4)
Let us simplify the given expression
p2 (p2 – 1/4) + 16 (p2 – 1/4)
p4 – 1/4p2 + 16p2 – 4
p4 + 63/4p2 – 4
2. Evaluate the following:
(i) 102 × 106
(ii) 109 × 107
(iii) 35 × 37
(iv) 53 × 55
(v) 103 × 96
(vi) 34 × 36
(vii) 994 × 1006
Solution:
(i) 102 × 106
We can express 102 as 100 + 2 and 106 as 100 + 6
Now let us simplify
102 × 106 = (100 + 2) (100 + 6)
= 100 (100 + 6) + 2 (100 + 6)
= 10000 + 600 + 200 + 12
= 10812
(ii) 109 × 107
We can express 109 as 100 + 9 and 107 as 100 + 7
Now let us simplify
109 × 107 = (100 + 9) (100 + 7)
= 100 (100 + 7) + 9 (100 + 7)
= 10000 + 700 + 900 + 63
= 11663
(iii) 35 × 37
We can express 35 as 30 + 5 and 37 as 30 + 7
Now let us simplify
35 × 37 = (30 + 5) (30 + 7)
= 30 (30 + 7) + 5 (30 + 7)
= 900 + 210 + 150 + 35
= 1295
(iv) 53 × 55
We can express 53 as 50 + 3 and 55 as 50 + 5
Now let us simplify
53 × 55 = (50 + 3) (50 + 5)
= 50 (50 + 5) + 3 (50 + 5)
= 2500 + 250 + 150 + 15
= 2915
(v) 103 × 96
We can express 103 as 100 + 3 and 96 as 100 – 4
Now let us simplify
103 × 96 = (100 + 3) (100 – 4)
= 100 (100 – 4) + 3 (100 – 4)
= 10000 – 400 + 300 – 12
= 10000 – 112
= 9888
(vi) 34 × 36
We can express 34 as 30 + 4 and 36 as 30 + 6
Now let us simplify
34 × 36 = (30 + 4) (30 + 6)
= 30 (30 + 6) + 4 (30 + 6)
= 900 + 180 + 120 + 24
= 1224
(vii) 994 × 1006
We can express 994 as 1000 – 6 and 1006 as 1000 + 6
Now let us simplify
994 × 1006 = (1000 – 6) (1000 + 6)
= 1000 (1000 + 6) – 6 (1000 + 6)
= 1000000 + 6000 – 6000 – 36
= 999964
Courtesy : CBSE