As students get familiar with numbers, their problem-solving and analytical skills improve considerably. We know that algebra plays an essential role in Maths. The algebraic expressions used in algebra consist of variables and basic operations such as addition, subtraction, multiplication and division. These operations are performed using certain laws and basic formulas which have to be remembered. Here, in RD Sharma Solutions for Class 8 Maths Chapter 6 – Algebraic Expressions and Identities, such problems are solved. Our expert team have solved the questions in a step-by-step format, which helps the students to understand the concepts better. Moreover, practising RD Sharma Class 8 Solutions will help students secure excellent scores in the annual exam.
Chapter 6 – Algebraic Expressions and Identities contains seven exercises, and the RD Sharma Solutions available on this page provide solutions for the questions in each exercise. Now, let us have a look at the concepts discussed in this chapter.
- Review of concepts and definitions
- Addition, subtraction and multiplication of algebraic expressions
- Multiplication of two monomials
- Multiplication of a monomial and a binomial
- Multiplication of two binomials
- Identities
RD Sharma Solutions for Class 8 Maths Chapter 6 Algebraic Expressions and Identities
Access Answers to RD Sharma Solutions for Class 8 Maths Chapter 6 – Algebraic Expressions and Identities
EXERCISE 6.1 PAGE NO: 6.2
1. Identify the terms and 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 and trinomials. Which polynomials do not fit into 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
The given expression contains one term, 1000
∴ It is Monomial
The given expression contains four terms
∴ It belongs to none of the categories
The given expression contains three terms
∴ It is Trinomial
The given expression contains two terms
∴ It is Binomial
The given expression contains three terms
∴ It is Trinomial
The given expression contains three terms
∴ It is Trinomial
The given expression contains two terms
∴ It is Binomial
The given expression contains four terms
∴ It belongs to none of the categories
The given expression contains one term
∴ It is Monomial
The given expression contains two terms
∴ It is Binomial
The given expression contains two terms
∴ It is Binomial
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
8a2b
(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
12x2y – 3x2y – 7x2y – 6xy2 + 5xy2 + 4xy2
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) 2/3y3 – 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
(-7a2) – 2a2
-7a2 – 2a2
-9a2
(iii) 2a-b from 3a-5b
Let us subtract the given expression
(3a – 5b) – (2a – b)
3a – 5b – 2a + b
a – 4b
(iv) 2x3 – 4x2 + 3x + 5 from 4x3 + x2 + x + 6
Let us subtract the given expression
(4x3 + x2 + x + 6) – (2x3 – 4x2 + 3x + 5)
4x3 + x2 + x + 6 – 2x3 + 4x2 – 3x – 5
2x3 + 5x2 – 2x + 1
(v) 2/3y3 – 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), and (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), and (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), and (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), and 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 – x2 + 5 and -4x – 3 + 7x2 from 5.
Solution:
Sum of 2x – x2 + 5 and -4x – 3 + 7x2 is
2x – x2 + 5 + (-4x – 3 + 7x2)
2x – x2 + 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 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
Verification
LHS = (3 × 1) × (4 × 1) × (-5 × 1)
= 3 × 4 × – 5
= – 60
RHS = -60 (1)3 = – 60
Therefore, LHS = RHS.
19. (4x2) × (-3x) × (4/5x3)
Solution:
Let us simplify the given expression
4 × -3 × 4/5 × x2 × x × x3
-48/5 × x2+1+3
-48/5x6
Verification
LHS = 4 × 12 × – 3 × 1 × 4/5 × 13
= – 48/5
RHS = – 48/5 × 16 = – 48/5
Therefore, LHS = RHS.
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
Verification
LHS = (5 × 14) × (12)3 × (2 × 1)2
= 5 × 4
= 20
RHS = 20 × 112 = 20
Therefore, LHS = RHS.
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
Verification
LHS = (12)3 × (2 × 1) × (-4 × 1) × 5
= – 40
RHS = – 40 × 18 = – 40
Therefore, LHS = RHS.
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
160 × 15.625
2500
For -8x2y6 and -20xy
-8 × 2.52 × 16 × -20 × 1 × 2.5
2500
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 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 × 17 × 27
5/32 × 128
5 × 4
20
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, and 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. 2a3 (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. -11y2 (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 (y2 + 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 (y2 + 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) – 6a2 (b-b2) -3b2 (2a2 -a) + 2ab (b-a)
(viii) x2 (x2 + 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) – 6a2 (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 (x2 + 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 – 3/2x2 + 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
9/8x3 + 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)2 = (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 = 119
∴ x2 + 1/x2 = 11
x4 + 1/x4 = 119
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 x2 + 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 9x2 + 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 + 25y2 + 2 (15xy) = 121 (since xy=2)
9x2 + 25y2 + 2(15(2)) = 121
9x2 + 25y2 + 60 = 121
9x2 + 25y2 = 121-60
9x2 + 25y2 = 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 x2 + y2 = 29 and xy = 2, find the value of
(i) x + y
(ii) x – y
(iii) x4 + 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) x4 + y4
We know that
x2 + 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 to each of the following expressions 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)
By 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] 1256x8 – 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
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