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Joy of Mathematics Class 8 Solutions Chapter 11 Algebraic Expressions
Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Joy of Mathematics Class 8 Book, Chapter 11 Algebraic Expressions. Here students can easily find step by step solutions of all the problems for Algebraic Expressions. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here students will find solutions for Exercise 11.1, 11.2, 11.3 and 11.4
Algebraic Expressions Exercise 11.1 Solution
Question no – (1)
Solution :
(a) Given, 1/2x2 + 3/7 x + 4
=> The expressions 1/2x2 + 3/7 x + 4 contains variables whose powers are whole numbers and not fractions.
Therefore this is a polynomial.
(b) Given, x3 + x7 + x9
=> The expressions x3 + x7 + x9 contains variables whose powers are whole numbers and not fractions.
Therefore this is a polynomial.
(c) Given, (√x + 2/√x)2
= (√x)2 + 2√x.2/√x + (2/√x)2 [∵ (a + b)2 = a2 + 2ab + b2]
= x + 4 + 4/x
= 4 + x + 4.x-1
The expressions x + 4 + 4x-1 contains negative powers,
Therefore this is not polynomials.
(d) Given, √x + 2/√x + 7
=> x1/2 + 2.x-1/2 + 7
The expressions x1/2 + 2x-1/2 + 7 contains negative power and fractions.
Therefore this is not polynomial.
(e) Given, x-3 + 1/√3
=> The expressions x-3 + 1/√3 contains negative powers.
Therefore this is not polynomial.
(f) Given, x1/2 + x3/2 + x5/2 + 3
=> The expressions x1/2 + x3/2 + x5/2 + 3 contains fractions power.
Therefore this is not polynomial.
(g) Given, (√x + 1/√x)3
= (√x)3 + 3(√x)2.1/√x + 3.√x (1/√x)2 + (1/√x)3
= x3/2 + 3√x + 3/√x + 1/x3/2
∴ This expressions contains negative power and fractions power.
Therefore this is not polynomial.
(h) Given, (x3 – 1)2 – (1 – x)3
= (x3)2 – 2.x3.1 + (1)2 – {13 – 3.x2.1 + 3.x.12 – x3}
= x6 – 2x3 + 1 – 1 + 3x2 – 3x + x3
= x6 – x3 + 3x2 – 3x
∴ This expression contains whole number power and non-negative powers.
Therefore this is a polynomial.
Question no – (2)
Solution :
(a) Given, x3 – 7x
= Degree of the polynomial = 3 [∵ highest power of the variable = 3]
(b) Given, x3 + 3x2 – 7
= Degree of the polynomial = 3 [Highest power of the variable = 3]
(c) Given, 5x2 + 7x + 1
= Degree of the polynomial = 2 [Highest power of the variable = 2]
(d) Given, 1 – 7x – 5x2
= Degree of the polynomial = 2 [Highest power of the variable = 2]
(e) Given, x + x7 + 3x3
= Degree of the polynomial = 7 [Highest power of the variable = 7]
(f) Given, 2x2 + 7x6 – x – 7x5
= Degree of the polynomial = 6 [∵ Highest power of the variable]
(g) Given, 20
= Degree of the polynomial = 0 [There is no variable]
(h) Given, 1/20
= Degree of the polynomial = 0 [There is no variable]
Question no – (3)
Solution :
(a) Every binomial is a polynomial.
= True
(b) A monomial is also a polynomial.
= True
(c) A binomial may have degree 5.
= True
[∵ Binomial contains 2 term not depended in degree]
(d) A polynomial can have negative powers.
= False
[∵ Polynomial always contains non-negative power]
(e) A polynomial can have degree zero.
= False
[∵ 6 this is one type of polynomial and its degree is zero]
(f) The degree of polynomial xy is 2.
= True
[∵ Sum of power of variable = 1 + 1 = 2, therefore this polynomials degree is 2]
Question no – (4)
Solution :
(a) Given, 3x2y
= Numerical coefficient = 3
Literal coefficient = x2y
(b) Given, √5 x2y3z
= Numerical coefficient = √5
∴ Literal coefficient = x2y3z
(c) Given, – 3/2 pq2
= Numerical coefficient = -3/2
∴ Literal coefficient = pq2
Question no – (5)
Solution :
(a) Given, 3x2y, √5/3 x2y
= This is pairs of like terms.
[∵ Literal coefficients are same]
(b) Given, 2/3a2b2, 7p2q2
= This is not pairs of like terms.
Since, literal coefficient is not same.
(c) Given, 17/6a2b2c2, 9a2bc
= This is not the pairs of like terms.
(d) Given, PQR, P × q × r
= This is the pairs of like terms.
Algebraic Expressions Exercise 11.2 Solution :
Question no – (1)
Solution :
(a) x + 5 and 3x – 7
= (x+ 5) + (3x – 7)
= x + 3x + 5 – 7 (By rearranging)
= 4x – 2
∴ Thus the required sum 4x – 2.
(b) 2xy – 5yz, 4xy – 4yz and – 3xy + 8yz
= (2xy – 5yz) + (4xy – 4yz) + (- 3xy + 8yz)
= 2xy + 4xy – 3xy – 5yz – 4yz + 8yz (By rearranging)
= 3xy – yz
∴ Thus the required sum (3xy – yz).
(c) a + b + c, a + b – c, a + c – b and b + c – a
= (a + b + c) + (a + b – c) + (a + c – b) + (b + c – a)
= a + b + c + a + b – c + a + x – b + b + c – a
= a + a + a – a + b + b – b + b + c – c + c + c (By rearranging)
= 2a + 2b + 2c
∴ Thus the required sum 2a + 2b + 2c.
(d) (x2 + y2 + z2), (2x2 – 3y2/5 + 2z2), and (5x2 + 2y2 – 5z2/7)
= (x2 + y2 + z2) + (2x2 – 3y2/5 + 2z2) + (5x2 + 2y2 – 5z2/7)
= x2 + y2 + z2 + 2x2 – 3y2/5 + 2z2 + 5x2 + 2y2 – 5z2/7
= x2 + 2x2 + 5x2 + y2 – 3y2/5 + 2y2 + z2+ 2z2 – 5z2/7 (By rearranging)
= 8x2 + 5y2 – 3y2 + 10y2/5 + 7z2 + 14z2 – 5z2/7
= 8x2 + 12y2/5 + 16z2/7
∴ Thus the required sum 8x2 + 12y2/5 + 16z2/7
(e) 3x + y – 4z/7 – 5, 2x – 3y + z + 5/6 and x + y + 2z- 4
= (3x – y – 4z/7 – 5) + (2x – 3y + z + 5/6) + (x+ y + 2z – 4)
= 3x + y – 4z/7 – 5 + 2x – 3y + z + 5/6 + x + y + 2z- 4
= 3x + 2x + x + y – 3y + y – 4z/7 + z + 2z – 5 + 5/6 – 4 (By rearranging)
= 6x – y – 4z + 7z + 14z/7 – 30 + 5 – 24/6
= 6x – y – 25z/7 – 11/6
∴ Thus the required sum 6x – y – 25z/7 – 11/6
(f) (x2 + xy + y2) + (2x2 – 2xy + y2) + (3x2 + xy/2 – 3y2)
= x2 + xy + y2 + 2x2 – 2xy + y2 + 3x2 + xy/2 – 3y2
= x2 + 2x2 + 3x2 + xy – 2xy + xy/2 + y2 + y2 – 3y2 (By rearranging)
= 6x2 – xy + xy/2 – y2
= 6x2 – 2xy + xy/2 – y2
= 6x2 – 3xy/2 – y2
∴ Thus the required sum 6x2 – 3xy/2 – y2
Question no – (2)
Solution :
(a) x3 + 2y3 – z3 + 6xyz, – x3 + 3y3 + z3 + 9xyz and 2x3 + 3y3 + z3 – 18xyz
= Column method
x3 + 2y3 – z3 + 6xyz
– x3 + 3y3 + z3 + 9xyz
+ 2x3 + 3y3 + z3 – 18xyz
_________________________
2x3 + 8y3 + z3 – 3xyz
∴ Thus the required sum 2x3 + 8y3 + z3 – 3xyz
(b) x2 + y2 + z2 – 2xy + 2yz + 2zx, x2 + y2 + 4z2 + 2xy + 4yz + 4zx and 9x2 + y2 + z2 – 6xy + 2yz – 6zx
= Column method
x2 + y2 + z2 – 2xy + 2yz + 2zx
x2 + y2 + 4z2 + 2xy + 4yz + 4zx
+ 9x2 + y2 + z2 – 6xy + 2yz – 6zx
__________________________________
12x2 + 3y2 + 6z2 – 6xy + 6yz – 0
∴ Thus the required sum 12x2 + 3y2 + 6z2 – 6xy + 6yz
(c) 7x + 8x3 – 3x2 – 5, 3 + 2x2 – 5x3 + 4x and 2x3 – 2x2 + 5x + 1
= Column method
7x + 8x3 – 3x2 – 5
4x – 5x3 + 2x2 + 3
+ 5x+ 2x3 – 2x2 + 1
_________________________
16x + 5x3 – 3x2 – 1
∴ Thus the required sum 16x + 5x3 – 3x2 – 1
(d) 5a4 – 2a3 + 5a2, 6a3 + 2a – 5 and – 5a3 – 2a + 4
= Column method
5a4 – 2a3 + 5a2
+ 6a3 + 2a – 5
+ – 5a3 – a3 + 5a2 + 0 – 1
∴ Thus the required sum 5a4 – a3 + 5a2 – 1
Question no – (3)
Solution :
(a) 2x2 – y + 2z from 3x2 – 2y + 3z
= (3x2 – 2y + 3z) – (2x2 – y + 2z)
= 3x2 – 2y + 3z – 2x2 + y – 2z
= 3x2 – 2x2 – 2y + y + 3z – 2z (By rearranging)
= x2 – y + z
∴ Thus the required difference x2 – y + z
(b) x3 – x2y2 + y3 from x4 – x2y2 + y4
= (x4 – x2y2 + y4) – (x3 – x2y2 + y3)
= x4 – x2y2 + y4 – x3 + x2y2 – y3
= x4 + y4 – x2y2 + x2y2 – x3 – y3 (By rearranging)
= x4 + y4 – 0 – x3 – y3
∴ Thus the required difference x4 + y4 – x3 – y3
(c) – 2a4 + 2a2b2 + b4/5 from – a4 – a3b + 3a2b2/2 – ab3 + 2b2/7
= (- a4 – a3b + 3a2b2/2 – ab3 + 2b2/7) – (2a4 + 2a2b2 + b4/5)
= – a4 – a2b + 3a2b2/2 – ab3 + 2b2/7 – 2a4 – 2a2b2 – b4/5
= – a4 – 2a4 – a2b – ab3 + 3a2b2/2 – 2a2b2 + 2b2/7 – b4/5 (By rearranging)
= – 3a4 – a3b – ab3 + 3a2b2 – 4a2b2/2 + 2b2/7 – b4/5
= – 3a4 – a3b – ab3 – a2b2/2 + 2b2/7 – b4/5
∴ Thus the required difference – 3a4 – a3b – ab3 – a2b2/2 + 2b2/7 – b4/5
Question no – (4)
Solution :
(a) 3a2 + 2b2 – c2 + 2ab + 1 from 5a2 + 7b2 + 2c2 + 5ab + 6
= Column method
5a2 + 7b2 + 2c2 + 5ab + 6
3a2 + 2b2 – c2 + 2ab + 1
(-) (-) (+) (-) (-)
_________________________
2a2 + 5b2 + 3c2 + 3ab + 5
∴ Thus the required difference 2a2 + 5b2 + 3c2 + 3ab + 5
(b) 2x2 + 3y2 + z2 – 2xyz from 3x2 – 2y2 – z2 + 5xyz
= Column method
3x2 – 2xy2 – z2 + 5xyz
2x2 + 3y2 + z2 – 2xyz
(-) (-) (-) (+)
___________________
x2 – 5y2 – 2z2 + 7xyz
∴ Thus the required difference x2 – 5y2 – 2z2 + 7xyz
(c) 5x + 2 – 3x2 + x3 + 3x4 from 5x4 + 2x3 + 2x2 – 3x + 7
= Column method
5x4 + 2x3 + 2x2 – 3x + 7
3x4 + x3 – 3x2 + 5x + 2
(-)(-) (+) (-) (-)
_________________________
2x4 – x3 + 5x2 – 8x + 5
∴ Thus the required difference 2x4 – x3 + 5x2 – 8x + 5
(d) 3x4 – 3xyz + 2xy + 4yz + y4 from 3y4 + 5yz+ 7xy + 2xyz + 5x4
= Column method
3y4 + 5yz + 7xy + 2xyz + 5x4
y4 + 4yz + 2xy – xyz + 3x4
(-) (-) (-) (+) (-)
_________________________
2y4 + 4yz + 5xy + 5xyz + 2x4
∴ Thus the required difference 2y4 + 4yz + 5xy + 5xyz + 2x4
Question no – (5)
Solution :
(a) 5x3 – 2x2 + 3x – 7 – 3x3 + 5x2 – 5x – 3 + 5x + 6
= 5x3 – 3x3 – 2x2 + 5x2 + 3x – 5x + 5x – 7 – 3 + 6 (By rearranging)
= 2x3 + 3x2 + 3x – 4
∴ Thus the required solution 2x3 + 3x2 + 3x – 4
(b) 2a4 – 3a3 + 2a2 – 8 + 5a3 – a + 5a2 + 10 – a4
= 2a4 – a4 – 3a3 + 5a3 + 2a2 + 5a2 – a – 8 + 10 (By rearranging)
= 3a4 + 2a3 + 7a2 – a + 2
∴ Thus the required solution 3a4 + 2a3 + 7a2 – a + 2
(c) 3a2 – 2ab + 4a – 3b + 7ab – 5a + 4a2 + 2b
= 3a2 + 4a2 – 2ab + 7ab + 4a – 5a – 3b + 2b
= 7a2 + 5ab – a – b
∴ Thus the required solution 7a2 + 5ab – a – b.
(d) (5x4 – 12x2y2 + y4) – (x4 – 8x2y2 + 6y4)
= 5x4 – 12x2y2 + y4 – x4 + 8x2y2 – 6y4
= 5x4 – x4 – 12x2y2 + 8x2y2 + y4 – 6y4 (By rearranging)
= 4x4 – 4x2y2 – 5y4
∴ Thus the required solution 4×4 – 4x2y2 – 5y4
(e) (7a2 + 3ab + 4b2) + (a2 – ab) + (- 4a2 + 5b2)
= 7a2 + 3ab + 4b2 + a2 – ab – 4a2 + 5b2
= 7a2 – 4a2 + 4b2 + 5b2 + 3ab – ab (By rearranging)
= 3a2 + 9b2 + 2ab
∴ Thus the required solution 3a2 + 9b2 + 2ab
(f) (3x3 – 4x2 – x + 5) – (x3 – x2 – 3x – 2) + (x3 – 2x2 + 7x + 1)
= 3x3 – 4x2 – x + 5 – x3 + x2 + 3x + 2 + x3 – 2x2 + 7x + 1
= 3x3 + x3 – x3 – 4x2 + x2 – 2x2 – x + 3x + 7x + 5 + 2 + 1 (By rearranging)
= 3x3 – 5x2 + 9x + 8
∴ Thus the required solution 3x3 – 5x2 + 9x + 8
(g) (3abc + 2ca + 2ab) – (- 4abc – 2ca + ab) – (abc + 4ca + ab)
= 3abc + 2ca + 2ab + 4abc + 2ca – ab – abc – 4ca – ab
= 3abc + 4abc – abc + 2ca + 2ab – ab – ab + 4ca [By rearranging]
= 6abc + 8ca
∴ Thus the required solution 6abc + 8ca
Question no – (6)
Solution :
Subtract x2 – y2 from 3 – 5xy/2 – 2y2
= (3 – 5xy/2 – 2y2) – (x2 – y2)
= 3 – 5xy/2 – 2y2 – x2 + y2
= 3 – 5xy/2 – y2 – x2 …….. (i)
Sum of (5xy – x2 – y2) and 2x2 + 3y2/2
= 5xy – x2 – y2 + 2x2 + 3y2/2
= 5xy + x2 + 3y2/2 – y2
= 5xy + x2 + 3y2 – 2y2/2
= 5xy + x2 + y2/2 …….. (ii)
(i) added to (ii)
3 – 5xy/2 – y2 – x2 + 5xy + x2 + y2/2
= 3 – 5xy/2 + 5xy – y2 + y2/2 – x2 + x2 (By rearranging)
= 3 – 5xy + 10xy/2 – 2y2 + y2/2 – 0
= 3 – 15xy/2 – y2/2
∴ Thus the required result 3 – 15xy/2 – y2/2
Question no – (7)
Solution :
Let, the expression be = E
According to the questions,
E + (8x2 – x + 1) = 2x2 + 1
=> E = (2x2 + 1) – (8x2 – x + 1)
= 2x2 + 1 – 8x2 + x – 1
∴ E = – 6x2 + x
∴ Thus the required expression – 6x2 + x
Question no – (8)
Solution :
Sum of (2a + b – c) and (3a – 5b + 7c)
= 2a + b – c + 3a – 5b + 7c
= 2a + 3a + b – 5b – c + 7c (By rearranging)
= 5a – 4b + 6c
take away (4a – 3b + 2c)
(5a – 4b + 6c) – (4a – 3b + 2c)
= 5a – 4b + 6c – 4a + 3b – 2c
= a – b + 4c
∴ Thus the required result a – b + 4c
Question no – (9)
Solution :
Let, the expression be = E
According to the question,
(5a – 2b + c) – E = 3a – b + 4c
= (5a – 2b + c) – (3a – b + 4c) = E
= E = 5a – 2b + c – 3a + b – 4c
= 5a – 3a – 2b + b + c – 4c (By rearranging)
=> E = 2a – b – 3c
∴ Thus the required expression 2a – b – 3c
Question no – (10)
Solution :
Let, the expression be = E
According to the question,
E + (6a2 – 4ab + b2) = 9a2 + 2ab – b2
=> E = (9a2 + ab – b2) – (6a2 – 4ab + b2)
= 9a2 + 2ab – b2 – 6a2 + 4ab – b2
= 9a2 – 6a2 + 2ab + 4ab – b2 – b2 (By rearranging)
=> E = 3a2 + 6ab – 2b2
∴ Thus the required expression 3a2 + 6ab – 2b2
Question no – (11)
Solution :
Sum of (3x2 + 5x – 7) and (5a2 + 2x + 9)
= 3x2 + 5x – 7 + 5x2 + 2x + 9
= 3x2 + 5x2 + 5x + 2x – 7 + 9 (By rearranging)
= 8x2 + 7x + 2 ……. (i)
Now,
Sum of (8x2 – 3x + 8) and (3x2 + 4x + 7)
= 2x2 – 3x + 8 + 3x2 + 4x + 7
= 2x2 + 3x2 – 3x + 4x + 8 + 7 (By rearranging)
= 5x2 + x + 15 ……. (ii)
Subtracted (ii) from (i) we get,
(5x2 + x + 15) – (8x2 + 7x + 2)
= 5x2 + x + 15 – 8x2 – 7x – 2
= – 3x2 – 6x + 13
∴ Thus the required difference – 3x2 – 6x + 12
Question no – (12)
Solution :
Let, the other expression be = E
According to the question,
(x2 – 3y + 4xy + 3y2 – 11) + E = x2 + y2 – {3y – 2 + 5 (7x – 2y2)}
= (x2 – 3y + 4xy + 3y2 – 11) + E = x2 + y2 – {3y + 2 + 5(7x – 2y2)}
= (x2 – 3y + 4xy + 3y2 – 11) + E = x2 + y2 – 3y – 2 – 35x + 10y2
= (x2 – 3y + 4xy + 3y2 – 11) + E = (x2 + 11y2 – 3y – 35x – 2)
=> E = (x2 + 11y2 – 3y – 35x – 2) – (x2 – 3y + 4xy + 3y2 – 11)
=> E = x2 + 11y2 – 3y – 35x – 2 – x2 + 3y – 4xy – 3y2 + 11
=> E = x2 – x2 + 11y2 – 3y2 – 4xy – 3y + 3y – 35x – 2 + 11 (By rearranging)
=> E = 0 + 8y2 – 4xy + 0 – 35x + 9
=> E = 8y2 – 4xy – 35x + 9
∴ Thus the required result 8y2 – 4xy – 35x + 9
Question no – (13)
Solution :
A = 6x2 + 3x- 7
B = 2x2 – 5x + 8
C = x2 + 9x – 6
(a) Given, A + B
Putting the value of A and B
= (6x2 + 3x – 7) + (2x2 – 5x + 8)
= 6x2 + 3x – 7 + 2x2 – 5x + 8
= 6x2 + 2x2 + 3x – 5x – 7 + 8 (By rearranging)
= 8x2 – 2x + 1
∴ A + B = 8x2 – 2x + 1
(b) Given, (A – C)
Putting the value of A and C
= (6x2 + 3x- 7) – (x2 + 9x – 6)
= 6x2 + 3x – 7 – x2 – 9x + 6
= 6x2 – x2 + 3x- 9x – 7 + 6 (By rearranging)
= 5x2 – 6x – 1
∴ A – C = 5x2 – 6x – 1
(c) Given, B – C
Putting the value of B and C
= (2x2 – 5x + 8) – (x2 + 9x – 6)
= 2x2 – 5x + 8 – x2 – 9x + 6
= 2x2 – x2 – 5x – 9x + 8 + 6 (By rearranging)
= x2 – 14x + 14
∴ B – C = x2 – 14x + 14
Question no – (14)
Solution :
a = x – 3y
b = 2x + 3y
and c = – 3x + 7
Now,
L.H.S. a + b + c
= (x – 3y) + (2x + 3y) + (- 3x + 7)
= x – 3y + 2x + 3y – 3x + 7
= 3x – 3x – 3y + 3y + 7
= 7 (R.H.S.)
∴ L.H.S. = R.H.S. (Proved)
Algebraic Expressions Exercise 11.3 Solution :
Question no – (1)
Solution :
(a) Given, 7xy2 by 3x2y2
= (7xy2) × (3x2y2)
= 21x3y4
∴ Thus the required result = 21x3y4
(b) Given, 8/7 x3y2 by 21/16 x2y
= 8/7x3y2 × 21/16 x2y
= 3/2 x5y3
∴ Thus the required result = 3/2 x5y3
(c) Given, 3x2y3 by 7x3y2
= 3x2y3 × 7x3y2
= 21x5y5
∴ Thus the required result = 21x5y5
(d) Given, 5/12xy by 4/15 x2y3
= 5/12xy × 4/15x2y3
= 1/9x3y4
∴ Thus the required result = 1/9x3y4
(e) Given, 5x2y3 by 2x3y2
= 5x2y3 × 2x3y2
= 10x5y5
∴ Thus the required result 10x5y5
(f) Given, 2x3y3 by 5x2y5
= 2x3y3 × 5x2y5
= 10x5y8
∴ Thus the required result 10x5y8
(g) Given, – 7xy2z by – 2xy3z2
= – 7xy2z × (- 2xy3z2)
= 14x2y5z3
∴ Thus the required result = 14x2y5z3
(h) Given, 9xyz2 by (- 3x2yz)
= 9xyz2 × (- 3x2yz)
= – 27x3y2z3
∴ Thus the required result = – 27x3y2z3
Question no – (2)
Solution :
(a) Given, – 3/5xy2 × 25x2y × (- 3 xyz2/10)
= – 3/5 × 25 × (- 3/10) × xy2 × x2y × xyz2
= 9/2 x4y4z2
∴ Thus the required product = 9/2x4y4z2
(b) Given, (- 3/5) × (2/7ab) × (- 14/3a2b)
= (- 3/5) × (2/7) × (- 14/3) × ab × a2b
= 4/5 a3b2
∴ Thus the required product = 4/5a3b2
(c) Given, (- 1/36x2y2) × (- 9/2x2yz3)
= – 1/36) × (- 9/2) × x2y2 × x2yz3
= 1/8 x4y3z3
∴ Thus the required product = 1/8x4y3z3
Question no – (3)
Solution :
(a) Given, – 8x/11 (2x + 33)
= – 8x/11 × 2x – 8x/11 × 33
= – 16x/11 – 24x
∴ Thus the required result – 16x/11 – 24x
(b) Given, 5x2y/9 (3xy + 7)
= 5x2y/9 × 3xy + 5x2y/9 × 7
= 5/3 x3y2 + 35x2y/9
∴ Thus the required result 5/3x3y2 + 35x2y/9
(c) Given, – 7xy2/12 (3x + 4y)
= – 7xy2/12 × 3x – 7xy2/12 × 4y
= – 7/4x2y2 – 7/3xy3
∴ Thus the required result – 7/4x2y2 – 7/3xy3
(d) Given, – 7/5x (x + 10)
= 7/5x × x – 7/5 × 10
= – 7x2/5 – 14
∴ Thus the required result – 7x2/5 – 14
(e) Given, 4xy/7 (3x + 14y)
= 4xy/7 × 3x + 4xy/7 × 14y
= 12x2y + 8xy2
∴ Thus the required result = 12x2y + 8xy2
(f) Given, abc (bc + ca – ab)
= abc × bc + abc × ca – abc × ab
= ab2c + a2bc2 – a2b2c
∴ Thus the required result = ab2c + a2bc2 – a2b2c
Question no – (4)
Solution :
(a) Given, (x + y) × (s + t)
= xs + xt + ys + yt
∴ Thus the required product xs + xt + ys + yt
(b) Given, (4x + 3y) × (z + 5)
= 4xz + 20x + 3yz + 15y
∴ Thus the required product 4xz + 20x + 3yz+ 15y
(c) Given, (a + b) × (2c – 3d)
= 2ac – 3ad + 2bc – 3db
∴ Thus the required product 2ac – 3ad + 2bc – 3db
(d) Given, (3a2 + b2) × (c2 + 3d2)
= 3a2c2 + 9a2d2 + b2c2 + 3b2d2
∴ Thus the required product = 3a2c2 + 9a2d2 + b2c2 + 3b2d2
(e) Given, (5x – 3y) × (4 – z)
= 20x – 5xz – 12y + 3yz
∴ Thus the required product 20x – 5xz – 12y + 3yz
(f) Given, (7ab2 – 2b) × (3 – 2cb)
= 21ab2 – 14ab3c – 6b + 4cb2
∴ Thus the required product 21ab2 – 14ab3c – 6b + 4cb2
Question no – (5)
Solution :
(a) Given, (9x2 – y2) × (2x + y)
= 10x3 + 5x2y – 2xy2 – y3
∴ Thus the required product 10x3 + 5x2y – 2xy2 – y3
(b) Given, {(4/7x2) – (2/5y2)} × {14y – 15x}
= 4/7x2 × 14y – 4/7x2 × 15x – 2/5y2 × 14y + 2/5y2 × 15y
= 8x2y – 60/7x3 – 28/5y3 + 6xy2
∴ Thus the required product 8x2y – 60/7x3 – 28/5y3 + 6xy
(c) Given, (3x2 – 2y2) × (2a + 3b)
= 6x2a + 9x2b – 4y2a – 6by2
∴ Thus the required product 6x2a + 9x2b – 4y2a – 6by2
(d) Given, (x + y) × (x – y)
= x2 – xy + xy – y2
= x2 – y2
∴ Thus the required product x2 – y2
(e) Given, (3/5x2 – 1/3y) × (15x – y)
= 3/5x2 × 15x – 3/5x2 × y – 1/3y × 15x + 1/3y × y
= 9x3 – 3/5x2y – 5xy + 1/3y2
∴ Thus the required product 9x3 – 3/5x2y – 5xy + 1/3y2
(f) Given, (x2 + y2) × (x2 – y2)
= x4 – x2y2 + x2y2 – y4
= x4 – y4
∴ Thus the required product x4 – y4
Question no – (6)
Solution :
(a) Given, (2x2 + 7x – 3) × (x + 3)
= 2x2 × x + 2x2 × 3 + 7x × x + 7x × 3 – 3 × x – 3 × 3
= 2x3 + 6x2 + 7x2 + 21x – 3x – 9
= 2x3 + 13x2 + 18x – 9
∴ Thus the required product 2x3 + 13x2 + 18x + 9
(b) Given, (3x2 – 5x – 7) × (x – 2)
= 3x2 × x – 3x2 × 2 – 5x × x + 5x × 2 – 7 × x + 7 × 2
= 3x3 – 6x2 – 5x2 + 10x – 7x + 14
= 3x3 – 11x2 + 3x + 14
∴ Thus the required product 3×3 – 11x2 + 3x + 14
(c) Given, (5x2 – 7x + 2) × (x + 5)
= 5x2 × x + 5x2 × 5 – 7x × x – 7x × 5 + 2 × x + 2 × 5
= 5x3 + 25x2 – 7x2 – 35x + 2x + 10
= 5x3 + 18x2 – 33x + 10
∴ Thus the required product 5x3 + 18x2 – 33x + 10
(d) Given, (a – b + 3c) × (a + 2b)
= a × a + a × 2b – b × a – b × 2b + 3c × a + 3c × 2b
= a2 + 2ab – ab – 2b2 + 3ac + 6bc
= a2 + ab – 2b2 + 3ac + 6bc
∴ Thus the required product a2 + ab – 2b2 + 3ac + 6bc
(e) Given, (x2 – 3x + 8) × (x – 5)
= x2 × x – x2 × 5 – 3x × x + 3x × 5 + 8 × x – 8 × 5
= x3 – 5x2 – 3x2 + 15x + 8x – 40
= x3 – 8x2 + 23x – 40
∴ Thus the required result x3 – 8x2 + 23x – 40
(f) Given, (x2 + 2x – 3) × (1 – x)
= x2 × 1 – x2 × x + 2x × 1 – 2x × x – 3 × 1 + 3 × x
= x2 – x3 + 2x – 2x2 – 3 + 3x
= – x3 – x2 + 5x – 3
∴ Thus the required product – x3 – x2 + 5x – 3
Question no – (7)
Solution :
(a) Given, (x2 + 2x + 1) × (x2 – 3x + 7)
= x2 × x2 – x2 × 3x + x2 × 7 + 2x × x2 – 2x × 3x + 2x × 7 + 1 × x2 – 1 × 3x + 1 × 7
= x4 – 3x3 + 7x2 + 2x3 – 6x2 + 14x + x2 – 3x + 7
= x4 – 3x3 + 2x3 + 7x2 – 6x2 + x2 + 14x – 3x + 7 (By rearranging)
= x4 – x3 + 2x2 + 11x + 7
∴ Thus the required product x4 – x3 + 2x2 + 11x + 7
(b) Given, (2x2 + 3x + 7) × (x2 + 7x + 5)
= 2x2 × x2 + 2x2 × 7x + 2x2 × 5 + 3x × x2 + 3x × 7x + 3x × 5 + 7x2 + 7 × 7x + 7 × 5
= 2x4 + 14x3 + 10x2 + 3x3 + 21x2 + 15x + 7x2 + 49x + 35
= 2x4 + 14x3 + 3x3 + 10x2 + 21x2 + 7x2 + 15x + 49x + 35 (By rearranging)
= 2x4 + 17x3 + 38x2 + 64x + 35
∴ Thus the required product 2x4 + 17x3 + 38x2 + 64x + 35
(c) Given, (x2 – 3x + 2) × (x2 – 5x + 6)
= x2 × x2 – x2 × 5x + x2 × 6 + 3x × 5x – 3x × x2 – 3x × 6 + 2 × x2 – 2 × 5x + 2 × 6
= x4 – 5x3 + 6x2 + 15x2 – 3x3 – 18x + 2x2 + 10x + 12
= x4 – 5x3 – 3x3 + 6x2 + 15x2 + 2x2 – 18x + 10x + 12 (By rearranging)
= x4 – 8x3 + 23x2 – 8x + 12
∴ Thus the required product x4 – 8x3 + 23x2 – 8x + 12
(d) Given, (5x2 + 7x + 2) (2x2 + 3x + 8)
= 5x2 × 2x2 + 5x2 × 3x+ 5x2 × 8 + 7x × 2x2 + 7x × 3x + 7x × 8 + 2 × 2x2 + 2 × 3x + 2 × 8
= 10x4 + 15x3 + 40x2 + 14x3 + 21x2 + 56x + 4x2 + 6x + 16
= 10x4 + 15x3 + 14x3 + 40x2 + 21x2 + 4x2 + 6x + 56x + 16 (By rearranging)
= 10x4 + 29x3 + 65x2 + 62x + 16
∴ Thus the required product 10x4 + 29x3 + 65x2 + 62x + 16
Question no – (8)
Solution :
(a) Given, (x3 – y3) × (x – y)
= x3 × x – x3 × y – y3 × x + y3 × y
= x4 – x3y – xy3 + y4
∴ Thus the required product x4 – x3y – xy3 + y4
(b) Given, (0.6x – 0.4y) × (0.5x – 0.2y)
= 0.6x × 0.5x – 0.6x × o.2y – 0.4y × 0.5x + 0.4y
= 0.30x2 – 0.12xy – 0.20xy + 0.8y2
= 0.30x2 – 0.32xy + 0.08y2
∴ Thus the required product 0.30x2 – 0.32xy + 0.08y2
(c) Given, (x4 – y4) × (x2 + y2)
= x4 × x2 + x4 × y2 – y4 × x2 – y4 × y2
= x6 + x4y2 – x2y4 – y6
∴ Thus the required product x6 + x4y2 – x2y4 – y6
(d) Given, (2.5x + 0.5y) × (0.5x + 0.2y)
= 2.5x × 0.5x + 2.5x × 0.2y + 0.5y × 0.5x + 0.5y × 0.2y
= 1.25x2 + 0.50xy + 0.25xy + 0.10y2
= 1.25x2 + 0.75xy + 0.10y2
∴ Thus the required product 1.25x2 + 0.75xy + 0.10y2
Algebraic Expressions Exercise 11.4 Solution :
Question no – (1)
Solution :
(a) 9x4 by 3x2
= 9x4/3x2
= 9/3 × x4-2
= 3x2
(b) 25x3y2 by 35y
= 25/35 x3y2/y
= 5/7 x3y2-1
= 5/7x3y
(c) 10a6 by – 5a4
= 10a6/- 5a4
= – 2a6-4
= – 2a2
(d) 12x3 by 16x
= 12/16 x3/x
= 3/4 x3-1
= 3/4x2
(e) – 25x9 by – 5x4
= – 25/- 5 × x9/x4
= 5x9-4
= 5x5
(f) 39m5 by 13m3
= 39/13 × m5/m3
= 3m5-3
= 3m2
Question no – (2)
Solution :
(a) 6x2 + 12x by 6x
= 6x2 + 12x/6x
= 6x2/6x + 12x/6x
= 1x2-1 + 2
= x + 2
(b) 25m7 – 15m3 + 5m2 by 5m2
= 25m7 – 15m3 + 5m2/5m2
= 25m7/5m2 – 15m3/5m2 + 5m2/5m2
= 5m7-2 – 3m3-2 + 1m2-2
= 5m5 – 3m + m0
= 5m5 – 3m + 1 [∵ a0 = 1]
(c) 3x3 – 12x5 + 6x2 by – 6x
= 3x3 – 12x5 + 6x2/- 6x
= 3x3/-6x – 15x5/- 6x + 6x2/6x
= – 1/2 x3-1 + 5/2x5-1 + x2-1
= – 1/2x2 + 5/2x4 + x
(d) 8a2 + 16a4 – 12a3 by 4a2
= 8a2 + 16a4 – 12a3/4a2
= 8a2/4a2 + 16a4/4a2 – 12a3/4a2
= 2a2-2 + 4a4-2 – 3a3-2
= 2a0 + 4a2 – 3a
= 2 + 4a2 – 3a [∵ a0 = 1]
(e) 6x7 – 12x5 + 36x3 by 6x2
= 6x7 – 12x5 + 36x3/6x2
= 6x7/6x2 – 12x5/6x2 + 36x3/6x2
= 1x7-2 – 2x5-2 + 6x3-2
= x5 – 2x3 + 6x
(f) 3z5 + z7 – 24z3 + 12z2 by – 3z2
= 3z5 + z7 – 24z3 + 12z2/- 3z2
= z5/- 3z2 + z7/- 3z2 – 24z3/- 3z2 + 12z2/- 3z2
= – z5-2 – 1/3z7-2 + 8z3-2 – 4z2-2
= – z3 – 1/3z5 + 8z – 4z0
= – z3 – 1/3z5 + 8z – 4z [∵ z0 = 1]
Question no – (3)
Solution :
(a) x3 + x2 – 5x – 2 by x – 2
∴ Thus the required quotient x2 + 3x + 1
(b) 2 – 2x3 – x + x4 by x – 2
= x4 – 2x3 – x + 2 by x – 2 (By rearranging)
∴ Thus the required quotient x3 – 1
(c) 2x3 – 9x2 + 3x + 14 by x – 2
∴ Thus the required quotient 2x2 – 5x – 7
(d) x3 – 7x2 + 10 by x – 2
∴ Thus the required quotient x2 – 5x
(e) x4 – 3x2 + 13x – 2x3 – 14 by x – 2
= x4 – 2x3 – 3x2 + 13x – 14 by (x – 2) (By rearranging)
∴ Thus the required quotient x3 – 3x + 7
(f) x3 – 8x + 8 by x – 2
Question no – (4)
Solution :
(a) 2x3 + 19x2 + 6x + 63 by x + 4
∴ Thus the required remainder 215 verifying division algorithm,
(2x2 + 11x – 38) (x + 4) + 215
= 2x2 × x + 2x2 × 4 + 11x × x + 11x × 4 – 38 × x – 38 × 4 + 215
= 2x3 + 8x2 + 11x2 + 44x – 38x – 152 + 215
= 2x3 + 19x2 + 6x + 63 (Verified)
(b) 7x3 + 13x2 – 16x + 23 by 7x – 1
∴ Thus the required remainder = 21
Now, 7x3 + 13x2 – 16x + 23 = (x2 + 2x – 2) (7x – 1) + 21
= x2 × 7x – x2 × 1 + 2x × 7x – 2x × 1 – 2 7x + 2 × 1 + 21
= 7x3 – x2 + 14x2 – 2x – 14x + 2 + 21
= 7x3 + 13x2 – 16x + 23
∴ Division algorithm verified.
(c) 6x6 – 3x2 + 7x – 152 by x2 – 3
= In order to make division easier, fill in the missing x5, x4, x3 terms in the dividend.
So we write 6x6 – 3x2 + 7x – 152 as 6x6 + 0x5 + 0x4 + 0x3 – 3x2 + 7x – 152
(d) x4 + 4x3 + 6x2 + 2 by x + 1
In order to make division easier fill in the missing x terms in the divided,
So we write x4 + 4x3 + 6x2 + 2 as x4 + 4x3 + 6x2 + 0x + 2
∴ Thus remainder = 5
∴ x4 + 4x3 + 6x2 + 0x + 2 = (x3 + 3x2 + 3x – 3) (x + 1) + 5
= x3 × x + x3 × 1 + 3x2 × x + 3x2 × 1 + 3x × x + 3x × 1.3 × x – 3 × 1 + 5
= x4+ x3 + 3x3 + 3x2 + 3x2 + 3x – 3x – 3 + 5
= x4 + 4x3 + 6x2 + 0x + 2
∴ Thus verified division algorithm.
Question no – (5)
Solution :
(a) (x – 3) is a factor of x2 + 2x – 15
Since, the remainder is zero, (x – 3) is a factor of a polynomial x2+ 2x – 15
(b) 2x + 3 is a factor of 2x2 – 11x – 21
Since, the remainder is zero, (2x + 3) is a factor of the polynomial 2x2 – 11x – 21
(c) x + 2 is a factor of 2x2 + 3x + 22
Using division algorithm
2x2 + 3x + 22 = (2x – 1) (x + 2) + 24
= 2x2 + 4x – x – 2 + 24
= 2x2 + 3x + 22
∴ Therefore (x + 2) is a factor of the polynomial 2x2 + 3x + 22
(d) x2 + 3 is factor of x5 – 9x
= In order to make division easier, fill in the missing x4, x3, x2 terms in the dividend.
So we write x5 – 9x = x5 + 0x4 + 0x3 + 0x2 – 9x + 0
Since, the remainder is zero, x2 + 3 is a factor of a polynomial x5 – 9x
(e) x -2 is a factor of 15x3 – 20x2 + 13x – 12
Using division algorithm,
15x3 – 20x2 + 13x – 12
= (15x2 + 10x + 33) (x – 2) + 54
= 15x2 × x – 15x2 × 2 + 10x × x – 10x × 2 + 33 × x – 33 × 2 + 54
= 15x3 – 30x2 + 10x2 – 20x2 + 33x – 66 + 54
= 15x3 – 20x2 + 13x – 12
∴ Therefore (x – 2) is a factor of the polynomial 15x3 – 20x2 + 13x – 12
(f) 3y2 + 5 is a factor of 12y5 + 17y3 + 9y2 – 5y + 12
= In order to make division easier fill in the missing x4, terms in the dividend.
So we write 12y5 + 17y3 + 9y2 – 5y + 12 – 12y5 + 0y4 + 17y3 + 9y2 – 5y + 12
Using division algorithm,
12y5 + 0y4 + 17y3 + 9y2 – 5y + 2 = (3y2 + 0y + 5) (4y3 – y + 3) + 2
= 3y2 × 4y3 – 3y2 × y + 0y × 4y3 – 0y × y + 5 × 4y3 – 5 × y + 9y2 + 15 – 13
= 12y5 – 3y3 + 20y3 – 5y + 9y2 + 2
= 12y5 + 17y3 + 9y2 – 5y + 2
∴ Thus (3y2 + 5) is a factor of a polynomial 12y5 + 0y4 + 17y3 + 9y2 – 5y + 2
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