Math Ncert Exemplar 2019 Solutions for Class 8 Maths Chapter 7 Algebraic Expression, Identities & Factorisation are provided here with simple step-by-step explanations. These solutions for Algebraic Expression, Identities & Factorisation are extremely popular among class 8 students for Maths Algebraic Expression, Identities & Factorisation Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Math Ncert Exemplar 2019 Book of class 8 Maths Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Math Ncert Exemplar 2019 Solutions. All Math Ncert Exemplar 2019 Solutions for class 8 Maths are prepared by experts and are 100% accurate.
Page No 224:
Question 1:
In questions ​there are four options out of which one is correct.
The product of a monomial and a binomial is a
(a) monomial
(b) binomial
(c) trinomial
(d) none of these
Answer:
Monomial consists of one term and Binomial consists of two terms. So, the multiplication of a binomial by a monomial will always be a binomial.
For example: Let 2x be a monomial and x – y be a binomial then their product is a binomial.
Hence, the correct answer is option (b).
Page No 224:
Question 2:
In questions ​there are four options out of which one is correct.
In a polynomial, the exponents of the variables are always
(a) integers
(b) positive integers
(c) non-negative integers
(d) non-positive integers
Answer:
In a polynomial, the exponent of the variables can be 0 or any positive integer.
For example: is a polynomial and is also = polynomial.
Thus, the exponents of the variables in a polynomial are always non-negative integers.
Hence, the correct answer is option (c).
Page No 224:
Question 3:
In questions ​there are four options out of which one is correct.
Which of the following is correct?
(a) (a – b)2 = a2 + 2ab – b2
(b) (a – b)2 = a2 – 2ab + b2
(c) (a – b)2 = a2 – b2
(d) (a + b)2 = a2 + 2ab – b2
Answer:
Hence, the correct answer is option (b).
Page No 224:
Question 4:
In questions ​there are four options out of which one is correct.
The sum of –7pq and 2pq is
(a) –9pq
(b) 9pq
(c) 5pq
(d) – 5pq
Answer:
The sum of –7pq and 2pq = –7pq + 2pq
= pq (–7 + 2)
= – 5 pq
Hence, the correct answer is option (d).
Page No 224:
Question 5:
In questions ​there are four options out of which one is correct.
If we subtract –3x2y2 from x2y2, then we get
(a) –4x2y2
(b) –2x2y2
(c) 2x2y2
(d) 4x2y2
Answer:
Hence, the correct answer is option (d).
Page No 224:
Question 6:
In questions ​there are four options out of which one is correct.
Like term as 4m3n2 is
(a) 4m2n2
(b) –6m3n2
(c) 6pm3n2
(d) 4m3n
Answer:
As like terms contain the same literal factor. So the like term as because it contains same literal factor i.e.
Hence, the correct answer is option (b).
Page No 225:
Question 7:
In questions ​there are four options out of which one is correct.
Which of the following is a binomial?
(a) 7 × a + a
(b) 6a2 + 7b + 2c
(c) 4a × 3b × 2c
(d) 6 (a2 + b)
Answer:
Binomials are algebraic expressions consisting of two unlike terms.
is a binomial as it consists of two unlike terms i.e.,
Hence, the correct answer is option (d).
Page No 225:
Question 8:
In questions ​there are four options out of which one is correct.
Sum of a – b + ab, b + c – bc and c – a – ac is
(a) 2c + ab – ac – bc
(b) 2c – ab – ac – bc
(c) 2c + ab + ac + bc
(d) 2c – ab + ac + bc
Answer:
Hence, the correct answer is option (a).
Page No 225:
Question 9:
In questions ​there are four options out of which one is correct.
Product of the following monomials 4p, –7q3, –7pq is
(a) 196 p2q4
(b) 196 pq4
(c) –196 p2q4
(d) 196 p2q3
Answer:
Hence, the correct answer is option (a).
Page No 225:
Question 10:
In questions ​there are four options out of which one is correct.
Area of a rectangle with length 4ab and breadth 6b2 is
(a) 24a2b2
(b) 24ab3
(c) 24ab2
(d) 24ab
Answer:
Hence, the correct answer is option (b).
Page No 225:
Question 11:
In questions ​there are four options out of which one is correct
Volume of a rectangular box (cuboid) with length = 2ab, breadth = 3ac and height = 2ac is
(a) 12a3bc2
(b) 12a3bc
(c) 12a2bc
(d) 2ab +3ac + 2ac
Answer:
Volume of a rectangular box (cuboid)
= Length × Breadth × Height
= 2 × 3 × 2 × a × b × a × c × a × c
= 12 × a3 × b × c2
= 12a3bc2
Hence, the correct answer is option (a).
Page No 226:
Question 12:
In questions ​there are four options out of which one is correct
Product of 6a2 – 7b + 5ab and 2ab is
(a) 12a3b – 14ab2 + 10ab
(b) 12a3b – 14ab2 + 10a2b2
(c) 6a2 – 7b + 7ab
(d) 12a2b – 7ab2 + 10ab
Answer:
Hence, the correct answer is option (b).
Page No 226:
Question 13:
In questions ​there are four options out of which one is correct
Square of 3x – 4y is
(a) 9x2 – 16y2
(b) 6x2 – 8y2
(c) 9x2 + 16y2 + 24xy
(d) 9x2 + 16y2 – 24xy
Answer:
(By using the identity of )
= 9x2 + 16y2 – 24xy
Hence, the correct answer is option (d).
Page No 226:
Question 14:
In questions ​there are four options out of which one is correct
Which of the following are like terms?
(a) 5xyz2, –3xy2z
(b) –5xyz2, 7xyz2
(c) 5xyz2, 5x2yz
(d) 5xyz2, x2y2z2
Answer:
The terms having same algebraic (literal) factors are called like terms.
Thus, –5xyz2 and 7 xyz2 are both like terms having the same literal factor as xyz2.
Hence, the correct answer is option (b).
Page No 226:
Question 15:
In questions ​there are four options out of which one is correct
Coefficient of y in the term is
(a) – 1
(b) – 3
(c)
(d)
Answer:
So, the coefficient of y in
Hence, the correct answer is option (c).
Page No 226:
Question 16:
In questions ​there are four options out of which one is correct
a2 – b2 is equal to
(a) (a – b)2
(b) (a – b) (a – b)
(c) (a + b) (a – b)
(d) (a + b) (a + b)
Answer:
Hence, the correct answer is option (c).
Page No 226:
Question 17:
In questions ​there are four options out of which one is correct
Common factor of 17abc, 34ab2, 51a2b is
(a) 17abc
(b) 17ab
(c) 17ac
(d) 17a2b2c
Answer:
Hence, the correct answer is option (b).
Page No 226:
Question 18:
In questions ​there are four options out of which one is correct
Square of 9x – 7xy is
(a) 81x2 + 49x2y2
(b) 81x2 – 49x2y2
(c) 81x2 + 49x2y2 –126x2y
(d) 81x2 + 49x2y2 – 63x2y
Answer:
Hence, the correct answer is option (c).
Page No 226:
Question 19:
In questions ​there are four options out of which one is correct
Factorised form of 23xy – 46x + 54y – 108 is
(a) (23x + 54) (y – 2)
(b) (23x + 54y) (y – 2)
(c) (23xy + 54y) (– 46x – 108)
(d) (23x + 54) (y + 2)
Answer:
Hence, the correct answer is option (a).
Page No 226:
Question 20:
In questions ​there are four options out of which one is correct
Factorised form of r2 – 10r + 21 is
(a) (r – 1) (r – 4)
(b) (r – 7) (r – 3)
(c) (r – 7) (r + 3)
(d) (r + 7) (r + 3)
Answer:
Hence, the correct answer is option (b).
Page No 226:
Question 21:
In questions ​there are four options out of which one is correct
Factorised form of p2 – 17p – 38 is
(a) (p – 19) (p + 2)
(b) (p – 19) (p – 2)
(c) (p + 19) (p + 2)
(d) (p + 19) (p – 2)
Answer:
Hence, the correct answer is option (a).
Page No 227:
Question 22:
In questions ​there are four options out of which one is correct
On dividing 57p2qr by 114pq, we get
(a)
(b)
(c)
(d) 2pr
Answer:
Hence, the correct answer is option (c).
Page No 227:
Question 23:
In questions ​there are four options out of which one is correct
On dividing p (4p2 – 16) by 4p (p – 2), we get
(a) 2p + 4
(b) 2p – 4
(c) p + 2
(d) p – 2
Answer:
Hence, the correct answer is option (c).
Page No 227:
Question 24:
In questions ​there are four options out of which one is correct
The common factor of 3ab and 2cd is
(a) 1
(b) –1
(c) a
(d) c
Answer:
The common factor between 3ab and 2cd is 1.
Hence, the correct answer is option (a).
Page No 227:
Question 25:
In questions ​there are four options out of which one is correct
An irreducible factor of 24x2y2 is
(a) x2
(b) y2
(c) x
(d) 24x
Answer:
A factor is said to be irreducible, if it cannot be factorised further.
We have, 24x2y2 = 2 × 2 × 2 × 3 × x × x × y × y
Now, as 'x' cannot be further factorised.
Therefore, x is an irreducible factor of 24x2y2.
Page No 227:
Question 26:
In questions ​there are four options out of which one is correct
Number of factors of (a + b)2 is
(a) 4
(b) 3
(c) 2
(d) 1
Answer:
Now, (a + b) cannot be further factorised.
Therefore, (a + b)2 has 2 factors.
Hence, the correct answer is option (c).
Page No 227:
Question 27:
In questions ​there are four options out of which one is correct
The factorised form of 3x – 24 is
(a) 3x × 24
(b) 3 (x – 8)
(c) 24 (x – 3)
(d) 3(x – 12)
Answer:
Hence, the correct answer is option (b).
Page No 227:
Question 28:
In questions ​there are four options out of which one is correct
The factors of x2 – 4 are
(a) (x – 2), (x – 2)
(b) (x + 2), (x – 2)
(c) (x + 2), (x + 2)
(d) (x – 4), (x – 4)
Answer:
Thus, (x + 2) and (x – 2) are the factors of x2 – 4.
Hence, the correct answer is option (b).
Page No 227:
Question 29:
In questions ​there are four options out of which one is correct
The value of (–27x2y) ÷ (–9xy) is
(a) 3xy
(b) –3xy
(c) –3x
(d) 3x
Answer:
Hence, the correct answer is option (c).
Page No 227:
Question 30:
In questions ​there are four options out of which one is correct
The value of (2x2 + 4) ÷ 2 is
(a) 2x2 + 2
(b) x2 + 2
(c) x2 + 4
(d) 2x2 + 4
Answer:
Hence, the correct answer is option (b).
Page No 227:
Question 31:
In questions ​there are four options out of which one is correct
The value of (3x3 +9x2 + 27x) ÷ 3x is
(a) x2 +9 + 27x
(b) 3x3 +3x2 + 27x
(c) 3x3 +9x2 + 9
(d) x2 +3x + 9
Answer:
Hence, the correct answer is option (d).
Page No 227:
Question 32:
In questions ​there are four options out of which one is correct
The value of (a + b)2 + (a – b)2 is
(a) 2a + 2b
(b) 2a – 2b
(c) 2a2 + 2b2
(d) 2a2 – 2b2
Answer:
Hence, the correct answer is option (c).
Page No 227:
Question 33:
In questions ​there are four options out of which one is correct
The value of (a + b)2 – (a – b)2 is
(a) 4ab
(b) – 4ab
(c) 2a2 + 2b2
(d) 2a2 – 2b2
Answer:
Hence, the correct answer is option (a).
Page No 227:
Question 34:
Fill in the blanks to make the statements true:
The product of two terms with like signs is a __________ term.
Answer:
The product of two terms with like signs is a positive term.
For example:
The product of –2x and –3x is 6x2 and the product of 2x and 3x is also 6x2.
Page No 227:
Question 35:
Fill in the blanks to make the statements true:
The product of two terms with unlike signs is a ________ term.
Answer:
The product of two terms with unlike signs is a negative term.
For example:
The product of 2x and –3x is –6x2 and the product of –4x2 and 2y is –8x2y.
Page No 228:
Question 36:
Fill in the blanks to make the statements true:
a (b + c) = a × ____ + a × _____.
Answer:
Page No 228:
Question 37:
Fill in the blanks to make the statements true:
(a – b) _________ = a2 – 2ab + b2
Answer:
Page No 228:
Question 38:
Fill in the blanks to make the statements true:
a2 – b2 = (a + b ) __________.
Answer:
Hence, a2 – b2 = (a + b )(a – b).
Page No 228:
Question 39:
Fill in the blanks to make the statements true:
(a – b)2 + ____________ = a2 – b2
Answer:
Hence, (a – b)2 + (–2ab + 2b2) = a2 – b2.
Page No 228:
Question 40:
Fill in the blanks to make the statements true:
(a + b)2 – 2ab = ___________ + ____________
Answer:
Hence, (a + b)2 – 2ab = a2 + b2.
Page No 228:
Question 41:
Fill in the blanks to make the statements true:
(x + a) (x + b) = x2 + (a + b) x + ________.
Answer:
Hence, (x + a) (x + b) = x2 + (a + b) x + ab.
Page No 228:
Question 42:
Fill in the blanks to make the statements true:
The product of two polynomials is a ________.
Answer:
The product of two polynomial is a polynomial.
Page No 228:
Question 43:
Fill in the blanks to make the statements true:
Common factor of ax2 + bx is __________.
Answer:
ax2 = a × x × x
bx = b × x
So, the common factor of ax2 and bx is x.
Hence, the common factor of ax2 + bx is x.
Page No 228:
Question 44:
Fill in the blanks to make the statements true:
Factorised form of 18mn + 10mnp is ________.
Answer:
Hence, factorised form of 18mn + 10mnp is 2mn(9 + 5p).
Page No 228:
Question 45:
Fill in the blanks to make the statements true:
Factorized form of 4y2 – 12y + 9 is ________.
Answer:
Page No 228:
Question 46:
Fill in the blanks to make the statements true:
38x3y2z ÷ 19xy2 is equal to _________.
Answer:
Hence, 38x3y2z ÷ 19xy2 is equal to 2x2z.
Page No 228:
Question 47:
Fill in the blanks to make the statements true:
Volume of a rectangular box with length 2x, breadth 3y and height 4z is _________.
Answer:
Volume of a rectangular box
= Length × Breadth × Height
= 2x × 3y × 4z
= 2 × 3 × 4 × x × y × z
= 24xyz
Hence, volume of a rectangular box with length 2x, breadth 3y and height 4z is 24xyz.
Page No 228:
Question 48:
Fill in the blanks to make the statements true:
672 – 372 = (67 – 37) × ________ = _________.
Answer:
672 – 372
= (67 + 37) × (67 – 37) [∵ a2 – b2) = (a + b) (a – b)]
= 104 × 30
= 3120
Hence, 672 – 372 = (67 – 37) × (67 + 37) = 3120
Page No 228:
Question 49:
Fill in the blanks to make the statements true:
1032 – 1022 = ________ × (103 – 102) = _________.
Answer:
1032 – 1022
= (103 + 102) (103 – 102) [∵ a2 – b2 = (a + b) (a – b)]
= 205 × 1
= 205
Hence, 1032 – 1022 = (103 + 102) (103 – 102) = 205
Page No 228:
Question 50:
Fill in the blanks to make the statements true:
Area of a rectangular plot with sides 4x2 and 3y2 is __________.
Answer:
Area of a rectangular plot
= Length × Breadth
= 4x2 × 3y2
= 4 × 3 × x2 × y2
= 12x2y2
Hence, area of a rectangular plot with sides 4x2 and 3y2 is 12x2y2.
Page No 228:
Question 51:
Fill in the blanks to make the statements true:
Volume of a rectangular box with l = b = h = 2x is _________.
Answer:
Volume of rectangular box
= Length × Breadth × Height
= 2x × 2x × 2x
= 2 × 2 × 2 × x × x × x
= 8x3
Hence, volume of a rectangular box with l = b = h = 2x is 8x3.
Page No 228:
Question 52:
Fill in the blanks to make the statements true:
The coefficient in – 37abc is __________.
Answer:
The constant term (with their sign) involved in term of an algebraic expression is called the numerical coefficient of that term.
Hence, the coefficient in –37abc is –37.
Page No 228:
Question 53:
Fill in the blanks to make the statements true:
Number of terms in the expression a2 + bc × d is ________.
Answer:
Consider: a2 + bc × d
So, a2 + bc × d = a2 + bcd
A term is a single mathematical expression. It may be a single number (positive or negative), a single variable, several variables multiplied but never added or subtract. It a2 + bcd, a2 and bcd are two terms.
Hence, the number of terms in the expression a2 + bc × d is 2.
Page No 228:
Question 54:
Fill in the blanks to make the statements true:
The sum of areas of two squares with sides 4a and 4b is _______.
Answer:
Area of a square = (side)2
So, the area of a square with side '4a' = (4a)2
= 16a2
= 16b2
Page No 228:
Question 55:
Fill in the blanks to make the statements true:
The common factor method of factorisation for a polynomial is based on ___________ property.
Answer:
In common factor method of factorisation for a polynomial, we regroup the terms in such a way that each term in the group contains a common variable or constant or both. We then take out the common part using distributive property.
Hence, the common factor method of factorisation for a polynomial is based on distributive property.
Page No 228:
Question 56:
Fill in the blanks to make the statements true:
The side of the square of area 9y2 is __________.
Answer:
Let the side of the square be x.
So, the area of the square is x2.
Given that the area of square is 9y2.
Therefore, x2 = 9y2.
⇒ x2 = (3y)2
⇒ x = 3y
Hence, the side of the square of area 9y2 is 3y.
Page No 228:
Question 57:
Fill in the blanks to make the statements true:
On simplification = _________.
Answer:
Page No 228:
Question 58:
Fill in the blanks to make the statements true:
The factorisation of 2x + 4y is __________.
Answer:
2x = 2 × x
4y = 2 × 2 × y
So, 2x + 4y = (2 × x) + (2 × 2 × y)
Page No 229:
Question 59:
State whether the statements are True (T) or False (F):
(a + b)2 = a2 + b2
Answer:
False,
(a + b)2 = a2 + b2 + 2ab ≠ a2 + b2
Page No 229:
Question 60:
State whether the statements are True (T) or False (F):
(a – b)2 = a2 – b2
Answer:
False,
(a – b)2 = a2 + b2 – 2ab ≠ a2 – b2
Page No 229:
Question 61:
State whether the statements are True (T) or False (F):
(a + b) (a – b) = a2 – b2
Answer:
True,
(a + b) (a – b) = a(a – b) + b(a – b)
= a2 – ab + ba – b2
= a2 – ab + ab – b2
= a2 – b2
Page No 229:
Question 62:
State whether the statements are True (T) or False (F):
The product of two negative terms is a negative term.
Answer:
False,
For example:
Let –4x and –2y be two negative terms then the product of –4x and –2y is 8xy, which is positive.
Page No 229:
Question 63:
State whether the statements are True (T) or False (F):
The product of one negative and one positive term is a negative term.
Answer:
True,
As, (–) × (+) = (–).
Therefore, the product of one negative and one positive term is a negative term.
Page No 229:
Question 64:
State whether the statements are True (T) or False (F):
The coefficient of the term – 6x2y2 is – 6.
Answer:
True,
The numerical coefficient of the term –6x2y2 is –6.
Therefore, the coefficient of the term –6x2y2 is –6.
Page No 229:
Question 65:
State whether the statements are True (T) or False (F):
p2q + q2r + r2q is a binomial.
Answer:
False,
p2q + q2r + r2q has three unlike terms.
Therefore, p2q + q2r + r2q is a trinomial.
Page No 229:
Question 66:
State whether the statements are True (T) or False (F):
The factors of a2 – 2ab + b2 are (a + b) and (a + b).
Answer:
False,
a2 – 2ab + b2 = (a – b)2
= (a – b) (a – b)
Hence, the factors of a2 – 2ab + b2 are (a – b) and (a – b).
Page No 229:
Question 67:
State whether the statements are True (T) or False (F):
h is a factor of 2π (h + r).
Answer:
False,
The only factors of 2π(h + r) are 2, 2π, 2(h + r), π, (h + r), 2π(h + r).
Hence, 'h' is not a factor of 2π(h + r).
Page No 229:
Question 68:
State whether the statements are True (T) or False (F):
Some of the factors of and (n + 1).
Answer:
True,
Hence, some of the factors of are , n and n + 1.
Page No 229:
Question 69:
State whether the statements are True (T) or False (F):
An equation is true for all values of its variables.
Answer:
False,
An equation may not be true for all value of its variables.
For example: x – 2 = 2 is true only for x = 4.
Page No 229:
Question 70:
State whether the statements are True (T) or False (F):
x2 + (a + b)x + ab = (a + b) (x + ab)
Answer:
False,
As, x2 + (a + b)x + ab
= x2 + ax + bx + ab
= x(x + a) +b(x + a)
= (x + a)(a + b)
≠ (a + b)(x + ab)
Page No 229:
Question 71:
State whether the statements are True (T) or False (F):
Common factor of 11pq2, 121p2q3, 1331p2q is 11p2q2.
Answer:
False,
11pq2 = 11 × p × q × q
121p2q3 = 11 × 11 × p × p × q × q × q
1331p2q =11 × 11 × 11 × p × p × q
∴ Common factor of 11pq2, 121p2q3, 1331p2q is 11 × p × q = 11pq
Page No 229:
Question 72:
State whether the statements are True (T) or False (F):
Common factor of 12a2b2, 4ab2, – 32 is 4.
Answer:
True,
12a2b2 = 3 × 2 × 2 × a × a × b × b
4ab2 = 2 × 2 × a × b × b
–32 = 2 × 2 × 2 × 2 × 2 × –1
Hence, common factors of 12a2b2 + 4ab2 – 32 is 2 × 2 = 4.
Page No 229:
Question 73:
State whether the statements are True (T) or False (F):
Factorisation of – 3a2 + 3ab + 3ac is 3a (–a – b – c).
Answer:
False,
–3a2 = –3 × a × a
3ab = 3 × a × b
3ac =3 × a × c
∴ Common factor of –3a2 + 3ab + 3ac is 3a.
Thus, –3a2 + 3ab + 3ac = 3a(–a + b + c)
Page No 229:
Question 74:
State whether the statements are True (T) or False (F):
Factorised form of p2 + 30p + 216 is (p + 18) (p – 12).
Answer:
False,
p2 + 30p +216 = p2 + 12p + 18p + 216 (By splitting the middle term)
= (p + 18) (p + 12)
≠ (p + 18) (p – 12)
Page No 229:
Question 75:
State whether the statements are True (T) or False (F):
The difference of the squares of two consecutive numbers is their sum.
Answer:
True,
Let 'n' be the number.
Then, n and n + 1 are the two consecutive numbers.
∴ The sum of the two consecutive numbers = n + (n + 1) = 2n + 1
Now, the difference of the square of two consecutive numbers
= (n + 1)2 – (n)2
= (n2 + 1 + 2n) – (n2) [∵ (a + b)2 + a2 + b2 + 2ab]
= n2 + 1 + 2n – n2
= 2n + 1
Hence, the difference of the squares of two consecutive numbers is their sum.
Page No 229:
Question 76:
State whether the statements are True (T) or False (F):
abc + bca + cab is a monomial.
Answer:
True,
abc + bca + cab = abc + abc + abc
Page No 229:
Question 77:
State whether the statements are True (T) or False (F):
On dividing the quotient is 9.
Answer:
False,
Page No 229:
Question 78:
State whether the statements are True (T) or False (F):
The value of p for 512 – 492 = 100p is 2.
Answer:
True,
512 – 492 = (51 + 49) (51 – 49) [∵ a2 – b2 = (a + b)
= (100) × 2 ...(1)
Given that 512 – 492 = 100p ...(2)
Therefore, p = 2.
Hence, the value of p for 512 – 492 = 100 p is 2.
Page No 229:
Question 79:
State whether the statements are True (T) or False (F):
(9x – 51) ÷ 9 is x – 51.
Answer:
False,
Page No 229:
Question 80:
State whether the statements are True (T) or False (F):
The value of (a + 1) (a – 1) (a2 + 1) is a4 – 1.
Answer:
True,
(a + 1) (a – 1)(a2 + 1)
= (a2 – 1) (a2 + 1) [∵ (a + b) (a – b) = a2 – b2]
= (a2)2 – (1)2 [∵ (a + b) (a – b) = a2 – b2]
= a4 – 1
Hence, the value of (a + 1) (a – 1) (a2 + 1) is a4 – 1.
Page No 230:
Question 81:
Add:
(i) 7a2bc, – 3abc2, 3a2bc, 2abc2
(ii) 9ax, + 3by – cz, – 5by + ax + 3cz
(iii) xy2z2 + 3x2y2z – 4x2yz2, – 9x2y2z + 3xy2z2 + x2yz2
(iv) 5x2 – 3xy + 4y2 – 9, 7y2 + 5xy – 2x2 + 13
(v) 2p4 – 3p3 + p2 – 5p +7, –3p4 – 7p3 – 3p2 – p – 12
(vi) 3a (a – b + c), 2b (a – b + c)
(vii) 3a (2b + 5c), 3c (2a + 2b)
Answer:
(i) 7a2bc + (–3abc2) + 3a2bc + 2abc2
= 7a2bc – 3abc2 + 3a2bc + 2abc2
= 7a2bc + 3a2bc – 3abc2 + 2abc2 [Grouping like terms]
= 10a2bc – abc2
(ii) (9ax + 3by – cz) + (–5by + ax + 3cz)
= 9ax + 3by – cz –5by + ax + 3cz
= 9ax + ax + 3by – 5by – cz + 3cz [Grouping like terms]
= 10x – 2by + 2cz
(iii) (xy2z2 + 3x2y2z – 4x2yz2) + (–9x2y2z + 3xy2z2 + x2yz2)
= xy2z2 + 3x2y2z – 4x2yz2 – 9x2y2z + 3xy2z2 + x2yz2
= xy2z2 + 3xy2z2 + 3x2y2z – 9x2y2z – 4x2yz2 + x2yz2 [Grouping like terms]
= 4xy2z2 – 6x2y2z – 3x2yz2
(iv) (5x2 – 3xy + 4y2 – 9) + (7y2 + 5xy – 2x2 + 13)
= 5x2 – 3xy + 4y2 – 9 + 7y2 + 5xy – 2x2 + 13
= 5x2 – 2x2 – 3xy + 5xy + 4y2 + 7y2 + 13 – 9 [Grouping like terms]
= 3x2 + 2xy + 11y2 + 4
(v) (2p4 – 3p3 + p2 – 5p + 7) + (–3p4 – 7p3 – 3p2 – p –12)
= 2p4 – 3p3 + p2 – 5p + 7 – 3p4 – 7p3 – 3p2 – p – 12
= 2p4 – 3p4 – 3p3 – 7p3 + p3 + 3p2 – 5p – p + 7 – 12 [Grouping like terms]
= –p4 – 10p3 – 2p2 – 6p – 5
(vi) 3a(a – b + c) + 2b(a – b + c)
= 3a2 – 3ab + 3ac + 2ba – 2b2 + 2bc
= 3a2 – 3ab + 2ab + 3ac + 2bc – 2b2 [Grouping like terms]
= 3a2 – ab + 3ac + 2bc2 – 2b2
(vii) 3a(2b + 5c) + 3c(2a + 2b)
= 6ab + 15ac + 6ca + 6bc
= 6ab + 15ac + 6ac + 6bc [∵ ca = ac]
= 6ab + 21ac + 6bc
Page No 230:
Question 82:
Subtract :
(i) 5a2b2c2 from – 7a2b2c2
(ii) 6x2 – 4xy + 5y2 from 8y2 + 6xy – 3x2
(iii) 2ab2c2 + 4a2b2c – 5a2bc2 from –10a2b2c + 4ab2c2 + 2a2bc2
(iv) 3t4 – 4t3 + 2t2 – 6t + 6 from – 4t4 + 8t3 – 4t2 – 2t + 11
(v) 2ab + 5bc – 7ac from 5ab – 2bc – 2ac + 10abc
(vi) 7p (3q + 7p) from 8p (2p – 7q)
(vii) –3p2 + 3pq + 3px from 3p (– p – a – r)
Answer:
(i) (–7a2b2c2) – (5a2b2c2)
= (–7 –5) a2b2c2
= –12a2b2c2
(ii) (8y2 + 6xy – 3x2) – (6x2 – 4xy + 5y2)
= 8y2 + 6xy – 3x2 – 6x2 + 4xy – 5y2
= 8y2 – 5y2 + 6xy + 4xy – 3x2 – 5x2 [Grouping like terms]
= 3y2 + 10xy – 9x2
(iii) (–10a2b2c + 4ab2c2 + 2a2bc2) – (2ab2c2 + 4a2b2c – 5a2bc2)
= –10a2b2c + 4ab2c2 + 2a2bc2 – 2ab2c2 – 4a2b2c + 5a2bc2
= –10a2b2c – 4a2b2c + 4ab2c2 – 2ab2c2 + 2a2bc2 + 5a2bc2 [Grouping like terms]
= –14a2b2c + 2ab2c2 +7a2bc2
(iv) (–4t4 + 8t3 – 4t2 – 2t + 11) – (3t4 – 4t3 + 2t2 – 6t + 6)
= –4t4 + 8t3 – 4t3 + 2t + 11 – 3t4 + 4t3 – 2t3 + 6t – 6
= –4t4 – 3t4 + 8t3 – 4t3 – 4t2 – 2t2 – 2t + 6t + 11 – 6 [Grouping like terms]
= –7t4 + 12t3 – 6t2 + 4t + 5
(v) (5ab – 2bc – 2ac + 10abc) – (2ab + 5bc –7ac)
= 5ab – 2bc – 2ac + 10abc – 2ab – 5bc + 7ac
= 5ab – 2ab – 2bc – 5bc – 2ac + 7ac + 10abc [Grouping like terms]
= 3ab – 7bc + 5ac + 10abc
(vi) 8p(2p – 7q) – 7p(3q + 7p)
= 16p2 – 56pq – 21pq – 49p2
= 16p2 – 77pq – 49p2
= 16p2 – 49p2 – 77pq [Grouping like terms]
= –33p2 – 77pq
(vii) 3p(–p – a – r) – (–3p2 + 3pq + 3px)
= 3p(–p – a – r) + 3p2 – 3pq – 3px
= –3p2 – 3ap – 3pr + 3p2 – 3pq – 3px
= –3p2 + 3p2 – 3ap – 3pr – 3pq – 3px [Grouping like terms]
= – 3ap – 3pr – 3pq – 3px
Page No 230:
Question 83:
Multiply the following:
(i) – 7pq2r3, – 13p3q2r
(ii) 3x2y2z2, 17xyz
(iii) 15xy2, 17yz2
(iv) –5a2bc, 11ab, 13abc2
(v) –3x2y, (5y – xy)
(vi) abc, (bc + ca)
(vii) 7pqr, (p – q + r)
(viii) x2y2z2, (xy – yz + zx)
(ix) (p + 6), (q – 7)
(x) 6mn, 0mn
(xi) a, a5, a6
(xii) –7st, –1, – 13st2
(xiii) b3, 3b2, 7ab5
(XIV)
(xv) (a2 – b2), (a2 + b2)
(xvi) (ab + c), (ab + c)
(xvii) (pq – 2r), (pq – 2r)
(xviii)
(xix)
(xx) (x2 – 5x + 6), (2x + 7)
(xxi) (3x2 + 4x – 8), (2x2 – 4x + 3)
(xxii) (2x – 2y – 3), (x + y + 5)
Answer:
(i) (–7pq2r3) × (–13p3q2r)
= (–7) × (–13) × p4 × q4 × r4
= 91p4q4r4
(ii) 3x2y2z2 × 17xyz
= 3 × 17 × x3 × y3 × z3
= 51x3y3z3
(iii) 15xy2 × 17yz2
= 15 × 17 × x × y3 × z2
= 255xy3z2
(iv) –5a2bc ×11ab × 13abc2
= –5 × 11 × 13 × a4 × b3c3
= –715a4b3c3
(v) – 3x2y × (5y – xy)
= –3x2y × 5y – (–3x2y) (xy)
= –15x2y2 + 3x3y2
(vi) abc × (bc + ca)
= abc × bc + abc × ca
= ab2c2 + a2bc2
(vii) 7pqr × (p – q + r)
= 7p2qr – 7pq2r +7pqr2
(viii) x2y2z2 × (xy – yz – zx)
= x2y2z2 × xy – x2y2z × yz – x2y2z × zx
= x3y3z2 – x2y3z3 – x3y2z3
(ix) (p + 6) × (q – 7)
= p(q – 7) + 6(q – 7)
= pq –7q + 6q –42
(x) 6mn × 0mn
= 6 × 0 × m2n2
= 0
(xi) a × a5 × a6
= a1 + 5 + 6
= a12
(xii) (–7st) × (–1) × (–13st2)
= –7 × –1 × 13 × s2t3
= –91s2t3
(xiii) b3 × 3b2 × 7ab5
= 1 × 3 × 7 × a × b10
= 21ab10
(xv) (a2 – b2) (a2 + b2)
= (a2)2 – (b2)2 [∵ (x – y) (x + y) = x2 – y2]
= a4 – b4
(xvi) (ab + c) (ab + c)
= ab(ab + c) + c(ab + c)
= a2b2 + abc + abc + c2
= a2b2 + 2abc + c2
(xvii) (pq – 2r) (pq – 2r)
= (pq – 2r)2
= (pr)2 + (2r)2 – 2 × pq × 2r [∵ (a – b)2 = a2 + b2 – 2ab]
= p2q2 + 4r2 – 4pqr
(xxii) (2x – 2y – 3) (x + y + 5)
= 2x(x + y + 5) –2y(x + y + 5) –3(x + y + 5)
= 2x × x + 2x – y + 2x × 5 – 2y × x – 2y × y – 2y × 5 – 3x – 3y – 15
= 2x2 + 2xy + 10x – 2xy – 2y2 – 10y – 3x – 3y – 15
= 2x2 + 2xy – 2xy + 10x – 3x – 2y2 – 10y – 3y – 15 [Grouping like terms]
= 2x2 + 7x – 2y2 – 13y –15
Page No 231:
Question 84:
Simplify
(i) (3x + 2y)2 + (3x – 2y)2
(ii) (3x + 2y)2 – (3x – 2y)2
(iii)
(iv)
(v) (1.5p + 1.2q)2 – (1.5p – 1.2q)2
(vi) (2.5m + 1.5q)2 + (2.5m – 1.5q)2
(vii) (x2 – 4) + (x2 + 4) + 16
(viii) (ab – c)2 + 2abc
(ix) (a – b) (a2 + b2 + ab) – (a + b) (a2 + b2 – ab)
(x) (b2 – 49) (b + 7) + 343
(xi) (4.5a + 1.5b)2 + (4.5b + 1.5a)2
(xii) (pq – qr)2 + 4pq2r
(xiii) (s2t + tq2)2 – (2stq)2
Answer:
(i) (3x + 2y)2 + (3x – 2y)2
= (3x)2 + (2y)2 + 2 × 3x × 2y + (3x)2 + (2y)2 – 2 × 3x × 2y [∵ (a + b)2 = a2 + b2 + 2ab and (a – b)2 = a2 + b2 – 2ab]
= 9x2 + 4y2 + 12xy + 9x2 + 4y2 – 12xy
= 9x2 + 9x2 + 4y2 + 12xy – 12xy
= 18x2 + 8y2
= 2(9x2 + 4y2)
(ii) (3x + 2y)2 – (3x – 2y)2
= [(3x + 2y) + (3x – 2y)] [(3x + 2y) – (3x – 2y] [∵ a2 – b2 = (a + b) (a – b)]
= (6x) × (3x + 2y – 3x + 2y)
= (6x) × (4y)
= 24xy
(v) (1.5p + 1.2q)2 – (1.5p – 1.2q)2
= [(1.5p)2 + (1.2q)2 + 2 × 1.5p × 1.2q] – [(1.5p)2 + (1.2q)2 – 2 × 1.5p × 1.2q]
= (1.5p)2 + (1.2q)2 + 2 × 1.5 × 1.2 × pq – (1.5p)2 – (1.2q)2 + 2 × 1.5 × 1.2 × pq
= (1.5p)2 – (1.5p)2 + (1.2q)2 – (1.2q)2 + 2 × 1.5 × 1.2 × pq + 2 × 1.5 × 1.2 × pq [Grouping like terms]
= 2 × 2 × 1.5 × 1.2 × pq
= 7.2pq
(vi) (2.5m + 1.5q)2 + (2.5m – 1.5q)2
= (2.5m)2 + (1.5q)2 + 2 × 2.5m × 1.5q + (2.5m)2 – (1.5q)2 – 2 × 2.5m × 1.5q [∵ (a + b)2 = a2 + b2 + 2ab and (a – b)2 = a2 + b2 – 2ab]
= (2.5m)2 + (2.5m)2 + (1.5q)2 + (1.5q)2 + 2 × 2.5m × 1.5q – 2 × 2.5m × 15q [Grouping like terms]
= 6.25m2 + 6.25m2 + 2.25q2 + 2.25q2
= 12.5m2 + 4.5q2
(vii) (x2 – 4) + (x2 + 4) + 16
= x2 – 4 + x2 + 4 + 16
= x2 + x2 – 4 + 4 + 16 (Grouping like terms)
= 2x2 + 16
(viii) (ab – c)2 + 2abc
= (ab)2 + (c)2 – 2abc + 2abc [∵ (a – b)2 = a2 + b2 – 2ab]
= a2b2 + c2
(ix) (a – b) (a2 + b2 + ab) – (a + b) (a2 + b2 – ab)
= a(a2 + b2 + ab) – b(a2 + b2 + ab) – a(a2 + b2 – ab) – b(a2 + b2 – ab)
= a3 + ab2 + a2b – ba2 – b3 – ab2 – a3 – ab2 + a2b – ba2 – b3 + ab2
= a3 – a3 + ab2 – ab2 – b3 – b3 + a2b – a2b + a2b – a2b (Grouping like terms)
= –2b3
(x) (b2 – 49) (b + 7) + 343
= b2 (b + 7) –49(b + 7) + 343
= b3 + 7b2 – 49b – 343 + 343
= b3 + 7b2 – 49b
(xi) (4.5a + 1.5b)2 + (4.5b + 1.5a)2
= (4.5a)2 + (1.5b)2 + 2 × 4.5a × 1.5b + (4.5b)2 + (1.5a)2 + 2 × 4.5b × 1.5a [∵ (a + b)2 = a2 + b2 + 2ab and (a – b)2 = a2 + b2 – 2ab]
= 20.25a2 + 2.25b2 + 13.5ab + 20.25b2 + 2.25a2 + 13.5ab
= 20.25a2 + 2.25a2 + 2.25b2 + 20.25b2 + 13.5ab + 13.5ab (Grouping like terms)
= 22.5a2 + 22.b2 + 27ab
(xii) (pq – qr)2 + 4pq2r
= (pq)2 + (qr)2 – 2 × pq × qr + 4pq2r [∵ (a – b)2 = a2 + b2 – 2ab]
= p2q2 + q2r2 – 2pq2r + 4pq2r
= p2q2 + q2r2 + 2pq2r
(xiii) (s2t + tq2)2 – (2stq)2
= (s2t)2 + (tq2) + 2 × s2t × tq2 – 4s2t2q2 [∵ (a + b)2 = a2 + b2 + 2ab]
= s4t2 + t2q4 + 2s2t2q2 – 4s2t2q2
= s4t2 + t2q4 – 2s2t2q2
Page No 232:
Question 85:
Expand the following, using suitable identities.
(i) (xy + yz)2
(ii) (x2y – xy2)2
(vi) (x + 3) (x + 7)
(vii) (2x + 9) (2x – 7)
(x) (2x – 5y) (2x – 5y)
.
(xi)
(xii) (x2 + y2) (x2 – y2)
(xiii) (a2 + b2)2
(xiv) (7x + 5)2
(xv) (0.9p – 0.5q)2
(xvi) x2y2 = (xy)2
Answer:
(i) (xy + yz)2
= (xy)2 + (yz)2 + 2 × xy × yz [∵ (a + b)2 = a2 + b2 + 2ab]
= x2y2 + y2z2 + 2xy2z
(ii) (x2y – xy)2
= (x2y)2 + (xy2)2 – 2 × x2y × xy2 [∵ (a – b)2 = a2 + b2 – 2ab]
= x4y2 + x2y4 – 2x3y3
(vi) (x + 3) (x + 7)
= x2 + (3 + 7)x + 3 × 7 [∵ (x + a) (x + b) = x + (a + b)x + ab]
= x2 + 10x + 21
(vii) (2x + 9) (2x – 7)
= (2x)2 + [9 + (–7)]2x + 9 × (–7) [∵ (x + a) (x + b) = x2 + (a + b)x + ab]
= 4x2 + 4x – 63
(x) (2x – 5y) (2x – 5y)
= (2x – 5y)2
= (2x)2 + (5y)2 –2 × 2x × 5y [∵ a – b)2 = a2 + b2 – 2ab]
= 4x2 + 25y2 – 20xy
(xii) (x2 + y2) (x2 – y2)
= (x2)2 – (y2)2 [∵ (a + b) (a – b) = a2 – b2]
= x4 – y4
(xiii) (a2 + b2)2
= (a2)2 + (b2)2 + 2 × a2 + b2 [∵ (a + b)2 = a2 + b2 + 2ab]
= a4 + b4 + 2a2b2
(xiv) (7x + 5)2
= (7x)2 + (5)2 + 2 × 7x × 5 [∵ (a + b)2 = a2 + b2 + 2ab]
= 49x2 + 25 + 70x
(xv) (0.9p – 0.5q)2
= (0.9p)2 + (0.5q)2 – 2 × 0.9p × 0.5q [∵ (a – b)2 = a2 + b2 – 2ab]
= 0.81p2 + 0.25q2 – 0.9pq
Page No 233:
Question 86:
Using suitable identities, evaluate the following.
(i) (52)2 (ii) (49)2
(iii) (103)2 (iv) (98)2
(v) (1005)2 (vi) (995)2
(vii) 47 × 53 (viii) 52 × 53
(ix) 105 × 95 (x) 104 × 97
(xi) 101 × 103 (xii) 98 × 103
(xiii) (9.9)2 (xiv) 9.8 × 10.2
(xv) 10.1 × 10.2 (xvi) (35.4)2 – (14.6)2
(xvii) (69.3)2 – (30.7)2 (xviii) (9.7)2 – (0.3)2
(xix) (132)2 – (68)2 (xx) (339)2 – (161)2
(xxi) (729)2 – (271)2
Answer:
(i) (52)2 = (50 + 2)2
= 2500 + 4 + 200
= 2704
= 2500 + 1 – 100
= 2401
= 10000 + 9 + 600
= 10609
= 10000 + 4 – 400
= 9604
= 1000000 + 25 + 10000
= 1010025
= 1000000 + 25 – 10000
= 990025
= (50 – 3) (50 + 3)
= (50)2 – (3)2 [∵ (a – b) (a + b) = a2 – b2]
= 2500 – 9
= 2490
(viii) 52 × 53
= (50 + 2) (50 + 3)
= (50)2 + (2 + 3) × 50 + 2 × 3 [∵ (x + a) (x + b) = x2 + (a + b)x + ab]
= 2500 + 250 + 6
= 2756
(ix) 105 × 95
= (100 + 5) × (100 – 5)
= (100)2 – (5)2 [∵ (a + b) (a – b) = a2 – b2]
= 10000 – 25
= 9975
(x) 104 × 97
= (100 + 4) (100 – 3)
= (100)2 + (4 – 3) × 100 + 4 × (–3) [∵ (x + a) (x + b) = x2 + (a + b)x + ab]
= 10000 + 100 – 12
= 10088
(xi) 101 × 103
= (100 + 1) (100 + 3)
= (100)2 + (1 + 3) × 100 × 3 × 1 [∵ (x + a) (x + b) = x2 + (a + b)x + ab]
= 10000 + 400 + 3
(xii) 98 × 103
= (100 – 2) (100 + 3)
= (100)2 + (–2 + 3) × 100 + (–2) × 3 [∵ (x + a) (x + b) = x2 + (a + b)x + ab]
= 10000 + 100 – 6
= 10094
(xiii) (9.9)2 = (10 – 0.1)2
= 100 + 0.01 – 2
= 98.01
= (10 – 0.2) × (10 + 0.2)
= (10)2 – (0.2)2 [∵ (a –b) (a + b) = a2 – b2)
= 100 – 0.04
= 98.96
(xv) 10.1 × 10.2
= (10 + 0.1) (10 + 0.2)
= (10)2 + (0.1 + 0.02) × 10 + 0.02 [∵ (x + a) (x + b) = x2 + (a + b)x + ab]
= 100 + 0.32 × 10 + 0.02
= 100 + 3 + 0.02
= 103.02
(xvi) (35.4)2 – (14.6)2
= (35.4 + 14.6) (35.4 – 14.6) [∵ a2b2 = (a + b) (a – b)]
= 50 × 20.8
= 1048
(xvii) (69.3)2 = (30.7)2
= (69.3 + 30.7) (69.3 – 30.7) [∵ a2 – b2 = (a + b) (a – b)]
= 100 × 38.6
= 3860
(xix) (132)2 – (68)2
= (135 + 68) (132 – 68) [∵ a2 – b2 = (a + b) (a + b)]
= 200 × 64
= 12800
(xx) (339)2 – (161)2
= (339 + 161) (339 – 161) [∵ a2 – b2 = (a + b) (a – b)]
= 500 × 178
= 89000
(xxi) (729)2 – (271)2
= (729 + 271) (729 – 271) [∵ a2 – b2) = (a + b) (a – b)]
= 1000 × 458
= 458000
Page No 233:
Question 87:
Write the greatest common factor in each of the following terms.
(i) – 18a2, 108a (ii) 3x2y, 18xy2, – 6xy
(iii) 2xy, –y2, 2x2y (iv) l2m2n, lm2n2, l2mn2
(v) 21pqr, –7p2q2r2, 49p2qr (vi) qrxy, pryz, rxyz
(vii) 3x3y2z, –6xy3z2, 12x2yz3
(viii) 63p2a2r2s, – 9pq2r2s2, 15p2qr2s2, – 60p2a2rs2
(ix) 13x2y, 169xy
(x) 11x2, 12y2
Answer:
(i) –18a2 = –18 × a × a
108a = 18 × 10 × a
∴ There greatest common factor is 18a.
(ii) 3x2y = 3 × x × x × y
18xy2 = 3 × 6 × x × y × y
–6xy = –1 × 3 × 2 × x × y
∴ The greatest common factor is 3xy.
(iii) 2xy = 2 × x × y
–y2 = –y × y
2x2y = 2 × x × x × y
∴ The greatest common factor is y.
(iv) l2m2n = l × l × m × m × n
lm2n2 = l × m × m × n × n
l2mn2 = l × l × m × n × n
∴ The greatest common factor = l × m × n = lmn.
(v) 21pqr = 7 × 3 × p × q × r
–7p2q2r = –7 × p × p × q × q × r × r
49p2qr = 7 × 7 × p × p × q × r
∴ The greatest common factor = 7pqr
(vi) qrxy = q × r × x × y
pry2 = p × r × y × z
rxy2 = r × x × y × z
∴ The greatest common factor = ry
(vii) 3x3y2z = 3 × x × x × x
–6xy3z2 = –3 × 2 × x × y × y × y × z × z
12x2y2z3 = 3 × 4 × x × x × y × z × z × z
∴ The greatest common factor = 3xyz
(viii) 63p2a2r2s = 3 × 3 × 7 × p × p × a × a × r × r × s
–9pq2r2s2 = –3 × 3 × p × q × q × r × r × s × s
15p2qr2s2 = 3 × 5 × p × p × q × r × r × s × s
–60p2a2rs2 = –2 × 2 × 3 × 5 × p × p × a × a × r
∴ The greatest common factor = 3 prs
(ix) 13x2y = 13 × x × x × y
169xy = 13 × 13 × x × y
∴ The greatest common factor = 13xy
(x) 11x2 = 11 × x × x
12y2 = 12 × y × y
∴ The greatest common factor = 1
Page No 233:
Question 88:
Factorise the following expressions.
(i) 6ab + 12bc (ii) –xy – ay
(iii) ax3 – bx2 + cx (iv) l2m2n – lm2n2– l2mn2
(v) 3pqr –6p 2q 2r2 – 15r2 (vi) x3y2 + x2y3 – xy4 + xy
(vii) 4xy2 – 10x2y + 16x2y2 + 2xy
(viii) 2a3 – 3a2b + 5ab2 – ab
(ix) 63p2q2r2s – 9pq2r2s2 + 15p2qr2s2 – 60p2q2rs2
(x) 24x2yz3 – 6xy3z2 + 15x2y2z – 5xyz
(xi) a3 + a2 + a + 1
(xii) lx + my + mx + ly
(xiii) a3x – x4 + a2x2 – ax3
(xiv) 2x2 – 2y + 4xy – x
(xv) y2 + 8zx – 2xy – 4yz
(xvi) ax2y – bxyz – ax2z + bxy2
(xvii) a2b + a2c + ab + ac + b2c + c2b
(xviii) 2ax2 + 4axy + 3bx2 + 2ay2 + 6bxy + 3by2
Answer:
(i) 6ab + 12bc
= 6ab + 6 × 2 × bc
= 6b(a + 2c)
(ii) –xy – ay
= –y (x + a)
(iii) ax3 – bx2 + cx
= x(ax2 – bx + c)
(iv) l2m2n2 – lm2n2 – l2mn2
= lmn (lm – mn – ln)
(v) 3pqr – 6p2q2r2 – 15r2
= 3r(pq – 2p2q2r – 5r)
(vi) x3y2 + x2y3 – xy4 + xy
= xy (x2y + xy2 – y3 + 1)
(vii) 4xy2 – 10x2y + 16x2y2 + 2xy
= 2 × 2xy2 – 2 × 5 × x2y + 2 × 8x2y2 + 2xy
= 2xy(2y – 5x + 8xy + 1) (Taking out 2xy)
(viii) 2a3 – 3a2b + 5ab2 – ab
= a(2a2 – 3ab + 5b2 – b
(ix) 63p2q2r2s – 9pq2r2s2 + 15p2qr2s2 – 60p2q2rs2
= 3 × 21p2q2r2s – 3 × 3pq2r2s2 + 3 × 5p2qr2s2 – 3 × 20p2q2rs2
= 3pqrs (21pqr – 3qrs + 5prs – 20pqs)
(x) 24x2yz3 – 6xy3z2 + 15x2y2z –5xyz
= xyz(24xz2 – 6y2z + 15xy – 5)
(xi) a3 + a2 + a + 1
= a2(a + 1) + 1(a + 1)
= (a2 + 1) (a + 1) [Taking out (a + 1)]
(xii) lx + my + mx + ly
= lx + mx + my + ly [Grouping like terms]
= (l + m)x + (m + l)y
= (l + m)x + (l + m)y
= (l + m)x + (l + m)y
= (l + m) (x + y) [Taking out (l + m)]
(xiii) a3x – a4 + a2x2 – ax3
= x(a3 – x3 + a2x – ax2)
= x(a3 + a2x – x3 – ax2)
= x[a2 (a + x) – x2(x + a)]
= x[a2(a + x) – x2(a + x)]
= x(x2 – x2) (a + x) [Taking out (a + x)]
= x(a + x) (a – x) (a + x) [∵ a2 – b2 = (a + b) (a – b)
(xiv) 2x2 – 2y + 4xy – x
= 2x2 – x – 2y + 4xy [Grouping like terms]
= x(2x – 1) + 2y(2x – 1)
= (x + 2y) (2x – 1) [Taking out 2x – 1)]
(xv) y2 + 8zx – 2xy – 4yz
= y2 – 2xy + 8zx – 4yz [Rearranging terms]
= y(y – 2x) + 4z(2x – y)
= y(y – 2x) – 4z(y – 2x)
= (y – 4z) (y – 2x) [Taking out (y – 2x)]
(xvi) ax2y – bxyz – ax2z + bxy2
= ax2y – ax2z – bxyz + bxy2 [Rearranging terms]
= ax2(y – z) – bxy(y – z)
= (ax2 – bx) (y – z)
= x(ax – by) (y – z)
(xvii) a2b + a2c + ab + ac + b2c + c2b
= a2(b + c) + a(b + c) + bc(b + c)
= (b + c) (a2 + a + bc)
(xviii) 2ax2 + 4axy + 3bx2 + 2ay2 + 6bxy + 3by2
= 2ax2 + 2ay2 + 4axy + 3bx2 + 3by2 + 6bxy [Rearranging terms]
= 2a(x2 + y2 + 2xy) + 3b(x2 + y2 + 2xy)
= (2a + 3b) (x2 + y2 + 2xy)
Page No 234:
Question 89:
Factorise the following, using the identity a2 + 2ab + b2 = (a + b)2
(i) x2 + 6x + 9 (ii) x2 + 12x + 36
(iii) x2 + 14x + 49 (iv) x2 + 2x + 1
(v) 4x2 + 4x + 1 (vi) a2x2 + 2ax + 1
(vii) a2x2 + 2abx + b2 (viii) a2x2 + 2abxy + b2y2
(ix) 4x2 + 12x + 9 (x) 16x2 + 40x +25
(xi) 9x2 + 24x + 16 (xii) 9x2 + 30x + 25
(xiii) 2x3 + 24x2 + 72x (xiv) a2x3 + 2abx2 + b2x
(xv) 4x4 + 12x3 + 9x2 (xvi)
(xvii)
Answer:
(i) x2 + 6x + 9
= x2 + 2 × 3 × x + (3)2
= (x + 3)2 [∵ a2 + 2ab + b2 = (a + b)2]
= (x + 3)2
= (x + 3) (x + 3)
(ii) x2 + 12x + 36
= x2 + 2 × 6 × x + (6)2
= (x + 6)2 [∵ a2 + 2ab + b2 = (a + b)2]
= (x + 6) (x + 6)
(iii) x2 + 14x + 49
= x2 + 2 × 7 × x + (7)2
= (x + 7)2 [∵ a2 + 2ab + b2 = (a + b)2]
= (x + 7) (x + 7)
(iv) x2 + 2x + 1
= x2 + 2 × 1 × x + (1)2
= (x + 1)2 [∵ a2 + 2ab + b2 = (a + b)2]
= (x + 1) (x + 1)
(v) 4x2 + 4x + 1
= (2x)2 + 2 × 2 × 1 × x + (1)2
= (2x + 1)2 [∵ a2 + 2ab + b2 = (a + b)2]
= (2x + 1) (2x + 1)
(vi) a2x2 + 2ax + 1
= (ax)2 + 2× ax × 1 + (1)2
= (ax + 1)2 [∵ a2 × 2ab + b2 = (a + b)2]
= (ax + 1)2
= (ax + 1) (ax + 1)
(vii) a2x2 + 2abx + b2
= (ax)2 + 2 × ax × b + (b)2
= (ax + b)2 [∵ a2 + 2ab + b2 = (a + b)2]
= (ax + b) (ax + b)
(viii) a2x2 + 2abxy + b2y2
= (ax)2 + 2 × ax × by + (by)2
= (ax + by)2 [∵ a2 + 2ab + b2 = (a + b)2]
(ix) 4x2 + 12x + 9
= (2x)2 + 2 × 2x × 3 + (3)2
= (2x + 3)2 [∵ a2 + 2ab + b2 = (a + b)2]
= (2x + 3) (2x + 3)
(x) 16x2 + 40x + 25
= (4x)2 + 2 × 4x × 5 + (5)2
= (4x + 5)2 [∵ a2 + 2ab + b2 = (a + b)2]
= (4x + 5) (4x + 5)
(xi) 9x2 + 24x + 16
= (3x)2 + 2 × 3x × 4 + (4)2
= (3x + 4)2 [∵ a2 + 2ab + b2 = (a + b)2]
= (3x + 4) (3x + 4)
(xii) 9x2 + 30x + 25
= (3x)2 + 2 × 3x × 5 + (5)2
= (3x + 5)2 [∵ a2 + 2ab + b2 = (a + b)2]
= (3x + 5) (3x + 5)
(xiii) 2x3 + 24x2 + 72x
= 2x(x2 + 12x + 36)
= 2x[(x)2 + 2 × x × 6 + (6)2]
= 2x(x + 6)2 [∵ a2 + 2ab + b2 = (a + b)2]
= 2x(x + 6) (x + 6)
(xiv) a2x3 + 2abx2 + b2x
= x(a2x2 + 2abx + b2)
= x[(ax)2 + 2 × ax × b + (b)2]
= x(ax + b)2 [∵ a2 + 2ab + b2 = (a + b)2]
(xv) 4x4 + 12x3 + 9x2
= x2(4x2 + 121x + 69)
= x2[(2x2 + 2 × 2x × 3 + (3)2]
= x2(2x + 3)2 [∵ a2 + 2ab + b2 = (a + b)2]
= x2(2x + 3) (2x + 3)
Page No 234:
Question 90:
Factorise the following, using the identity a2 – 2ab + b2 = (a – b)2.
(i) x2 – 8x + 16 (ii) x2 – 10x + 25
(iii) y2 – 14y + 49 (iv) p2 – 2p + 1
(v) 4a2 – 4ab + b2 (vi) p2y2 – 2py + 1
(vii) a2y2 – 2aby + b2 (viii) 9x2 – 12x + 4
(ix) 4y2 – 12y + 9
(x)
(xi) a2y3 – 2aby2 + b2y
(xii) 9y2 – 4xy +
Answer:
(i) x2 – 8x + 16
= (x)2 – 2 × x × 4 + (4)2
= (x – 4)2 [∵ a2 – 2ab + b2 = (a – b)2]
= (x – 4) (x – 4)
(ii) x2 – 10x + 25
= (x)2 – 2 × x × 5 + (5)2
= (x – 5)2 [∵ a2 – 2ab + b2 = (a – b)2]
= (x – 5) (x – 5)
(iii) y2 – 14y + 49
= (y)2 – 2 × y × 7 + (7)2
= (y – 7)2 [∵ a2 – 2ab + b2 = (a – b)2]
= (y – 7) (y – 7)
(iv) p2 – 2p + 1
= (p)2 – 2 × p × 1 + (1)2
= (p – 1)2 [∵ a2 – 2ab + b2 = (a – b)2]
= (p – 1) (p – 1)
(v) 4a2 – 4ab + b2
= (2a)2 – 2 × 2a × b + (b)2
= (2a – b)2 [∵ a2 – 2ab + b2 = (a – b)2]
= (2a – b) (2a – b)
(vi) p2y2 – 2py + 1
= (py)2 – 2 × py × 1 + (1)2
= (py – 1)2 [∵ a2 – 2ab + b2 = (a – b)2]
= (py – 1) (py – 1)
(vii) a2y2 – 2aby + b2
= (ay)2 – 2 × ay × b + (b)2
= (ay – b)2 [∵ a2 – 2ab + b2 = (a – b)2]
= (ay – b) (ay – b)
(viii) 9x2 – 12x + 4
= (3x)2 – 2 × 3x × 2 + (2)2
= (3x – 2)2 [∵ a2 – 2ab + b2 = (a – b)2]
= (3x – 2) (3x – 2)
(ix) 4y2 – 12y + 9
= (2y)2 – 2 × 2y × 3 + (3)2
= (2y – 3)2 [∵ a2 – 2ab + b2 = (a – b)2]
= (2y – 3) (2y – 3)
(xi) a2y3 – 2ab2 + b2y
= y(a2y2 – 2aby + b2)
= y[(ay)2 – 2 × ay × b + (b)2]
= y(ay – b) [∵ a2 – 2ab + b2 = (a – b)2]
Page No 235:
Question 91:
Factorise the following.
(i) x2 + 15x + 26 (ii) x2 + 9x + 20
(iii) y2 + 18x + 65 (iv) p2 + 14p + 13
(v) y2 + 4y – 21 (vi) y2 – 2y – 15
(vii) 18 + 11x + x2 (viii) x2 – 10x + 21
(ix) x2 = 17x + 60 (x) x2 + 4x – 77
(xi) y2 + 7y + 12 (xii) p2 – 13p – 30
(xiii) a2 – 16p – 80
Answer:
(i) x2 + 15x + 26
= x2 + 2x + 13x + 26 (By splitting the middle term)
= x(x + 2) + 13(x + 2)
(ii) x2 + 9x + 20
= x2 + 5x + 4x + 20 (By splitting the middle term)
= x(x + 5) +4(x + 5)
= (x + 4) (x + 5)
(iii) y2 + 18y + 65
= y2 + 13y + 5y + 65 (By splitting the middle term)
= y(y + 13) + 5(y + 13)
= (y + 5) (y + 13)
(iv) p2 + 14p + 13
= p2 + 13p + p + (13 × 1) (By splitting the middle term)
= p2 (p + 13) + 1(p + 13)
= (p + 13) (p + 1)
(v) y2 + 4y – 21
= y2 + 7y – 3y – 21 (By splitting the middle term)
= y(y + 7) –3(y + 7)
= (y – 3) (y + 7)
(vi) y2 – 2y – 15
= y2 + 3 – 5y – 15
= y2 + 3y – 5y – 15 (By splitting the middle term)
= y(y + 3) – 5(y + 3)
= (y – 5) (y + 3)
(vii) 18 + 11x + x2
= x2 + 11x + 18
= x2 + 9x + 2x + 18 (By splitting the middle term)
= x(x + 9) + 2(x + 9)
= (x + 2) (x + 9)
(viii) x2 – 10x + 21
= x2 – 7x – 3x – 21 (By splitting the middle term)
= x(x – 7) –3(x – 7)
= (x – 3) (x – 7)
(ix) x2 – 17x + 60
= x2 – 12x – 5x + 60 (By splitting the middle term)
= x(x – 12) –5(x – 12)
= (x – 5) (x – 12)
(x) x2 + 4x – 77
= x2 + 11x – 7x – 77 (By splitting the middle term)
= x(x + 11) – 7(x + 11)
= (x – 7) (x + 11)
(xi) y2 + 7y + 12
= y2 + 4y + 3y + 12 (By splitting the middle term)
= y(y + 4) + 3(y + 4)
= (y + 3) (y + 4)
(xii) p2 – 13p – 30
= p2 – 15p + 2p – 30 (By splitting the middle term)
= p(p – 15) + 2(p – 15)
= (p + 2) (p – 15)
(xiii) p2 – 16p – 80
= p2 – 20p + 4p – 80 (By splitting the middle term)
= p(p – 20) + 4(p – 20)
= (p + 4) (p – 20)
Page No 235:
Question 92:
Factorise the following using the identity a2 – b2 = (a + b) (a – b).
(i) x2 – 9 (ii) 4x2 – 25y2
(iii) 4x2 – 49y2 (iv) 3a2b3 – 27a4b
(v) 28ay2 – 175ax2 (vi) 9x2 – 1
(vii) 25ax2 – 25a (viii)
(ix) (x) 49x2 – 36y2
(xi) (xii)
(xiii) (xiv)
(xv) (xvi) 1331x3y – 11y3x
(xvii) (xviii) a4 – (a – b)4
(xix) x4 – 1 (xx) y4 – 625
(xxi) p5 – 16p (xxii) 16x4 – 81
(xxiii) x4 – y4 (xxiv) y4 – 81
(xxv) 16x4 – 625y4 (xxvi) (a – b)2 – (b – c)2
(xxvii) (x + y)4 – (x – y)4 (xxviii) x4 – y4 + x2 – y2
(xxix) 8a3 – 2a
(xxx)
(xxxi) 9x2 – (3y + z)2
Answer:
(i) x2 – 9
= (x)2 – (3)2
= (x + 3) (x – 3) [∵ a2 – b2 = (a + b) (a – b)]
(ii) 4x2 – 25y2
= (2x)2 – (5y)2
= (2x + 5y) (2x – 5y) [∵ a2 – b2 = (a + b) (a – b)]
(iii) 4x2 – 49y2
= (2x)2 – (7y)2
= (2x + 7y) (2x – 7y) [∵ a2 – b2 = (a + b) (a – b)]
(iv) 3a2b3 – 27a4b
= 3a2b(b2 – 9a2)
= 3a2b[(b)2 – (3a)2]
= (3a2b) (b + 3a) (b – 3a) [∵ a2 – b2 = (a + b) (a – b)]
(v) 28ay2 – 175ax2
= 7a(4y2 – 25x2)
= 7a[(2y)2 – (5x)2]
= 7a(2y + 5x) (2y – 5x) [∵ a2 – b2 = (a + b) (a – b)]
(vi) 9x2 – 1
= (3x)2 – (1)2
= (3x + 1) (3x – 1) [∵ a2 – b2 = (a + b) (a – b)]
(vii) 25a2 – 25a
= 25a[x2 – (1)2]
= 25a(x + 1) (x – 1) [∵ a2 – b2 = (a + b) (a – b)]
(x) 49x2 – 36y2
= (7x)2 – (6y)2
= (7x + 6y) (7x – 6y) [∵ a2 – b2 = (a + b) (a – b)]
(xvi) 1331x3y – 11y3x
= (11)3x3y – (11)y3x
= 11xy[(11x)2 – (y)2]
= 11xy(11x + y) (11x – y) [∵ a2 – b2 = (a + b) (a – b)]
(xviii) a4 – (a – b)4
= (a2)2 – ((a – b)2)2
= [a2 + (a – b) [a2 – (a – b)2]
= a2 + a2 + b2 – 2ab [a2 – (a2 + b2 – 2ab)] [∵ a2 – b2 = (a + b) (a – b)]
= (2a2 + b2 – 2ab) (a2 – a2 – b2 + 2ab)
= (2a2 + b2 – 2ab (–b2 + 2ab)
(xix) x4 – 1
= (x2)2 – (1)2
= (x2 + 1) (x2 – 1) [∵ a2 – b2 = (a + b) (a – b)]
= (x2 + 1) (x + 1) (x – 1) [∵ a2 – b2 = (a + b) (a – b)]
(xx) y4 – 625
= (y2)2 – (25)2
= (y2 + 25) (y2 – 25) [∵ a2 – b2 = (a + b) (a – b)]
= (y2 + 25) [y2 – (5)2]
= (y2 + 25) (y + 5) (y – 5) [∵ a2 – b2 = (a + b) (a – b)]
(xxi) p5 – 16p
= p(p4 – 16)
= p[(p2)2 – (4)2]
= p(p2 + 4) (p2 – 4) [∵ a2 – b2 = (a + b) (a – b)]
= p(p2 + 4) (p + 2) (p – 2) [∵ a2 – b2 = (a + b) (a – b)]
(xxii) 16x4 – 81
= (4x2)2 – (9)2
= (4x2 + 9) (4x2 – 9)
= (4x2 + 9) [(2x)2 – (3)2]
= (4x2 + 9) (2x + 3) (2x – 3) [∵ a2 – b2 = (a + b) (a – b)]
(xxiii) x4 – y4
= (x2)2 – (y2)2
= (x2 + y2) (x2 – y2) [∵ a2 – b2 = (a + b) (a – b)]
= (x2 + y2) (x + y) (x – y) [∵ a2 – b2 = (a + b) (a – b)]
(xxiv) y4 – 81
= (y2)2 – (9)2
= (y2 + 9) (y2 – 9) [∵ a2 – b2 = (a + b) (a – b)]
= (y2 + 9) [y2 – (3)2]
= (y2 + 9) (y + 3) (y – 3) [∵ a2 – b2 = (a + b) (a – b)]
(xxv) 16x4 – 625y4
= (4x2)2 – (25y2)2
= (4x2 + 25y2) (4x2 – 25y2) [∵ a2 – b2 = (a + b) (a – b)]
= (4x2 + 25y2) [(2x)2 – (5y)2]
= (4x2 + 25y2 (2x + 5y) (2x – 5y) [∵ a2 – b2 = (a + b) (a – b)]
(xxvi) (a – b)2 – (b – c)2
= (a – b + b – c) [(a – b) – (b – c)] [∵ a2 – b2 = (a + b) (a – b)]
= (a – c) (a – b – b + c)
= (a – c) (a – 2b + c)
(xxvii) (x + y)4 – (x – y)4
= [(x + y)2 – (x – y)2]2
= (x + y)2 + (x – y)2] (x + y)2 – (x + y)2] [∵ a2 – b2 = (a + b) (a – b)]
= (x2 + y2 + 2xy + x2 + y2 – 2xy) [(x + y)2 – (x – y)2] [∵ (a + b)2 = a2 + b2 + 2ab and (a – b)2 = a2 + b2 – 2ab]
= (2x2 + 2y2) (x + y + x – y) [(x + y) – (x – y) [∵ a2 – b2 = (a + b) (a – b)]
= (2x2 + 2y2) (2x) (x + y – x + y)
= (2x2 + 2y2) (2x) (2y)
= 8xy(x2 + y2)
(xxviii) x4 – y4 + x2 – y2
= (x2)2 – (y2)2 + x2 – y2
= (x2 – y2) (x2 + y2) + (x2 – y2) [∵ a2 – b2 = (a + b) (a – b)]
= (x2 – y2) (x2 + y2 + 1)
= (x + y) (x – y) (x2 + y2 + 1) [∵ a2 – b2 = (a + b) (a – b)]
(xxix) 8a3 – 2a
= 2a (4a2 – 1)
= 2a [(2a)2 – (1)2]
= 2a (2a + 1) (2a – 1) [∵ a2 – b2 = (a + b) (a – b)]
(xxxi) 9x2 – (3y + 2)2
= (3x)2 – (3y + 2)2
= (3x + 3y + 2) [3x – (3y + 2)] [∵ a2 – b2 = (a + b) (a – b)]
= (3x + 3y + 2) (3x – 3y – 2)
Page No 236:
Question 93:
The following expressions are the areas of rectangles. Find the possible lengths and breadths of these rectangles.
(i) x2 – 6x + 8 (ii) x2 – 3x + 2
(iii) x2 – 7x + 10 (iv) x2 + 19x – 20
(v) x2 + 9x + 20
Answer:
(i) x2 – 6x + 8
= x2 – 4x – 2x + 8 (By splitting the middle term)
= x(x – 4) –2(x – 4)
= (x – 2) (x – 4)
As, area of rectangle is length × breadth
∴ Possible length and Breadth of the rectangle are (x – 2) and (x – 4)
(ii) x2 – 3x + 2
= x2 – 2x – x + 2 (By splitting the middle term)
= x(x – 2) –1(x – 2)
= (x – 1) (x – 2)
As, are of the rectangle is length × breadth
∴ Possible length and breadth of the rectangle are (x –1) and (x – 2)
(iii) x2 – 7x + 10
= x2 – 5x – 2x + 10 (By splitting the middle term)
= x(x – 5) –2(x – 5)
= (x – 2) (x – 5)
As, area of the rectangle is length × breadth
∴ Possible length and breadth of the rectangle are (x – 2) and (x – 5).
(iv) x2 + 19x – 20
= x2 + 20x – x – 20 (By splitting the middle term)
= x(x + 20) – (x + 20)
= (x – 1) (x + 20)
As, area of a rectangle is length × breadth
∴ Possible length nad breadth of the rectangle are (x + 20) and (x – 1).
(v) x2 + 9x + 20
= x2 + 5x + 4x + 20 (By splitting the middle term)
= x(x + 5) + 4(x + 5)
= (x + 4) (x + 5)
As, are of a rectangle is length × breadth.
∴ Possible length and breadth of the rectangle are (x + 5) and (x + 4).
Page No 236:
Question 94:
Carry out the following divisions:
(i) 51x3y2z ÷ 17xyz (ii) 76x3yz3 ÷ 19x2y2
(iii) 17ab2c3 ÷ (–abc2) (iv) –121p3q3r3 ÷ (–11xy2z3)
Answer:
Page No 236:
Question 95:
Perform the following divisions:
(i) (3pqr – 6p2q2r2) ÷ 3pq
(ii) (ax3 – bx2 + cx) ÷ (– dx)
(iii) (x3y3 + x2y3 – xy4 + xy) ÷ xy
(iv) (– qrxy + xryz – rxyz) ÷ (– xyz)
Answer:
Page No 236:
Question 96:
Factorise the expressions and divide them as directed:
(i) (x2 – 22x + 117) ÷ (x – 13)
(ii) (x3 + x2 – 132x) ÷ x (x – 11)
(iii) (2x3 – 12x2 + 16x) ÷ (x – 2) (x – 4)
(iv) (9x2 – 4) ÷ (3x + 2)
(v) (3x2 – 48) ÷ (x – 4)
(vi) (x4 – 16) ÷ x3 + 2x2 + 4x + 8
(vii) (3x4 – 1875) ÷ (3x2 – 75)
Answer:
Page No 236:
Question 97:
The area of a square is given by 4x2 + 12xy + 9y2. Find the side of the square.
Answer:
Area of the square = 4x2 + 12xy + 9y2
So, 4x2 + 12xy + 9y2
= (2x)2 + 2 × 2x × 3y + (3y)2
= (2x + 3y)2 [∵ a2 + 2ab + b2 = (a + b)2]
Since, the area of a square having side length 'a' is a2. Hence, side of the given square is 2x + 3y.
Page No 236:
Question 98:
The area of a square is 9x2 + 24xy + 16y2. Find the side of the square.
Answer:
Area of the square is 9x2 + 24xy + 16y2.
So, 9x2 + 24xy + 16y2
= (3x)2 + 2 × (3x) × (4y) + (4y)2
= (3x + 4y)2 [∵ a2 + 2ab + b2 = (a + b)2
Since, the area of a square having side length 'a' is a2. Hence, side of the given square is 3x + 4y.
Page No 237:
Question 99:
The area of a rectangle is x2 + 7x + 12. If its breadth is (x + 3), then find its length.
Answer:
Given: the area of a rectangle is x2 + 7x + 12 and its breadth is (x + 3).
Since, the area of the rectangle = Length × Breadth
∴ Length of the rectangle =
Hence, the length of the rectangle is x + 4.
Page No 237:
Question 100:
The curved surface area of a cylinder is 2π (y2 – 7y + 12) and its radius is (y – 3). Find the height of the cylinder (C.S.A. of cylinder =2πrh).
Answer:
Given: The curved surface area of a cylinder is 2π(y2 – 7y + 12) and its radius is (y – 3).
As, the CSA of cylinder = 2πrh
∴ 2πrh = 2π(y2 – 7y + 12)
⇒ 2π × (y – 3)h = 2π(y2 – 7y + 12)
Page No 237:
Question 101:
The area of a circle is given by the expression πx2 + 6πx + 9π. Find the radius of the circle.
Answer:
Given: The area of a circle is πx2 + 6πx + 9π
As, the area of a circle = πr2
Hence, the radius of the circle is x + 3.
Page No 237:
Question 102:
The sum of first n natural numbers is given by the expression . Factorise this expression.
Answer:
Given: The sum of first 'n' natural numbers
Consider:
Hence, factorisation of the expression
Page No 237:
Question 103:
The sum of (x + 5) observations is x4 – 625. Find the mean of the observations.
Answer:
Given: The sum of (x + 5) observations = x4 – 625.
As, the mean of the 'n' observations
∴ The mean of (x + 5) observations
Page No 237:
Question 104:
The height of a triangle is x4 + y4 and its base is 14xy. Find the area of the triangle.
Answer:
Given: The height of a triangle and its base are x4 + y4 and 14xy, respectively.
The area of the triangle
Page No 237:
Question 105:
The cost of chocolate is ₹(x + 4) and Rohit bought (x + 4) chocolates. Find the total amount paid by him in terms of x. If x = 10, find the amount paid by him.
Answer:
Given: The cost of a chocolate = ₹(x + 4)
Rohit bought (x + 4) chocolates.
∴ The cost of (x + 4) chocolates
= Cost of one chocolate × Number of chocolates
= (x + 4) (x + 4)
= (x + 4)2
= x2 + (4)2 + 2 × x × 4 [∵ (a + b)2 = a2 + b2 + 2ab]
= x2 + 16 + 8x
∴ The total amount paid by Rohit = ₹(x2 + 8x + 16)
Now, if x = 10.
Then, the amount paid by Rohit
Page No 237:
Question 106:
The base of a parallelogram is (2x + 3 units) and the corresponding height is (2x – 3 units). Find the area of the parallelogram in terms of x. What will be the area of parallelogram of x = 30 units?
Answer:
Given: The base and the height of a parallelogram are (2x + 3) units and (2x – 3) units, respectively.
∴ The area of a parallelogram = Base × Height
Now, if x = 30 units then, the area of the parallelogram = 4 × (30)2 – 9 = 3591 sq. units.
Page No 237:
Question 107:
The radius of a circle is 7ab – 7bc – 14ac. Find the circumference of the circle.
Answer:
Given: The radius of the circle is 7ab – 7bc – 14ac
∴ The circumference of the circle
Page No 237:
Question 108:
If p + q = 12 and pq = 22, then find p2 + q2.
Answer:
Given: p + q = 12 and pq = 22
As, (p + q)2 = p2 + q2 + 2pq
(12)2 = p2 + q2 + (2 × 22)
p2 + q2 = (12)2 – 44
Page No 237:
Question 109:
If a + b = 25 and a2 + b2 = 225, then find ab.
Answer:
Given: and
As,
Page No 237:
Question 110:
If x – y = 13 and xy = 28, then find x2 + y2.
Answer:
Given: and
As
Page No 237:
Question 111:
If m – n = 16 and m2 + n2 = 400, then find mn.
Answer:
Given: m – n = 16 and m2 + n2 = 400
As, (m – n)2 = m2 + n2 – 2 mn
(16)2 = 400 – 2 mn
mn = 12
Page No 237:
Question 112:
If a2 + b2 = 74 and ab = 35, then find a + b.
Answer:
Given:
As
(a + b) = 74 + 70
(a + b)2 = 144
(a + b)2 = (12)2
a + b = 12
Page No 238:
Question 113:
Verify the following:
(i) (ab + bc) (ab – bc) + (bc + ca) (bc – ca) + (ca + ab) (ca – ab) = 0
(ii) (a + b + c) (a2 + b2 + c2 – ab – bc – ca) = a3 + b3+ c3 – 3abc
(iii) (p – q) (p2 + pq + q2) = p3 – q3
(iv) (m + n) (m2 – mn + n2) = m3 + n3
(v) (a + b) (a + b) (a + b) = a3 + 3a2b + 3ab2 + b3
(vi) (a – b) (a – b) (a – b) = a3 – 3a2b + 3ab2 – b3
(vii) (a2 – b2) (a2 + b2) + (b2 – c2) (b2 + c2) + (c2 – a2) + (c2 + a2) = 0
(viii) (5x + 8)2 – 160x = (5x – 8)2
(ix) (7p – 13q)2 + 364pq = (7p + 13q)2
(x)
Answer:
(i) LHS = (ab + bc) (ab – bc) + (bc + ca) (bc – ca) + (ca + ab) (ca – ab)
= (ab)2 – (bc)2 + (bc)2– (ca)2 + (ca)2 – (ab)2
[ (a + b) (a – b) = a2 – b2]
= a2b2 – b2c2 + b2c2 – c2a2 + c2a2 – a2b2
(Grouping like terms)
= 0
= RHS
Hence verified
(ii) LHS =
(Grouping like terms)
= RHS
= Hence verified
(iii) LHS
(Rearranging like terms)
= RHS
Hence verified
(iv) LHS =
= RHS
Hence verified
(v) LHS
Hence verified
(vi) LHS
= RHS
Hence verified.
(vii) LHS =
(Grouping like terms)
= 0
= RHS
Hence verified.
(viii)
= RHS
Hence verified
(ix)
Hence verified
(x)
= RHS
Hence verified
Page No 238:
Question 114:
Find the value of a, if
(i) 8a = 352 – 272
(ii) 9a = 762 – 672
(iii) pqa = (3p + q)2 – (3p – q)2
(iv) pq2a = (4pq + 3q)2 – (4pq – 3q)2
Answer:
(i)
(ii)
(iii)
(iv)
Page No 238:
Question 115:
What should be added to 4c (– a + b + c) to obtain 3a (a + b + c) – 2b (a – b + c)?
Answer:
Let x be added to the given expression
to obtain
i.e.
Page No 238:
Question 116:
Subtract b (b2 + b – 7) + 5 from 3b2 – 8 and find the value of expression obtained for b = – 3.
Answer:
Required difference
Now for b = –3, the value of the above
expression will be
i.e. 27 + 18 – 21 – 13 = 45 – 34 = 11
Page No 238:
Question 117:
If then find the value of
Answer:
Given:
Now,
Page No 238:
Question 118:
Factorise
Answer:
Page No 238:
Question 119:
Factorise p4 + q4 + p2q2.
Answer:
Page No 238:
Question 120:
Find the value of
(i)
(ii)
Answer:
(ii)
Page No 239:
Question 121:
The product of two expressions is x5 + x3 + x. If one of them is x2 + x + 1, find the other.
Answer:
Given: The product of two expressions is
If one of the numbers is
then, the other one is given by:
Other expression
Hence, the other expression is x (x2 – x + 1).
Page No 239:
Question 122:
Find the length of the side of the given square if area of the square is 625 square units and then find the value of x.
Answer:
Area of the square = (side)2
Thus, the side of the square is 25 units.
Hence, the value of x is 5.
Page No 239:
Question 123:
Take suitable number of cards given in the adjoining diagram [G(x × x) representing x2, R (x × 1) representing x and Y (1 × 1) representing
1] to factorise the following expressions, by arranging the cards inthe form of rectangles: (i) 2x2 + 6x + 4 (ii) x2 + 4x + 4. Factorise 2x2 + 6x + 4 by using the figure.
Calculate the area of figure.
Answer:
Disclaimer:
The given information is incomplete for solution of this question.
Page No 239:
Question 124:
The figure shows the dimensions of a wall having a window and a door of a room. Write an algebraic expression for the area of the wall to be painted.
Answer:
Length of the window = 2x
Breadth of the window = x
The area of the window = 2x × x = 2x2 square units
Length of the door = 3x
Breadth of the door = x
The area of the door = 3x × x
= 3x2 square units
Length of the wall = 5x + 2
Breadth of the wall = 5x
The area of the wall = (5x + 2) × 5x
= 25x2 + 10x square units
Now, area of the required part of wall that is to be painted = Area of the wall – (Area of the window + Area of the door)
Page No 240:
Question 125:
Match the expressions of column I with that of column II:
Column I Column II
(1) (21x + 13y)2 (a) 441x2 – 169y2
(2) (21x – 13y)2 (b) 441x2 + 169y2 + 546 xy
(3) (21x – 13y) (21x + 13y) (c) 441x2 + 169y2 – 546xy
(d) 441x2 – 169y2 + 546xy
Answer:
(i)
(ii)
(iii)
Hence, (i) (a), (ii) (c), (iii) (a).
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