Q1: Find the common factors of the following terms.
(a) 25x2y, 30xy2
(b) 63m3n, 54mn4
Sol:
(a) 25x2y, 30xy2
25x2y = 5 × 5 × x × x × y
30xy2 = 2 × 3 × 5 × x × y × y
Common factors are 5× x × y = 5 xy
(b) 63m3n, 54mn4
63m3n = 3 × 3 × 7 × m × m × m × n
54mn4 = 2 × 3 × 3 × 3 × m × n × n × n × n
Common factors are 3 × 3 × m × n = 9mn
Q2: Factorise the following expressions.
(a) 54m3n + 81m4n2
(b) 15x2y3z + 25x3y2z + 35x2y2z2
Sol:
(a) 54m3n + 81m4n2
= 2 × 3 × 3 × 3 × m × m × m × n + 3 × 3 × 3 × 3 × m × m × m × m × n × n
= 3 × 3 × 3 × m × m × m × n × (2 + 3 mn)
= 27m3n (2 + 3mn)
(b) 15x2y3z + 25 x3y2z + 35x2y2z2 = 5x2y2z ( 3y + 5x + 7)
Q3: Factorise the following polynomials.
(a) 6p(p – 3) + 1 (p – 3)
(b) 14(3y – 5z)3 + 7(3y – 5z)2
Sol: (a) 6p(p – 3) + 1 (p – 3) = (p – 3) (6p + 1)
(b) 14(3y – 5z)3 + 7(3y – 5z)2
= 7(3y – 5z)2 [2(3y – 5z) +1]
= 7(3y – 5z)2 (6y – 10z + 1)
Q4: Factorise the following:
(a) p2q – pr2 – pq + r2
(b) x2 + yz + xy + xz
Sol: (a) p2q – pr2 – pq + r2
= (p2q – pq) + (-pr2 + r2)
= pq(p – 1) – r2(p – 1)
= (p – 1) (pq – r2)
(b) x2 + yz + xy + xz
= x2 + xy +xz + yz
= x(x + y) + z(x + y)
= (x + y) (x + z)
Q5: Factorise the following polynomials.
(a) xy(z2 + 1) + z(x2 + y2)
(b) 2axy2 + 10x + 3ay2 + 15
Sol:
(a) xy(z2 + 1) + z(x2 + y2)
= xyz2 + xy + 2x2 + zy2
= (xyz2 + zx2) + (xy + zy2)
= zx(yz + x) + y(x + yz)
= zx(x + yz) + y(x + yz)
= (x + yz) (zx + y)
(b) 2axy2 + 10x + 3ay2 + 15
= (2axy2 + 3ay2) + (10x + 15)
= ay2(2x + 3) +5(2x + 3)
= (2x + 3) (ay2 + 5)
Q6: Factorise the following expressions.
(а) x2 + 4x + 8y + 4xy + 4y2
(b) 4p2 + 2q2 + p2q2 + 8
Sol: (a) x2 + 4x + 8y + 4xy + 4y2
= (x2 + 4xy + 4y2) + (4x + 8y)
= (x + 2y)2 + 4(x + 2y)
= (x + 2y)(x + 2y + 4)
(b) 4p2 + 2q2 + p2q2 + 8
= (4p2 + 8) + (p2q2 + 2q2)
= 4(p2 + 2) + q2(p2 + 2)
= (p2 + 2)(4 + q2)
Q7: Factorise:
(a) a2 + 14a + 48
(b) m2 – 10m – 56
Sol:
(a) a2 + 14a + 48
= a2 + 6a + 8a + 48
[6 + 8 = 14 ; 6 × 8 = 48]
= a(a + 6) + 8(a + 6)
= (a + 6) (a + 8)
(b) m2 – 10m – 56
= m2 – 14m + 4m – 56
[14 – 4 = 10; 4 × 4 = 56]
= m(m – 14) + 6(m – 14)
= (m – 14) (m + 6)
Q8: Factorise:
(a) x4 – (x – y)4
(b) 4x2 + 9 – 12x – a2 – b2 + 2ab
Sol: (a) x4 – (x – y)4
= (x2)2 – [(x – y)2]2
= [x2 – (x – y)2] [x2 + (x – y)2]
= [x + (x – y] [x – (x – y)] [x2 + x2 – 2xy + y2]
= (x + x – y) (x – x + y)[2x2 – 2xy + y2]
= (2x – y) y(2x2 – 2xy + y2)
= y(2x – y) (2x2 – 2xy + y2)
(b) 4x2 + 9 – 12x – a2 – b2 + 2ab
= (4x2 – 12x + 9) – (a2 + b2 – 2ab)
= (2x – 3)2 – (a – b)2
= [(2x – 3) + (a – b)] [(2x – 3) – (a – b)]
= (2x – 3 + a – b)(2x – 3 – a + b)
Q9: Factorise the following polynomials.
(a) 16x4 – 81
(b) (a – b)2 + 4ab
Sol: (a) 16x4 – 81
= (4x2)2 – (9)2
= (4x2 + 9)(4x2 – 9)
= (4x2 + 9)[(2x)2 – (3)2]
= (4x2 + 9)(2x + 3) (2x – 3)
(b) (a – b)2 + 4ab
= a2 – 2ab + b2 + 4ab
= a2 + 2ab + b2
= (a + b)2
Q10: Factorise:
(а) 14m5n4p2 – 42m7n3p7 – 70m6n4p3
(b) 2a2(b2 – c2) + b2(2c2 – 2a2) + 2c2(a2 – b2)
Sol:
(a) 14m5n4p2 – 42m7n3p7 – 70m6n4p3
= 14m5n3p2(n – 3m2p5 – 5mnp)
(b) 2a2(b2 – c2) + b2(2c2 – 2a2) + 2c2(a2 – b2)
= 2a2(b2 – c2) + 2b2(c2 – a2) + 2c2(a2 – b2)
= 2[a2(b2 – c2) + b2(c2 – a2) + c2(a2 – b2)]
= 2 × 0
= 0
Q11: Factorise:
(a) (x + y)2 – 4xy – 9z2
(b) 25x2 – 4y2 + 28yz – 49z2
Sol: (a) (x + y)2 – 4xy – 9z2
= x2 + 2xy + y2 – 4xy – 9z2
= (x2 – 2xy + y2) – 9z2
= (x – y)2 – (3z)2
= (x – y + 3z) (x – y – 3z)
(b) 25x2 – 4y2 + 28yz – 49z2
= 25x2 – (4y2 – 28yz + 49z2)
= (5x)2 – (2y – 7)2
= (5x + 2y – 7) [5x – (2y – 7)]
= (5x + 2y – 7) (5x – 2y + 7)
Q12: If one of the factors of (5x2 + 70x – 160) is (x – 2). Find the other factor.
Sol: Let the other factor be m.
(x – 2) × m = 5x2 + 70x – 160
Q13: Express the following as in the form of (a+b)(a-b)
(i) a2 – 64
(ii) 20a2 – 45b2
(iii) 32x2y2 – 8
(iv) x2 – 2xy + y2 – z2
(v) 49x2 – 1
Sol: For representing the expressions in (a+b)(a-b) form, use the following formula
a2 – b2 = (a+b)(a-b)
(i) a2 – 64 = a2 – 82 = (a + 8)(a – 8)
(ii) 20a2 – 45b2 = 5(4a2 – 9b2) = 5(2a + 3b)(2a – 3b)
(iii) 32x2y2 – 8 = 8( 4x2y2 – 1) = 8(2xy + 1)(2xy – 1)
(iv) x2 – 2xy + y2 – z2 = (x – y)2 – z2 = (x – y – z)(x – y + z)
(v) 49x2 – 1 = (7x)2 – (1)2 = (7x + 1)(7x – 1)
Q14: Verify whether the following equations are correct. Rewrite the incorrect equations correctly.
(i) (a + 6)2 = a2 + 12a + 36
(ii) (2a)2 + 5a = 4a + 5a
Sol:
(i) (a + 6)2 = a2 + 12a + 36
Here, LHS = (a + 6)2 = a2 + 12a + 36
Now, RHS = a2 + 12a + 36
Hence, LHS = RHS.
(ii) (2a)2 + 5a = 4a + 5a
Here, LHS = (2a)2 + 5a = 4a2 + 5a
Now, RHS = 4a + 5a
So, LHS ≠ RHS
Correct equation: (2a)2 + 5a = 4a2 + 5a
Q15: For a = 3, simplify a2 + 5a + 4 and a2 – 5a
Sol:
Substitute the value of a = 3 in the given equations.
a2 + 5a + 4 = 32 + 5(3) + 4 = 9 + 15 + 4 = 28
And,
a2 – 5a = 32 – 5(3) = 9 – 15 = -6
Q16: Find the common factors of the following:
(i) 6 xyz, 24 xy2 and 12 x2y
(ii) 3x2 y3, 10x3 y2 and 6x2 y2 z
Sol:
(i) 6 xyz = 2 × 3 × x × y × z
24 xy2 = 2 × 2 × 2 × 3 × x × y × y
12 x2y = 2 × 2 × 3 × x × x × y
Thus, the common factors are common factors of 6 xyz, 24 xy2 and 12 x2y are 2, 3, x, y and, (2 × 3 × x × y) = 6xy
(ii) 3x2 y3 = 3 × x × x × y × y × y
10x3 y2 = 2 × 5 × x × x × x × y × y
6 x2 y2 z = 3 × 2 × x × x × y × y × z
Now, the common factors of 3x2 y3, 10x3 y2 and 6x2 y2 z are x2, y2 and, (x2 × y2) = x2 y2
Q17: Factorize the following expressions:
(i) 54x3y + 81x4y2
(ii) 14(3x – 5y)3 + 7(3x – 5y)2
(iii) 15xy + 15 + 9y + 25x
Sol:
(i) 54x3y + 81x4y2
= 2 × 3 × 3 × 3 × x × x × x × y + 3 × 3 × 3 × 3 × x × x × x × x × y × y
= 3 × 3 × 3 × x × x × x × y × (2 + 3 xy)
= 27x3y (2 + 3 xy)
(ii) 14(3x – 5y)3 + 7(3x – 5y)2
= 7(3x – 5y)2 [2(3x – 5y) +1]
= 7(3x – 5y)2 (6x – 10y + 1)
(iii) 15xy + 15 + 9y + 25x
Rearrange the terms as:
15xy + 25x + 9y + 15
= 5x(3y + 5) + 3(3y + 5)
Or, (5x + 3)(3y + 5)
Q18: Factorize (x + y)2 – 4xy
Sol:
To solve this expression, expand (x + y)2
Use the formula:
(x + y)2 = x2 + 2xy + y2
(x + y)2 – 4xy = x2 + 2xy + y2 – 4xy
= x2 + y2 – 2xy
We know, (x – y)2 = x2 + y2 – 2xy
So, factorization of (x + y)2 – 4xy = (x – y)2
Q19: When we factorise an expression, we write it as a ______ of factors.
Sol: When we factorise an algebraic expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions.
Q20: Find the common factors of 12x, 36
(a) 12
(b) 36
(c) x
(d) 12x
Ans: (a)
Sol: Here,
12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3
So, common factor = 2 × 2 × 3
⇒ 4 x 3 = 12
⇒ 4 × 3 = 12
Hence the correct option is (a).
Q22: Factorise using identity x2 + 10x + 25
Sol:
x2 + 10x + 25 = x2 + 2 × 5 × x + 52
Using the identity, (a + b)2 = a2 + 2ab + b2
= x2 + 2 × 5 × x + 52
= (x + 5)2= (x + 5)2
= (x + 5)(x + 5)
Q23: Simplify [102 - 18 × 10 + 81]
Sol: [102-18 × 10 + 81] = [102 − 2 × 9 × 10 + 92]
Using the identity, (a−b)2 = a2− 2ab+b2
= [102 − 2 × 9 × 10 + 92]
= [10 − 9]2
= [1]2
=1
Q24: Simplify
Sol:
Using the identity, x2−y2 = (x + y)(x − y)
Q25: Find x if
Sol: Ans: Taking 4 common from the numerator we get
Using the identity, x2 − y2 = (x + y)(x − y) in the numerator
On comparing both side we can see that x = 4.
Q26: Find the remainder in the following (x4 - a4) ÷ (x2 + a2)
Sol: For dividing both the equation first we have to simplify it.
So, (x4 − a4) can be written as (x2)2−(a2)2.
Now by applying the identity a2 − b2 = (a + b)(a − b)
Hence the remainder is (x2 - a2).
Q27: Using identity (a - b)2 = a2 - 2ab + b2 find the value of (98)2 .
Sol: (98)2 can be further written as (100 - 2)2
By comparing the above equation with the identity, we get
a = 100 and b = 2
(98)2 = (100)2 − 2(100)(2) + (2)2
=10000−400+4
=9604
Q28: Simplify
Sol: Identities to be used in the question are;
a2 − b2 = (a + b)(a − b)
Q29: The area of a rectangle is 6a2 + 36a and its width is 36a. Find its length.
Sol: Let the length of the rectangle is x
Breath = 36a
Area of rectangle = Length × Breath
6a2 + 36a = x × 36a
Hence the length of rectangle is
Q30: The combined area of two squares is 20cm2. Each side of one square is twice as long as a side of the other square. Find the length of the sides of each square.
Sol: Let the side of the smaller square be S and that of the bigger square be 2S.
Combined area of the two squares = 20cm2
S2 + (2S)2 = 20
S2 + 4S2 = 20
5S2 = 20
S2 = 4
S = 2cm
Hence the side of the smaller square is 2cm and that of the bigger square is 4cm.
Q31: Find the factors of 25x2 - 4y2 + 28yz - 49z2.
Sol:
25x2 - 4y2 + 28yz - 49z2 = 25x2 - (4y2 - 28yz + 49z2)
= 25x2- [ (2y)2- 2 × 2y × 7z + (7z)2]
Using the identity , (a − b)2 = a2 − 2ab + b2
(5x)2 − (2y − 7z)2
Now using the identity , a2 − b2 = (a + b)(a − b)
(5x + 2y − 7z)(5x − 2y + 7y)
Q32: Factorise 6xy – 4y + 6 – 9x
(a) (3x – 2)
(b) (3x – 2)(2y – 3)
(c) (2y – 3)
(d) (2x – 3)(3y – 2)
Ans: (b)
Sol: 6xy − 4y + 6 − 9x = 2 × 3 × x × y − 2 × 2 × y + 2 × 3 − 3 × 3 × x
= 2y(3x − 2) − 3(3x − 2)
= (3x − 2)(2y − 3)
Hence the correct option is (b).
Q33: Simplify
Sol:
(p2 + 11p + 28) = (p2 + 7p + 4p + 28)
= p(p + 7) + 4(p + 7)
= (p + 7)(p + 4)
Now,
= (p + 7)
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