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x² + 4x + 2x + 8 

x(x + 4) + 2(x + 4) 

(x + 4)(x + 2)

(i) lx² + mx

= lx² + mx

= l × x × x + m × x 

= x(lx + m)

(ii) 7y² + 35z²

= 7y²+ 35z²

= 7 × y² + 7 × 5 × z²

= 7(y² + 5z²)

(iii) 3x⁴ + 6x³y + 9x²Z

= 3x⁴ + 6x³y + 9x²Z

= 3 × x² × x² + 3 × 2 × x × x² × y + 3 × 3 × x² × z

= 3x² (x² + 2xy + 3z)

First find the two numbers whose sum =10 and product = 24

Clearly, the numbers are 6 and 4

∴ we get, y² +10y + 24 

= y² + 6y + 4y + 24

= y(y+6) + 4(y+6)

= (y+4) (y+6)

i) x² – 36

= x² – 36

⇒ (x)² – (6)² is in the form of a² – b²

a² – b² = (a + b) (a – b)

∴ x² – 36 = (x + 6) (x – 6)

ii) 49x² – 25y²

= (7x)² – (5y)²

= (7x + 5y) (7x – 5y)

iii) m² – 121

= m² -121

= (m)² – (11)²

= (m + 11) (m – 11)

iv) 81 – 64x²

= 81 – 64x²

= (9)² – (8x)²

= (9 + 8x) (9 – 8x)

(i) x⁴ – y⁴

= (x²)² – (y²)² is in the form of a² – b²

a² – b² = (a + b) (a – b)

x⁴ – y⁴ = (x² + y²)(x² – y²)

= (x² + y²)(x + y)(x – y)

(ii) a⁴ – (b + c)⁴

a⁴ – (b + c)⁴

= (a²)² – [(b + c)²]²

= [a² + (b + c)²] [a² – (b + c)²] ,

= [a² + (b + c)²] (a + b + c) [a – (b + c)]

= [a² + (b + c)²] (a + b + c) (a – b – c)

(iii) l² – (m – n)²

l² – (m – n)²

= (l)² – (m – n)²

= [l + m – n] [l – (m – n)]

= [l + m -n] [l – m + n]

B) 5ab (3a² – 7b²) 

Correct option is (B) 5ab (3a² – 7b²)

\(15a^3b-35ab^3\) \(=5ab(3a^2-7b^2)\)

First find the two numbers whose sum=6 and product= 8

Clearly, the numbers are 4 and 2

∴ we get, p² +6p + 8 

= p² + 4p + 2p + 8

= p (p+4) + 2(p+4)

= (p+2) (p+4)

i) 81x⁴ – 121x²

= 81x⁴ – 121x²

= x²(81² – 121)

= x²[(9x)² – (11)²]

= x²(9x + 11) (9x -11)

ii) (p² – 2pq + q²) - r²

= (p² – 2pq + q²) – r²

= (p – q)² – (r)²  [∵ p² – 2pq + q² = (p – q)²]

= (p – q + r) (p – q – r)

iii) (x + y)² – (x – y)²

(x + y)² – (x – y)²

It is in the form of a² – b²

a = x + y, b = x - y

∴ a² – b² = (a + b)(a-b)

= (x + y + x – y) [(x + y) - (x – y)]

= 2x [x + y - x + y]

= 2x x 2y = 4xy

i) x²y² – 64

= (xy)² – (8)²

= (xy + 8)(xy – 8)

ii) 6x² – 54

= 6x² – 54

= 6x² – 6 x 9 ‘

= 6(x² – 9)

= 6[(x)² – (3)²]

= 6(x + 3) (x – 3)

iii) x² – 81

= x² – 81

= x² – 9²

= (x + 9 )(x – 9)

iv) 2x – 32x⁵

= 2x – 32x⁵

= 2x – 2x x 16x⁴

= 2x(1 – 16x⁴)

= 2x[(1²) – (4x²)²]

= 2x(1 + 4x²)(1 – 4x²)

= 2x(1 + 4x²) [(1² – (2x)²]

= 2x(1 + 4x²) (1 + 2x) (1 – 2x)

First find the two numbers whose sum =12 and product = 27

Clearly, the numbers are 9 and 3

∴ we get, z² +12z + 27 

= z² + 9z + 3z + 27

= z (z+9) + 3(z+9)

= (z+3) (z+9)

let’s take HCF of above equation By taking 9a²b² as a common factor for the above equation we get,

18a³b³ – 27a²b³ + 36a³b² = 9a²b² (2ab – 3b + 4a)

By simplifying further the above equation is 3(4x² – 9)

By using the formula a² – b² = (a+b) (a-b)

Now solving for the above equation

3(4x² – 9) 

= 3((2x)² – (3)²)

= 3(2x+3) (2x-3)

let’s take HCF of above equation By taking 3x as a common factor for the above equation we get,

9x³ – 6x² + 12x = 3x (3x² – 2x + 4)

x³ – 12x² + 36x
Identify a² – 2ab + b² =(a – b)²
=x [x² -12x + 36]
= x[x² – (2)(x)(6) + 6²]
= x(x – 6)²

let’s take HCF of above equation By taking 4x as a common factor for the above equation we get,

8x² – 72xy + 12x = 4x (2x -18y +3)

12x² − 7x + 1

We can find two numbers such that pq = 12 × 1 = 12 and p + q = −7. They are p = −4 and q = −3.

Here, 12x² − 7x + 1 = 12x² − 4x − 3x + 1

= 4x (3x − 1) − 1 (3x − 1)

= (3x − 1) (4x − 1)

L.H.S = 3x/(3x + 2) ≠ R.H.S

The correct statement is  3x/(3x + 2) =  3x/(3x + 2)

2a²b² – 98b⁴ = 2b²(a² - 49b²)

= 2b²[(a)² - (7b)²]

= 2b²[(a + 7b) (a - 7b)

(i) a² + 8a + 16 

= (a)² + 2 × a × 4 + (4)²

= (a + 4)²    [(x + y)² = x² + 2xy + y²]

(ii) p² − 10p + 25 

= (p)² − 2 × p × 5 + (5)²

= (p − 5)²    [(a − b)² = a² − 2ab + b²]

(iii) 25m² + 30m + 9 = (5m)² + 2 × 5m × 3 + (3)²

= (5m + 3)²    [(a + b)²  = a² + 2ab + b² ]

(i) y² + 2y – 48

y² + 2y – 48
= y² + 8y - 6y – 48
= y(y + 8) – 6(y + 8)
= (y + 8) (y – 6)

(ii) d² – 4d – 45

d² – 4d – 45
= d² – 9d + 5 d – 45
= d (d – 9) + 5 (d – 9)
= (d – 9) (d + 5)

(iii) m² + 16m + 63

m² + 16m + 63
= m² + 9m + 7m + 63
= m (m + 9) + 7 (m + 9)
= (m + 9) (m + 7)

(iv) n² – 19n – 9²

n² – 19n – 92
= n² – 23n + 4n – 92
= n (n – 23) + 4 (n – 23)
= (n – 23) (n + 4)

(v) p² – 10p + 16

p² – 10p + 16
= p² – 8p – 2p + 16
= p (p – 8) – 2(p – 8)
= (p – 8) (p – 2)

(vi) x² + 4x – 45

x² + 4x – 45
= x² + 9x – 5x – 45
= x (x + 9) – 5(x + 9)
= (x + 9) (x – 5)

9a²b² – 25 can be written as (3ab)² – (5)²

By using the formula a² – b² = (a+b) (a-b)

Now solving for the above equation

(3ab)² – (5)² = (3ab+5) (3ab-5)

(4.5)² – (1.5)² 

= (4.5 + 1.5) (4.5 - 1.5)

= 6 x 3 = 18

a² + 3a + 2a + 6 

a(a + 3) + 2(a + 3) 

(a + 3)(a + 2)

a² + 5a + 6 = a² + 3a + 2a + 6

= a(a+3) + 2(a+3)

= (a+3) (a+2)

2a² + 7a + 6 = 2a² + 4a + 3a + 6

= 2a(a+2) + 3(a+2)

= (a+2) (2a+3)

3a² - 5a + 2 = 3a² - 3a - 2a + 2

= 3a(a-1) -2(a-1)

= (a – 1)(3a – 2)

x² - 10xy + 24y² = x² - 6xy - 4xy + 24y² 

= x(x-6y) - 4y(x-6y)

= (x-6y) (x-4y)

2x² – 24x + 72 

= 2(x² – 12x + 36) 

Identity a² – 2ab + b² = (a – b)² 

= 2(x² – (2)(x)(6) + 6²) 

= 2(x - 6)²

x² - x - 72 = x² - 9x + 8x - 72

= x(x-9) + 8(x-9)

= (x-9) (x+8)

ab² - (a-c) b - c

ab² - ab + bc - c

= ab (b - 1) + c(b - 1)

= (b - 1) (ab + c)

a(b-c) - d(c-b)

= a(b-c) + d(b-c)

= (b-c) (a+d)

x² - (b-2)x - 2b

x² - bx + 2x - 2b

= x(x-b) + 2(x-b)

= (x-b) (x+2)

(i) a² + 8a + 16

Answer: This equation can be facorised by using the identity; (a + b)² = a² + 2ab + b²
Factors = (a + 4)² = (a + 4)(a + 4)

(ii) p² – 10 p + 25

Answer: This equation can be factorised by using the identity; (a – b)² = a² – 2ab + b²
Factors = (p – 5)²

(iii) 25m² + 30m + 9

Answer: = (5m – 3)²

(iv) 49y² + 84yz + 36z²

Answer: (7y + 6z)²

(v) 4x² – 8x + 4

Answer: (2x – 2)²

(vi) 121b² – 88bc + 16c²

Answer: (11b – 4c)²

(vii) (l + m)² – 4lm

Answer: l² + m² + 2lm - 4lm
l² + m² - 2lm = (l + m)²

(viii) a⁴ + 2a²b² + b⁴

Answer: This can be solved using (a+b)² = a²+ 2ab + b²
Hence, a⁴ + 2a²b² + b⁴
=(a²+b²)²