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2ab² - aby + 2cby - cy² 

= 2b (ab + cy) - y(ab + cy)

= (ab + cy) (2b - y)

4x² – 9y² can be written as (2x)² – (3y)²

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

Now solving for the above equation

(2x)² – (3y)² = (2x+3y) (2x-3y)

Correct option is  D) None

Correct option is (D) None

\(\frac{x^4-1}{x-3}=\frac{(x^2-1)(x^2+1)}{x-3}\)

\(=\frac{(x-1)(x+1)(x^2+1)}{x-3}\)

Correct option is  A) – 3ab

Correct option is (A) –3ab

\(18\,a^2b^2c\div(-6\,abc)=\frac{18\,a^2b^2c}{-6\,abc}\)

\(=\frac{6abc\times3ab}{-6\,abc}\)

= -3ab

Correct option is  B) 3x – 6

Correct option is (B) 3x – 6

\((9x^2–12x–12)\div(3x+2)=\frac{9x^2–12x–12}{3x+2}\)

\(=\frac{3(3x^2-4x-4)}{3x+2}\)

\(=\frac{3(3x^2-6x+2x-4)}{3x+2}\)

\(=\frac{3(3x(x-2)+2(x-2))}{3x+2}\)

\(=\frac{3(3x+2)(x-2)}{3x+2}\)

= 3 (x - 2)

= 3x – 6

Correct option is  D) 6

x² - (a-3)x-3a

x² - ax + 3x - 3a

= x(x-a) + 3(x-a)

= (x-a) (x+3)

2x³b² - ax⁵b⁴ = 2x³b² (1 - 2x²b²)

Given equation is x² + 3xy + x + my – m ……….(1)

Let the two linear equations in x and y be (x + 3y + a) and (x + 0y + b).

Then (x + 3y + a) (x + 0y + b)

= x² + 0xy + bx + 3xy + 0y² + 3by + ax + 0y + ab

= x² + bx + ax + 3xy + 3by + ab ………….. (2)

Comparing equation (2) with (1),

x² + 3xy + x + my – m = x² + (a + b)x + 3xy + 3by + ab

Equating the like terms on both sides,

ab = – m ………….. (3)

(a + b)x = x 

⇒ a + b = 1…………. (4)

3by = my ⇒ 3b = m ⇒ b = m/3

Substitute ‘b’ value in equation (4),

a = 1−m/3 = (3 − m)/3

ab = -m

[ ∵ from (3)]

put a & b value then ,

((3−m)/3)(m/3) = -m

(3m − m²)/9 = -m

⇒ 3m – m² = – 9m

⇒ m² – 12m = 0

⇒ m(m – 12) = 0

⇒ m = 0 (or) m = 12

lf m = 12

∴ b = 12/3 = 4&a = (3−m)/3 = (3−12)/3

= −9/3 = -3

∴ Linear factors are (x + 3y – 3), (x + 4)  If m = 0

b = 0/3 = 0 & a = (3−0)/3 = 3/3 = 1

∴ Linear factors are (x + 3y + 1), x.

B) 3 (x – 4) = 3x – 3

Correct option is  C) 96

Correct option is (C) 96

\(2^5\times3=32\times3\) = 96

Correct option is  D) None

Correct option is (D) None

(4x + 51y + 1) (4x + 51y – 1) \(=(4x+51y)^2-1^2\)

\(=(4x+51y)^2-1\)

Correct option is  B) 21

Correct option is (B) 21

3 + 2 (9) = 3+18 = 21

Correct option is  B) 48

Correct option is (B) 48

\(2^4\times3=16\times3\)

= 48

Correct option is  A) √x + √y

Correct option is (A) √x + √y

\(\frac{x-y}{\sqrt x-\sqrt y}=\frac{(\sqrt x+\sqrt y)(\sqrt x-\sqrt y)}{\sqrt x-\sqrt y}\)

= \(\sqrt x+\sqrt y\)

36x² + 96xy + 64y²

36x² + 96xy + 64y²

= (6x)² + 2 × 6x × 8y + (8y)²

It is in the form of a² + 2ab + b²

a² + 2ab + b² = (a + b)²

∴ 36x² + 96xy + 64y²

= (6x + 8y)² = (6x + 8y) (6x + 8y)

L.H.S. = (3x + 2)² 

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

= 9x² + 12x + 4 ≠ R.H.S

The correct statement is (3x + 2)² = 9x² + 12x + 4

Correct option is  D) x/3 

Correct option is (D) x/3

\(14x^3\div42x^2=\frac{14x^3}{42x^2}\)

= \(\frac x3\)

L.H.S = (2x)² + 4(2x) + 7 = 4x² + 8x + 7 ≠ R.H.S

The correct statement is (2x)² + 4(2x) + 7 = 4x² + 8x + 7

L.H.S. = (a − 4) (a − 2) = (a)² + [(− 4) + (− 2)] (a) + (− 4) (− 2)

= a² − 6a + 8 ≠ R.H.S.

The correct statement is (a − 4) (a − 2) = a² − 6a + 8

14x² + 7x - 6x - 3

7x(2x+1) -3(2x+1)

= (2x+1) (7x-3)

L.H.S = (2x)² + 5x = 4x² + 5x ≠ R.H.S.

The correct statement is (2x)² + 5x = 4x² + 5x

L.H.S = x + 2x + 3x =1x + 2x + 3x = x(1 + 2 + 3) 

= 6x ≠ R.H.S.

The correct statement is x + 2x + 3x = 6x

x² + 5xy + ay² = x² + 4xy + xy + 4y² 

= x(x+4y) + y(x+4y)

= (x+4y) (x+y)

16x⁵ -144x³ can be written as 16x³(x² – 9)

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

Now solving for the above equation

16x³(x² – 9) 

= 16x³((x) ² – (3)²)

= 16x³(x+3) (x-3)

(i) 16x⁵ − 144x³ 

= 16x³(x² − 9)

= 16 x³ [(x)² − (3)²]

= 16 x³(x − 3) (x + 3)  [a² − b² = (a − b) (a + b)]

(ii) (l + m)² − (l − m)² 

= [(l + m) − (l − m)] [(l + m) + (l − m)] 

[Using identity a² − b² = (a − b) (a + b)]

= (l + m − l + m) (l + m + l − m)

= 2m × 2l

= 4ml

= 4lm

(iii) 9x²y² − 16 = (3xy)² − (4)²

= (3xy − 4) (3xy + 4) [a² − b² = (a − b) (a + b)]

(i) ax² + bx 

= a × x × x + b × x 

= x(ax + b)

(ii) 7p² + 21q² 

= 7 × p × p + 3 × 7 × q × q 

= 7(p² + 3q²)

(iii) 2x³ + 2xy² + 2xz² 

= 2x(x² + y² + z²)

We have,

4y² + 12y + 5

4(y² + 3y + 5/4)

Coefficient of y² is unity. So, we add and subtract square of half of coefficient of y.

4(y² + 3y + 5/4) = 4 [y² + 3y + (3/2)² – (3/2)² + 5/4] (Adding and subtracting (3/2)²)

= 4 [(y + 3/2)² – 1²] (Completing the square)

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

= 4 (y + 3/2 + 1) (y + 3/2 – 1)

= 4 (y + 1/2) (y + 5/2) (by taking LCM)

= 4 [(2y + 1)/2] [(2y + 5)/2]

= (2y + 1) (2y + 5)

We have,

4x² – 12x + 5

4(x² – 3x + 5/4)

Coefficient of x² is unity. So, we add and subtract square of half of coefficient of x.

4(x² – 3x + 5/4) = 4 [x² – 3x + (3/2)² – (3/2)² + 5/4] (Adding and subtracting (3/2)²)

= 4 [(x – 3/2)² – 1²] (Completing the square)

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

= 4 (x – 3/2 + 1) (x – 3/2 – 1)

= 4 (x – 1/2) (x – 5/2) (by taking LCM)

= 4 [(2x-1)/2] [(2x – 5)/2]

= (2x – 5) (2x – 1)

We have,

a² + 2a – 3

Coefficient of a² is unity. So, we add and subtract square of half of coefficient of a.

a² + 2a – 3 = a² + 2a + 1² – 1² – 3 (Adding and subtracting 1²)

= (a + 1)² – 2² (Completing the square)

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

= (a + 1 + 2) (a + 1 – 2)

= (a + 3) (a – 1)

5 + 3a - 14a² = 5 + 10a - 7a - 14a² 

= 5(1 + 2a) -7a(1 + 2a)

= (1 + 2a) (5 - 7a)

We have,

z² – 4z – 12

Coefficient of z² is unity. So, we add and subtract square of half of coefficient of z.

z² – 4z – 12 = z² – 4z + 2² – 2² – 12 [Adding and subtracting 2²]

= (z – 2)² – 4² (Completing the square)

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

= (z – 2 + 4) (z – 2 – 4)

= (z – 6) (z + 2)

We have,

y² – 7y + 12

Coefficient of y² is unity. So, we add and subtract square of half of coefficient of y.

y² – 7y + 12 = y² – 7y + (7/2)² – (7/2)² + 12 [Adding and subtracting (7/2)²]

= (y – 7/2)² – (7/2)² (Completing the square)

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

= (y – (7/2- 1/2)) (y – (7/2 + 1/2))

= (y – 3) (y – 4)

We have,

q² – 10q + 21

Coefficient of q² is unity. So, we add and subtract square of half of coefficient of q.

q² – 10q + 21 = q² – 10q+ 5² – 5² + 21 (Adding and subtracting 5²)

= (q – 5)² – 2² (By completing the square)

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

= (q – 5 – 2) (q – 5 + 2)

= (q – 3) (q – 7)

7lm – 21lmn 

= 7 × l × m – 7 × 3 × m × n × l 

= 7 × l × m [1 – 3n] 

= 7lm [1 – 3n]

We have,

36a² + 12abc – 15b²c²

The coefficient of a² is 36

The coefficient of a is 12bc

Constant term is -15b²c²

So, we express the middle term 12abc as 30abc – 18abc

36a² –12abc– 15b²c² = 36a² + 30abc – 18abc – 15b²c²

= 6a (6a + 5bc) – 3bc (6a + 5bc)

= (6a + 5bc) (6a – 3bc)

= (6a + 5bc) 3(2a – bc)

15a³b – 35ab³

= 3 × 5 × a × a × a × b – 7 × 5 × a × b × b × b 

= 5 × a × b [3 × a × a – 7 × b × b] 

= 5ab [3a² – 7b²]

A) 5ab (5a+ 7b)

Correct option is (A) 5ab (5a + 7b)

\(25a^2b+35ab^2\) \(=5ab\times5a+5ab\times7b\)

= 5ab (5a + 7b)

We have,

p² + 6p + 8

Coefficient of p² is unity. So, we add and subtract square of half of coefficient of p.

p² + 6p + 8 = p² + 6p + 3² – 3² + 8 (Adding and subtracting 3²)

= (p + 3)² – 1² (By completing the square)

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

= (p + 3 – 1) (p + 3 + 1)

= (p + 2) (p + 4)

First find the two numbers whose sum=15 and product=56

Clearly, the numbers are 7 and 8

∴ we get, x² +15x + 56 

= x² + 7x +8x + 56

= x(x+7) + 8(x+7)

= (x+7) (x+8)

2(a+3b) – 3(a+3b) ²

Now by taking (a+3b) as a common factor for the above equation we get,

2(a+3b) – 3(a+3b)² 

= (a+3b) (2 – 3(a+3b))

= (a+3b) (2- 3a -9b)

First find the two numbers whose sum=6 and product= -16

Clearly, the numbers are 8 and 2

∴ we get, p² +6p – 16 

= p² +8p – 2p -16

= p (p+8) – 2(p+8)

= (p+8) (p-2)

Correct option is  A) 3, y

Correct option is (A) 3, y

Common factor of 9 & 6 is 3.

\(\therefore\) Common factors of 6xy & \(9y^2\) are 3 & y.

B) (x + y) (a + b)

Correct option is (B) (x + y) (a + b)

ax + ay + bx + by \(=a(x+y)+b(x+y)\)

= (a + b) (x + y)

We solve by using the formula (a+b)² = a² + b² + 2ab We get, 

(6a)² + (3)² + 2 (6a) (3) = (6a+3) ²