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4z² – 8z + 3 

Factorizing the equation and taking 2z and – 1 as common, 

= 4z² – 6z – 2z + 3 

= 2z(2z – 3) – 1(2z – 3) 

= (2z – 1)(2z – 3).

The numerical coefficients of given numerical are 2, 12

Greatest common factor of 2, 12 is 2

Common literals appearing in given numerical is x

Smallest power of x in two monomials = 2

Monomials of common literals with smallest power= x²

Hence, the greatest common factor = 2x²

The numerical coefficients of given numerical are 6,18

Greatest common factor of 6, 18 is 6

Common literals appearing in given numerical are x and y

Smallest power of x in both monomials= 2

Smallest power of y in both monomials = 1

Binomials of common literals with smallest power= x²y

Hence, the greatest common factor = 6x²y

The numerical coefficients of given numerical are 0

Common literals appearing in given numerical are a and b

Smallest power of a in two monomials = 2

Smallest power of b in two monomials = 2

Monomials of common literals with smallest power= the greatest common factor = a²b²

The numerical coefficients of given numerical are 6, 9, 3

Greatest common factor of 6, 9, 3 is 3.

Common literals appearing in given numerical are x and y

Smallest power of x in three monomials = 1

Smallest power of y in three monomials = 2

Monomials of common literals with smallest power= xy²

Hence, the greatest common factor = 3xy²

The highest common factor of three terms = 5a²

=5a²(a² + 2a - 3)

The numerical coefficients of given numerical are 15, -45, -150

Greatest common factor of 15, -45, -150 is 15.

Common literals appearing in given numerical is smallest power of a in three monomials = 1

Monomials of common literals with smallest power= a

Hence, the greatest common factor = 15a

Greatest common factor of the two terms namely 20a¹²b² and -15a⁸b⁴ of expression 20a¹²b² -15a⁸b⁴ is 5a⁸b²

20a¹²b² = 5a⁸b² (4a⁴) and - 15a⁸b⁴= 5a⁸b² (-3b²)

20a¹²b² - 15a⁸b⁴ = 5a⁸b² (4a⁴ - 3b²) = 5a⁸b²((2a)² - (b√3)²) = 5a⁸b²(2a + b√3)(2a - b√3)

We have, 

20a² – 45b² 

= 5(4a² – 9b²) 

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

We get, 

20a² – 45b² = 5(4a² – 9b²) 

= 5{(2a)² – (3b)²} 

= 5(2a + 3b)(2a – 3b)

We have, 

16a² – 144 

= (4a)² – (12)² 

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

We get, 

16a² – 144 = (4a)² – (12)² 

= (4a + 12)(4a – 12) 

= 4(a + 3) 4(a – 3) 

= 16(a + 3)(a – 3)

16a² -144 can be written as (4a)² – (12)²

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

Now solving for the above equation

(4a)² – (12)² = (4a+12) (4a-12)

= 4(a+3) 4(a-3)

= 16(a+3) (a-3)

We have, 

16a² – 225b² 

= (4a)² – (15b)² 

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

We get, 

16a² – 225b² = (4a)² – (15b)² 

= (4a + 15b)(4a – 15b)

Given, 

9a² – b² + 4b – 4 

= 9a² – (b² – 4b + 4) 

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

We get, 

9a² – b² + 4b – 4 = 9a² – (b² – 4b + 4) 

= (3a)² – (b – 2)² 

= {3a + (b – 2)}{3a – (b – 2)} 

= (3a + b - 2)(3a – b + 2)

(i) Let’s take HCF of 16a² – 24ab

Taking 8a as common from the whole, we get,

16a² – 24ab = 8a(2a – 3b).

(ii) 15ab² – 20a²b,

Taking 5ab as common from the whole, we get,

15ab² – 20a²b = 5ab(3b – 4a)

(iii) 12x²y³ – 21x³y²,

Taking 3x²y² as common from the whole, we get,

12x²y³ – 21x³y² = 3x²y²(4y – 7x)

(i) 12x + 15 

Taking 3 as common from the whole, we get, 

12x + 15 = 3(4x + 5).

(ii) 14m – 21, 

Taking 7 as common from the whole, we get, 

14m – 21 = 7(2m – 3)

(iii) 9n – 12n², 

Taking 3n as common from the whole, we get, 

9n – 12n² = 3n (3 – 4n).

q² – 10q + 21 Coefficient of q² is 1 so we add and subtract square of half of coefficient of q

Therefore,

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

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

= (q – 5 – 2) (q – 5 + 2)

= (q – 7) (q – 3)

x² – 2 (x) (2z) + (2z)² – y²

As (a-b)² = a² + b² – 2ab

= (x – 2z)² – y²

As a² – b² = (a+b)(a-b)

= (x – 2z + y) (x – 2z – y)