Question

factorize (16z^4- 72z^2+81)​

Answer 1

Answer:

The factors of 16$z^{4}$ - 72z² + 81 are (2z+3) (2z-3)(2z+3) (2z-3) or  

                                                        (2z+3)² (2z-3)²  

Step-by-step explanation:

Given : 16$z^{4}$ - 72z² + 81

Solution : Equation 16$z^{4}$ - 72z² + 81 can be written as

                                (4z²)² - 2 x 4z² x 9 + 9²

             This is in the form a²-2ab+b² = (a-b)² ----------1

In the given equation a=4z² and b=9

Substituting the values of a & b in 1

we get the factors of 16$z^{4}$ - 72z² + 81 = (4z² -9)²

                                                             = (4z² -9) (4z²-9)

(4z² - 9) is in the form x²-y² = (x+y)(x-y) -----------2

∴ (4z² - 9)  = ((2z)² - 3²)

here x = 2z and y = 3

Substituting the values of x & y in 2

    (4z² - 9) = (2z+3) (2z-3)            

∴   the factors of 16$z^{4}$ - 72z² + 81  are  (2z+3) (2z-3)(2z+3) (2z-3)