Question

find the square root of 256 (x-a)8 (x-b)4 (x-c)16 (x-d)20 with expansion

Answer 1

Given  : Expression :

256 (x-a)8 (x-b)4 (x-c)16 (x-d)20

To Find : Square root

Solution:

256(x-a)⁸(x-b)⁴(x-c)¹⁶(x-d)²⁰

256 = 16²

(x-a)⁸ = (x-a)⁴ˣ²

using   (x)ᵃˣᵇ =  (xᵃ)ᵇ

Hence  (x-a)⁴ˣ²  =  ((x -a)⁴)²

=> (x-a)⁸ =  ((x -a)⁴)²

Similarly

(x-b)⁴  =  ((x -b)²)²

(x-c)¹⁶ =  ((x -c)⁸)²

(x-d)²⁰  =  ((x -d)¹⁰)²

256(x-a)⁸(x-b)⁴(x-c)¹⁶(x-d)²⁰

= 16² ((x -a)⁴)²((x -a)⁴)² ((x -b)²)² ((x -c)⁸)²  ((x -d)¹⁰)²

Using aⁿ * bⁿ = (a b)ⁿ

= ( 16 (x -a)⁴(x -b)²(x -c)⁸(x -d)¹⁰)²

√a² = a

Hence square root of   ( 16 (x -a)⁴(x -b)²(x -c)⁸(x -d)¹⁰)²

=  16 (x -a)⁴(x -b)²(x -c)⁸(x -d)¹⁰

Learn More:

find the value of x by factorization, x2-8x+1280=0 - Brainly.in

brainly.in/question/7343350

Using factor theorem, factorize each of the following polynomial ...

brainly.in/question/15904651